Comprehensive University Study Notes: Advanced Multivariable Calculus, Vector Analysis, and Ordinary Differential Equations

Dimensional Analysis, Similitude, and Physical Modelling

  • Foundations of Mathematical Modelling:

    • Mathematical modelling is the process of creating, analysing, and validating mathematical representations of physical, biological, chemical, economic, and engineering phenomena.

    • The modelling cycle proceeds from identifying key variables and interactions in the natural world to formulating equations, solving and analysing those equations, and validating and interpreting predictions against experimental data.

  • Units and Base Physical Dimensions:

    • Postulate 1: There exist base quantities providing units in terms of which the units of all other physical quantities can be expressed.

    • In classical mechanics, the three primary base dimensions are:

    • Mass: [M][M]

    • Length: [L][L]

    • Time: [T][T]

    • Additional fundamental base quantities in the International System of Units (SI) include temperature (Kelvin\text{Kelvin}, [K][K]), electric current (Ampere\text{Ampere}, [A][A]), amount of substance (mole\text{mole}, [mol][mol]), and luminous intensity (candela\text{candela}, [cd][cd]).

    • The dimensions of any derived physical variable yy are denoted [y][y], expressed as a product of integer powers of the base dimensions: [y]=MaLbTc[y] = M^a L^b T^c where a,b,c∈Za, b, c \in \mathbb{Z}.

    • For composite quantities, the dimension operator satisfies the product property: [yz]=[y][z][yz] = [y][z].

    • Fundamental mechanical examples:

    • Velocity: [v]=[dxdt]=LT−1[v] = \left[\frac{dx}{dt}\right] = LT^{-1}

    • Acceleration: [a]=[d2xdt2]=LT−2[a] = \left[\frac{d^2x}{dt^2}\right] = LT^{-2}

    • Force: [F]=[ma]=[m][a]=MLT−2[F] = [ma] = [m][a] = MLT^{-2}

    • A quantity is defined as dimensionless if its dimensional representation is M0L0T0M^0 L^0 T^0.

  • Dimensional Homogeneity and Complete Sets:

    • Postulate 2: Equations representing physical laws are dimensionally homogeneous.

    • An equation is dimensionally homogeneous if and only if the dimensions of every additive term on both the left-hand side and right-hand side are identical: [LHS]=[RHS][\text{LHS}] = [\text{RHS}].

    • In transcendental functions (such as exp⁡(z)\exp(z), ln⁡(z)\ln(z), sin⁡(z)\sin(z), cos⁡(z)\cos(z)), the argument zz must be strictly dimensionless ([z]=M0L0T0[z] = M^0 L^0 T^0) for dimensional compatibility, as established by their Maclaurin series expansions (e.g., sin⁡(x)=x−x33!+x55!−…\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots).

    • A complete set of dimensionless products is a maximal set of independent dimensionless combinations formed from the variables of a model.

  • The Buckingham Π\Pi-Theorem:

    • An equation is dimensionally homogeneous if and only if it can be expressed in the form:     f(Π1,Π2,…,Πn)=0f(\Pi_1, \Pi_2, \dots, \Pi_n) = 0     where ff is a function and {Π1,Π2,…,Πn}\{\Pi_1, \Pi_2, \dots, \Pi_n\} constitutes a complete set of independent dimensionless products corresponding to the mathematical model.

    • The set of exponents yielding dimensionless products is obtained by determining a basis for the null space of the dimensional matrix representing the system of linear equations for the exponents.

  • Worked Case Studies in Dimensional Analysis:

    • A Falling Mass:

    • An object of mass mm falls from rest in a vacuum near Earth's surface over distance xx in time tt.

    • In metric units (Australia), experiments yield x=4.9t2x = 4.9 t^2, whereas in imperial units (USA), experiments yield x=16.1t2x = 16.1 t^2.

    • Dimensional consistency requires x=ct2x = c t^2, where [x]=[c][t]2  ⟹  L=[c]T2  ⟹  [c]=LT−2[x] = [c][t]^2 \implies L = [c] T^2 \implies [c] = LT^{-2}, revealing that c=12gc = \frac{1}{2}g.

    • The Simple Pendulum:

    • Period τ\tau depends on pendulum length rr, mass mm, initial release angle θ\theta, and gravitational acceleration gg: τ=f(r,m,θ,g)\tau = f(r, m, \theta, g).

    • Dimensions: [τ]=T[\tau] = T, [r]=L[r] = L, [m]=M[m] = M, [g]=LT−2[g] = LT^{-2}, [θ]=1[\theta] = 1 (M0L0T0M^0 L^0 T^0).

    • Product: Π=τarbmcgeθd\Pi = \tau^a r^b m^c g^e \theta^d.

    • Dimensional balance: TaLbMc(LT−2)e1d=McLb+eTa−2e=M0L0T0T^a L^b M^c (LT^{-2})^e 1^d = M^c L^{b+e} T^{a-2e} = M^0 L^0 T^0.

    • Linear system for exponents:

      • M:c=0M: c = 0

      • L:b+e=0  ⟹  b=−eL: b + e = 0 \implies b = -e

      • T:a−2e=0  ⟹  e=a2,b=−a2T: a - 2e = 0 \implies e = \frac{a}{2}, b = -\frac{a}{2}

      • Exponent dd is free.

    • Independent basis choices (a,d)=(1,0)(a, d) = (1, 0) and (0,1)(0, 1) generate the complete set:       Π1=τgr,Π2=θ\Pi_1 = \tau \sqrt{\frac{g}{r}}, \quad \Pi_2 = \theta

    • Applying the Buckingham Π\Pi-theorem yields f(τgr,θ)=0  ⟹  τgr=h(θ)  ⟹  τ=rgh(θ)f\left(\tau \sqrt{\frac{g}{r}}, \theta\right) = 0 \implies \tau \sqrt{\frac{g}{r}} = h(\theta) \implies \tau = \sqrt{\frac{r}{g}} h(\theta).

    • Archimedes' Buoyancy Law:

    • A box of mass mm displacing fluid volume VV with fluid density ρ\rho ([ρ]=ML−3[\rho] = ML^{-3}) under gravity gg.

    • Testing a proposed equation mx¨=mg−mVgm\ddot{x} = mg - mVg:

      • [mx¨]=[mg]=MLT−2[m\ddot{x}] = [mg] = MLT^{-2}

      • [mVg]=M⋅L3⋅LT−2=ML4T−2[mVg] = M \cdot L^3 \cdot LT^{-2} = ML^4 T^{-2}

      • Since MLT−2≠ML4T−2MLT^{-2} \ne ML^4 T^{-2}, the proposed law is dimensionally invalid.

    • Testing an alternative equation mx¨=mg−ρVgm\ddot{x} = mg - \rho V g:

      • [ρVg]=(ML−3)(L3)(LT−2)=MLT−2[\rho V g] = (ML^{-3})(L^3)(LT^{-2}) = MLT^{-2}

      • The equation is dimensionally homogeneous and physically viable.

    • By the Buckingham Π\Pi-theorem, any valid relationship between total force FF, ρ\rho, gg, VV, and mm must satisfy:       F=ρgVh(mρV)F = \rho g V h\left(\frac{m}{\rho V}\right)       Choosing h(u)=u−1h(u) = u - 1 directly recovers F=mg−ρVgF = mg - \rho V g.

    • Stokes' Drag Law:

    • Drag force FF ([F]=MLT−2[F] = MLT^{-2}) on a sphere of radius rr ([r]=L[r] = L) moving at velocity vv ([v]=LT−1[v] = LT^{-1}) in fluid of dynamic viscosity η\eta ([η]=ML−1T−1[\eta] = ML^{-1}T^{-1}).

    • Dimensionless combination Faηbvcrd=Ma+bLa−b+c+dT−2a−b−c=M0L0T0F^a \eta^b v^c r^d = M^{a+b} L^{a-b+c+d} T^{-2a-b-c} = M^0 L^0 T^0.

    • Setting a=1a = 1 isolates the dependent variable: b=−1b = -1, c=−1c = -1, d=−1d = -1.

    • The solitary dimensionless group is Π=Fηvr\Pi = \frac{F}{\eta v r}, implying Fηvr=k  ⟹  F=kηvr\frac{F}{\eta v r} = k \implies F = k \eta v r.

    • Stokes' exact theoretical solution yields k=6πk = 6\pi, giving F=6πηvrF = 6\pi \eta v r.

    • Fluid and Solute Mixing Dynamics:

    • A tank initially contains v0v_0 litres of water with m0m_0 grams of dissolved salt. Brine containing n g/Ln\,g/L enters at x L/minx\,L/min, and well-stirred solution drains at y L/miny\,L/min.

    • Let s(t)s(t) be the mass of salt at time tt. The volume is V(t)=v0+(x−y)tV(t) = v_0 + (x - y)t, and solute density is ρ(t)=s(t)v0+(x−y)t\rho(t) = \frac{s(t)}{v_0 + (x - y)t}.

    • Rate of mass inflow: nx g/minnx\,g/min. Rate of mass outflow: yρ(t)=ys(t)v0+(x−y)t g/miny\rho(t) = \frac{ys(t)}{v_0 + (x - y)t}\,g/min.

    • Governing Initial Value Problem:       dsdt=nx−ysv0+(x−y)t,s(0)=m0\frac{ds}{dt} = nx - \frac{ys}{v_0 + (x - y)t}, \quad s(0) = m_0

    • Dimensional consistency check:

      • [dsdt]=MT−1\left[\frac{ds}{dt}\right] = MT^{-1}

      • [nx]=(ML−3)(L3T−1)=MT−1[nx] = (ML^{-3})(L^3 T^{-1}) = MT^{-1}

      • [v0]=[(x−y)t]=L3[v_0] = [(x - y)t] = L^3

      • [ysv0+(x−y)t]=(L3T−1)(M)L3=MT−1\left[\frac{ys}{v_0 + (x - y)t}\right] = \frac{(L^3 T^{-1})(M)}{L^3} = MT^{-1}

  • Scaling and Similitude:

    • Laboratory scale models preserve dimensionless numbers to match the physical behaviour of full-scale prototypes.

    • Fundamental dimensionless numbers in fluid mechanics:

    • Reynolds number: Re=ρlvμRe = \frac{\rho l v}{\mu}

    • Froude number: Fr=vglFr = \frac{v}{\sqrt{gl}}

    • Mach number: M=vcM = \frac{v}{c}     where ρ\rho is fluid density, ll is characteristic length, vv is velocity, μ\mu is dynamic viscosity, gg is gravitational acceleration, and cc is speed of sound.

    • Ship Scaling Example:

    • Full-scale vessel: length l=20 ml = 20\,m, cruising velocity v=10 m/sv = 10\,m/s.

    • Model scale: length l∗=0.2 ml^* = 0.2\,m (scale ratio l∗/l=1/100l^*/l = 1/100).

    • Maintaining dynamic similitude via the Froude number:       Fr=vgl=v∗gl∗  ⟹  v∗=vl∗l=100.220=101100=1 m/sFr = \frac{v}{\sqrt{gl}} = \frac{v^*}{\sqrt{gl^*}} \implies v^* = v \sqrt{\frac{l^*}{l}} = 10 \sqrt{\frac{0.2}{20}} = 10 \sqrt{\frac{1}{100}} = 1\,m/s

Multivariable Functions, Limits, and Continuity

  • Functions of Multiple Variables:

    • A scalar-valued function of several variables is a mapping f:D⊂Rn→Rf: D \subset \mathbb{R}^n \to \mathbb{R}, where the domain DD is a subset of Rn\mathbb{R}^n.

    • The graph of a two-variable function z=f(x,y)z = f(x, y) represents a surface in R3\mathbb{R}^3.

  • Multivariate Limits and Definitions:

    • Formal ϵ\epsilon-δ\delta Definition: Let f:D⊂R2→Rf: D \subset \mathbb{R}^2 \to \mathbb{R} be defined on an open domain DD containing points arbitrarily close to (a,b)(a, b). The limit of f(x,y)f(x, y) as (x,y)→(a,b)(x, y) \to (a, b) is LL, written:     lim⁡(x,y)→(a,b)f(x,y)=L\lim_{(x, y) \to (a, b)} f(x, y) = L     if for every ϵ>0\epsilon > 0, there exists a corresponding δ>0\delta > 0 such that:     if (x,y)∈D and 0<(x−a)2+(y−b)2<δ,then ∣f(x,y)−L∣<ϵ\text{if } (x, y) \in D \text{ and } 0 < \sqrt{(x - a)^2 + (y - b)^2} < \delta, \quad \text{then } |f(x, y) - L| < \epsilon

    • Here, ∣f(x,y)−L∣|f(x, y) - L| denotes the absolute distance in R\mathbb{R}, while (x−a)2+(y−b)2=∥(x,y)−(a,b)∥\sqrt{(x - a)^2 + (y - b)^2} = \|(x, y) - (a, b)\| denotes the Euclidean distance in R2\mathbb{R}^2.

    • Generalisation to Rn\mathbb{R}^n: for f:D⊂Rn→Rf: D \subset \mathbb{R}^n \to \mathbb{R}, lim⁡x→af(x)=L\lim_{\mathbf{x} \to \mathbf{a}} f(\mathbf{x}) = L if for every ϵ>0\epsilon > 0, there exists δ>0\delta > 0 such that 0<∥x−a∥<δ  ⟹  ∣f(x)−L∣<ϵ0 < \|\mathbf{x} - \mathbf{a}\| < \delta \implies |f(\mathbf{x}) - L| < \epsilon.

  • The Path-Dependent Non-Existence Criterion:

    • For lim⁡(x,y)→(a,b)f(x,y)\lim_{(x, y) \to (a, b)} f(x, y) to exist, f(x,y)f(x, y) must approach the exact same value LL along every path in DD terminating at (a,b)(a, b).

    • If f(x,y)→L1f(x, y) \to L_1 along path C1C_1 and f(x,y)→L2f(x, y) \to L_2 along path C2C_2 with L1≠L2L_1 \ne L_2, the limit does not exist.

  • Case Evaluations:

    • Function f(x,y)=x2x2+y2f(x, y) = \frac{x^2}{x^2 + y^2} at (0,0)(0, 0):

    • Along y=0y = 0 (x≠0x \ne 0): f(x,0)=x2x2+0=1  ⟹  lim⁡x→0f(x,0)=1f(x, 0) = \frac{x^2}{x^2 + 0} = 1 \implies \lim_{x \to 0} f(x, 0) = 1.

  • Along x=0x = 0 (y≠0y \ne 0): f(0,y)=00+y2=0  ⟹  lim⁡y→0f(0,y)=0f(0, y) = \frac{0}{0 + y^2} = 0 \implies \lim_{y \to 0} f(0, y) = 0.

  • Since 1≠01 \ne 0, the limit does not exist.

  • Function f(x,y)=x2−y2x2+y2f(x, y) = \frac{x^2 - y^2}{x^2 + y^2} at (0,0)(0, 0):

    • Along y=0y = 0: f(x,0)=1f(x, 0) = 1.

    • Along x=0x = 0: f(0,y)=−1f(0, y) = -1.

    • Since 1≠−11 \ne -1, the limit does not exist.

  • Function f(x,y)=xyx2+y2f(x, y) = \frac{xy}{x^2 + y^2} at (0,0)(0, 0):

    • Along coordinate axes (x=0x = 0 or y=0y = 0): f(x,y)=0f(x, y) = 0.

    • Along y=xy = x (x≠0x \ne 0): f(x,x)=x22x2=12f(x, x) = \frac{x^2}{2x^2} = \frac{1}{2}.

  • Since 0≠120 \ne \frac{1}{2}, the limit does not exist.

  • Techniques for Establishing Multivariate Limits:

    • Polar Coordinate Substitution:

    • Evaluate lim⁡(x,y)→(0,0)3x2yx2+y2\lim_{(x, y) \to (0, 0)} \frac{3x^2 y}{x^2 + y^2}.

    • Set x=rcos⁡(θ)x = r\cos(\theta), y=rsin⁡(θ)y = r\sin(\theta), with r2=x2+y2r^2 = x^2 + y^2:       f(rcos⁡(θ),rsin⁡(θ))=3(r2cos⁡2(θ))(rsin⁡(θ))r2=3rcos⁡2(θ)sin⁡(θ)f(r\cos(\theta), r\sin(\theta)) = \frac{3(r^2 \cos^2(\theta))(r\sin(\theta))}{r^2} = 3r\cos^2(\theta)\sin(\theta)

    • Since ∣cos⁡2(θ)sin⁡(θ)∣≤1|\cos^2(\theta)\sin(\theta)| \le 1, we have:       ∣f(x,y)−0∣≤3r=3x2+y2|f(x, y) - 0| \le 3r = 3\sqrt{x^2 + y^2}

    • For any ϵ>0\epsilon > 0, select δ=ϵ3\delta = \frac{\epsilon}{3}. Then 0<x2+y2<δ  ⟹  ∣f(x,y)−0∣<3(ϵ3)=ϵ0 < \sqrt{x^2 + y^2} < \delta \implies |f(x, y) - 0| < 3\left(\frac{\epsilon}{3}\right) = \epsilon.

    • Hence, the limit exists and equals 00.

    • Algebraic Factorisation:

    • Evaluate lim⁡(x,y)→(0,0)x3−y3x2+xy+y2\lim_{(x, y) \to (0, 0)} \frac{x^3 - y^3}{x^2 + xy + y^2}.

    • Factorise numerator: x3−y3=(x−y)(x2+xy+y2)x^3 - y^3 = (x - y)(x^2 + xy + y^2).

    • For (x,y)≠(0,0)(x, y) \ne (0, 0), cancel common factors:       lim⁡(x,y)→(0,0)(x−y)(x2+xy+y2)x2+xy+y2=lim⁡(x,y)→(0,0)(x−y)=0−0=0\lim_{(x, y) \to (0, 0)} \frac{(x - y)(x^2 + xy + y^2)}{x^2 + xy + y^2} = \lim_{(x, y) \to (0, 0)} (x - y) = 0 - 0 = 0

  • Multivariate Continuity and Properties:

    • A function f:D⊂R2→Rf: D \subset \mathbb{R}^2 \to \mathbb{R} is continuous at (a,b)∈D(a, b) \in D if:     lim⁡(x,y)→(a,b)f(x,y)=f(a,b)\lim_{(x, y) \to (a, b)} f(x, y) = f(a, b)

    • All polynomial functions of nn variables are continuous across all of Rn\mathbb{R}^n.

    • All rational functions are continuous everywhere across their defined domains.

Differential Calculus of Several Variables: Tangents, Gradients, and Chain Rules

  • Partial Derivatives:

    • The first partial derivatives of f(x,y)f(x, y) at (a,b)(a, b) represent the instantaneous rates of change along cross-sections where the other coordinate is held constant:     fx(a,b)=∂f∂x(a,b)=lim⁡h→0f(a+h,b)−f(a,b)hf_x(a, b) = \frac{\partial f}{\partial x}(a, b) = \lim_{h \to 0} \frac{f(a + h, b) - f(a, b)}{h}     fy(a,b)=∂f∂y(a,b)=lim⁡h→0f(a,b+h)−f(a,b)hf_y(a, b) = \frac{\partial f}{\partial y}(a, b) = \lim_{h \to 0} \frac{f(a, b + h) - f(a, b)}{h}

    • Higher-Order Partial Derivatives and Clairaut's Theorem:

    • Second partial derivatives are denoted fxx=∂2f∂x2f_{xx} = \frac{\partial^2 f}{\partial x^2}, fyy=∂2f∂y2f_{yy} = \frac{\partial^2 f}{\partial y^2}, fxy=∂2f∂y∂xf_{xy} = \frac{\partial^2 f}{\partial y \partial x}, fyx=∂2f∂x∂yf_{yx} = \frac{\partial^2 f}{\partial x \partial y}.

    • Clairaut's Theorem: If the mixed second partial derivatives fxyf_{xy} and fyxf_{yx} are continuous on an open disc containing (a,b)(a, b), then fxy(a,b)=fyx(a,b)f_{xy}(a, b) = f_{yx}(a, b).

    • Example: For f(x,y)=xsin⁡(y)+ycos⁡(x)f(x, y) = x\sin(y) + y\cos(x), we find:       fx=sin⁡(y)−ysin⁡(x),fy=xcos⁡(y)+cos⁡(x)f_x = \sin(y) - y\sin(x), \quad f_y = x\cos(y) + \cos(x)       fxy=cos⁡(y)−sin⁡(x),fyx=cos⁡(y)−sin⁡(x)  ⟹  fxy=fyxf_{xy} = \cos(y) - \sin(x), \quad f_{yx} = \cos(y) - \sin(x) \implies f_{xy} = f_{yx}       fxx=−ycos⁡(x),fyy=−xsin⁡(y)f_{xx} = -y\cos(x), \quad f_{yy} = -x\sin(y)

    • Pathology of Partial Derivatives:

    • The existence of partial derivatives fx(a,b)f_x(a, b) and fy(a,b)f_y(a, b) does not imply continuity at (a,b)(a, b).

    • For f(x,y)=xyx2+y2f(x, y) = \frac{xy}{x^2 + y^2} ((x,y)≠(0,0)(x, y) \ne (0, 0)) and f(0,0)=0f(0, 0) = 0:       fx(0,0)=lim⁡h→0f(h,0)−f(0,0)h=0,fy(0,0)=lim⁡h→0f(0,h)−f(0,0)h=0f_x(0, 0) = \lim_{h \to 0} \frac{f(h, 0) - f(0, 0)}{h} = 0, \quad f_y(0, 0) = \lim_{h \to 0} \frac{f(0, h) - f(0, 0)}{h} = 0       Both partial derivatives exist at the origin, but the function itself is discontinuous at (0,0)(0, 0).

  • The Tangent Plane and Differentiability:

    • Equation of the Tangent Plane: For a smooth surface z=f(x,y)z = f(x, y) at (a,b,f(a,b))(a, b, f(a, b)), the tangent plane has equation:     z=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)

    • Vector parametric representation:     (xyz)=(abf(a,b))+λ(10fx(a,b))+μ(01fy(a,b)),λ,μ∈R\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a \\ b \\ f(a, b) \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ f_x(a, b) \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 1 \\ f_y(a, b) \end{pmatrix}, \quad \lambda, \mu \in \mathbb{R}

    • Definition of Multivariable Differentiability: Let V⊂RnV \subset \mathbb{R}^n. A function f:V→Rf: V \to \mathbb{R} is differentiable at a\mathbf{a} if there exists a vector ∇f(a)∈Rn\nabla f(\mathbf{a}) \in \mathbb{R}^n such that:     lim⁡h→0f(a+h)−f(a)−∇f(a)⋅h∥h∥=0\lim_{\mathbf{h} \to \mathbf{0}} \frac{f(\mathbf{a} + \mathbf{h}) - f(\mathbf{a}) - \nabla f(\mathbf{a}) \cdot \mathbf{h}}{\|\mathbf{h}\|} = 0

    • Theorems on Differentiability:

    • If ff is differentiable at a point, it is continuous at that point.

    • If all first-order partial derivatives of ff exist and are continuous in a neighbourhood of a\mathbf{a}, then ff is differentiable at a\mathbf{a}.

  • Linear Approximations, Total Differentials, and Error Estimation:

    • The linear (first-order) approximation near (a,b)(a, b) is:     f(x,y)≈f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)f(x, y) \approx f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)

    • The total differential of z=f(x,y)z = f(x, y) is:     dz=df=fx(a,b) dx+fy(a,b) dydz = df = f_x(a, b)\,dx + f_y(a, b)\,dy

    • If independent variables x,yx, y have maximum absolute errors ∣Δx∣≤E1|\Delta x| \le E_1 and ∣Δy∣≤E2|\Delta y| \le E_2, the estimated worst-case error in ff is:     ∣E∣≈∣fx(a,b)E1∣+∣fy(a,b)E2∣|E| \approx |f_x(a, b)E_1| + |f_y(a, b)E_2|

    • Example (Cylindrical Barrel): Volume V(r,h)=πr2hV(r, h) = \pi r^2 h. Nominal dimensions r=1 mr = 1\,m, h=2 mh = 2\,m. Nominal volume V(1,2)=2π m3V(1, 2) = 2\pi\,m^3. Partial derivatives: Vr=2πrh  ⟹  Vr(1,2)=4πV_r = 2\pi r h \implies V_r(1, 2) = 4\pi; Vh=πr2  ⟹  Vh(1,2)=πV_h = \pi r^2 \implies V_h(1, 2) = \pi.

    • Given a 5%5\% error in radius (Er=0.05 mE_r = 0.05\,m) and height (Eh=0.1 mE_h = 0.1\,m):     ∣E∣≈∣4π(0.05)∣+∣π(0.1)∣=0.2π+0.1π=0.3π m3|E| \approx |4\pi(0.05)| + |\pi(0.1)| = 0.2\pi + 0.1\pi = 0.3\pi\,m^3     Relative percentage error: ∣E∣V(1,2)×100%=0.3π2π×100%=15%\frac{|E|}{V(1, 2)} \times 100\% = \frac{0.3\pi}{2\pi} \times 100\% = 15\%.

  • Directional Derivatives and the Gradient Vector:

    • Directional Derivative: For a unit vector u\mathbf{u} (∥u∥=1\|\mathbf{u}\| = 1), the directional derivative of ff at a\mathbf{a} in direction u\mathbf{u} is:     Duf(a)=fu(a)=lim⁡t→0f(a+tu)−f(a)tD_{\mathbf{u}} f(\mathbf{a}) = f_{\mathbf{u}}(\mathbf{a}) = \lim_{t \to 0} \frac{f(\mathbf{a} + t\mathbf{u}) - f(\mathbf{a})}{t}

    • When ff is differentiable:     Duf(a)=∇f(a)⋅uD_{\mathbf{u}} f(\mathbf{a}) = \nabla f(\mathbf{a}) \cdot \mathbf{u}

  • For an arbitrary non-zero direction vector v\mathbf{v}:   Dvf(a)=∇f(a)⋅v∥v∥D_{\mathbf{v}} f(\mathbf{a}) = \nabla f(\mathbf{a}) \cdot \frac{\mathbf{v}}{\|\mathbf{v}\|}

  • Properties of the Gradient Vector ∇f\nabla f:

  • ∇f=(∂f∂x1,∂f∂x2,…,∂f∂xn)\nabla f = \left(\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \dots, \frac{\partial f}{\partial x_n}\right).

  • By Duf=∥∇f∥∥u∥cos⁡(θ)=∥∇f∥cos⁡(θ)D_{\mathbf{u}} f = \|\nabla f\| \|\mathbf{u}\| \cos(\theta) = \|\nabla f\| \cos(\theta), the directional derivative is maximized when θ=0\theta = 0 (u\mathbf{u} parallel to ∇f\nabla f). Hence, ∇f(a)\nabla f(\mathbf{a}) points in the direction of maximal increase of ff, with maximum rate of change equal to ∥∇f(a)∥\|\nabla f(\mathbf{a})\|.

  • The directional derivative is minimized when θ=π\theta = \pi, pointing in the direction of steepest descent (−∇f(a)-\nabla f(\mathbf{a})) with rate −∥∇f(a)∥-\|\nabla f(\mathbf{a})\|.

  • When θ=π2\theta = \frac{\pi}{2}, Duf=0D_{\mathbf{u}} f = 0. Thus, ∇f(a)\nabla f(\mathbf{a}) is orthogonal to the level curve (or level surface) of ff passing through a\mathbf{a}.

  • Multivariable Chain Rules and Implicit Differentiation:

    • Chain Rule: Let z=f(x1,…,xn)z = f(x_1, \dots, x_n) where each xi=xi(t)x_i = x_i(t) is differentiable. Then:     dfdt=∑i=1n∂f∂xidxidt=∇f⋅dxdt\frac{df}{dt} = \sum_{i=1}^n \frac{\partial f}{\partial x_i} \frac{dx_i}{dt} = \nabla f \cdot \frac{d\mathbf{x}}{dt}

  • If each xix_i depends on multiple variables (s,t)(s, t), then:   ∂z∂t=∑i=1n∂z∂xi∂xi∂t\frac{\partial z}{\partial t} = \sum_{i=1}^n \frac{\partial z}{\partial x_i} \frac{\partial x_i}{\partial t}

  • Implicit Differentiation: Let a plane curve be defined implicitly by F(x,y)=0F(x, y) = 0. Differentiating with respect to xx via the chain rule:   ∂F∂xdxdx+∂F∂ydydx=0  ⟹  Fx+Fydydx=0  ⟹  dydx=−FxFy(Fy≠0)\frac{\partial F}{\partial x} \frac{dx}{dx} + \frac{\partial F}{\partial y} \frac{dy}{dx} = 0 \implies F_x + F_y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{F_x}{F_y} \quad (F_y \ne 0)

  • Example (Folium of Descartes): Curve x3+y3−6xy=0x^3 + y^3 - 6xy = 0. Define F(x,y)=x3+y3−6xyF(x, y) = x^3 + y^3 - 6xy. Then Fx=3x2−6yF_x = 3x^2 - 6y, Fy=3y2−6xF_y = 3y^2 - 6x. Thus:   dydx=−3x2−6y3y2−6x=−x2−2yy2−2x\frac{dy}{dx} = -\frac{3x^2 - 6y}{3y^2 - 6x} = -\frac{x^2 - 2y}{y^2 - 2x}

  • Implicit Function Theorem: Let U⊂R2U \subset \mathbb{R}^2 be an open set and f:U→Rf: U \to \mathbb{R} be continuously differentiable. Let (x0,y0)∈U(x_0, y_0) \in U satisfy f(x0,y0)=0f(x_0, y_0) = 0 and ∂f∂y(x0,y0)≠0\frac{\partial f}{\partial y}(x_0, y_0) \ne 0. Then there exists an interval I=(x0−r,x0+r)I = (x_0 - r, x_0 + r) and a unique continuously differentiable function g:I→Rg: I \to \mathbb{R} such that g(x0)=y0g(x_0) = y_0 and f(x,g(x))=0f(x, g(x)) = 0 for all x∈Ix \in I.

Multivariable Optimisation, Critical Points, and Constrained Extrema

  • Quadratic Approximations:

    • The second-order Taylor approximation of f(x,y)f(x, y) about (a,b)(a, b) is:     Q(x,y)=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)+12fxx(a,b)(x−a)2+fxy(a,b)(x−a)(y−b)+12fyy(a,b)(y−b)2Q(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) + \frac{1}{2}f_{xx}(a, b)(x - a)^2 + f_{xy}(a, b)(x - a)(y - b) + \frac{1}{2}f_{yy}(a, b)(y - b)^2

  • Classification of Critical Points and the Second Derivative Test:

    • A point (a,b)(a, b) is a critical point of f(x,y)f(x, y) if ∇f(a,b)=(0,0)\nabla f(a, b) = (0, 0) or if ∇f(a,b)\nabla f(a, b) is undefined.

  • The Second Derivative Test: Let f(x,y)f(x, y) have continuous second partial derivatives in a neighbourhood of a critical point (a,b)(a, b) where ∇f(a,b)=(0,0)\nabla f(a, b) = (0, 0). Define the Hessian determinant (discriminant):     D=det⁡(fxx(a,b)fxy(a,b)fyx(a,b)fyy(a,b))=fxx(a,b)fyy(a,b)−[fxy(a,b)]2D = \det \begin{pmatrix} f_{xx}(a, b) & f_{xy}(a, b) \\ f_{yx}(a, b) & f_{yy}(a, b) \end{pmatrix} = f_{xx}(a, b)f_{yy}(a, b) - [f_{xy}(a, b)]^2

    • If D>0D > 0 and fxx(a,b)>0f_{xx}(a, b) > 0, then ff has a local minimum at (a,b)(a, b).

    • If D>0D > 0 and fxx(a,b)<0f_{xx}(a, b) < 0, then ff has a local maximum at (a,b)(a, b).

    • If D<0D < 0, then ff has a saddle point at (a,b)(a, b).

    • If D=0D = 0, the test is inconclusive.

  • Global Extrema on Closed and Bounded Sets:

    • By the Extreme Value Theorem, a continuous function ff on a closed and bounded (compact) set D⊂R2D \subset \mathbb{R}^2 attains an absolute maximum and an absolute minimum.

    • Procedure: (1) Find values of ff at all critical points inside DD; (2) Find extreme values of ff on the boundary of DD using single-variable calculus or Lagrange multipliers; (3) The largest and smallest values from these steps are the absolute maximum and minimum.

  • Constrained Optimisation via Lagrange Multipliers:

    • Single Constraint: To maximize or minimize f(x)f(\mathbf{x}) subject to g(x)=0g(\mathbf{x}) = 0, where ∇g≠0\nabla g \ne \mathbf{0}, solve:     ∇f(x)=λ∇g(x),g(x)=0\nabla f(\mathbf{x}) = \lambda \nabla g(\mathbf{x}), \quad g(\mathbf{x}) = 0     where λ\lambda is the Lagrange multiplier. Geometrically, extrema occur where level sets of ff are tangent to the constraint manifold g=0g = 0.

  • Multiple Constraints: To optimize f(x1,…,xL)f(x_1, \dots, x_L) subject to NN constraints gj(x1,…,xL)=0g_j(x_1, \dots, x_L) = 0 (j=1,…,Nj = 1, \dots, N with N<LN < L):   ∇f=∑j=1Nλj∇gj,gj=0(j=1,…,N)\nabla f = \sum_{j=1}^N \lambda_j \nabla g_j, \quad g_j = 0 \quad (j = 1, \dots, N)     This is equivalent to rendering stationary the Lagrangian function:     L(x1,…,xL,λ1,…,λN)=f(x1,…,xL)−∑j=1Nλjgj(x1,…,xL)\mathcal{L}(x_1, \dots, x_L, \lambda_1, \dots, \lambda_N) = f(x_1, \dots, x_L) - \sum_{j=1}^N \lambda_j g_j(x_1, \dots, x_L)

    • For inequality constraints g(x)≤cg(\mathbf{x}) \le c, the formulation generalises to the Karush-Kuhn-Tucker (KKT) conditions.

  • Envelopes and Economic Applications:

    • Envelope of a Family of Curves: Given a family of curves y(t)(x)=f(x,t)y^{(t)}(x) = f(x, t) parameterized by tt, the envelope is a curve y=E(x)y = E(x) tangent to each member of the family.

    • The envelope satisfies the dual conditions: ft(x,t)=0  ⟹  t=g(x)f_t(x, t) = 0 \implies t = g(x), giving the envelope as E(x)=f(x,g(x))E(x) = f(x, g(x)).

    • Economic Profit Maximisation (Hotelling's Lemma):

    • Let f(x)f(x) be a production function (f(0)=0f(0) = 0, f(x)≥0f(x) \ge 0, f′(x)>0f'(x) > 0, f′′(x)<0f''(x) < 0) yielding output for input xx.

    • Selling price is pp, unit input cost is cc. Profit is P(p,x)=pf(x)−cxP(p, x) = pf(x) - cx.

    • Optimal input x=g(p)x = g(p) satisfies Px=pf′(x)−c=0  ⟹  f′(x)=cpP_x = pf'(x) - c = 0 \implies f'(x) = \frac{c}{p}.

    • The maximized profit function Pmax⁡(p)=pf(g(p))−cg(p)P_{\max}(p) = pf(g(p)) - cg(p) represents the envelope of the family of profit functions with respect to xx, possessing the envelope property:       dPmax⁡dp=f(g(p))\frac{dP_{\max}}{dp} = f(g(p))

    • Example: For f(x)=4xf(x) = 4\sqrt{x}, f′(x)=2x=cp  ⟹  x=g(p)=4p2c2f'(x) = \frac{2}{\sqrt{x}} = \frac{c}{p} \implies x = g(p) = \frac{4p^2}{c^2}.       Maximized profit: Pmax⁡(p)=p(4⋅2pc)−c(4p2c2)=4p2cP_{\max}(p) = p\left(4 \cdot \frac{2p}{c}\right) - c\left(\frac{4p^2}{c^2}\right) = \frac{4p^2}{c}.       For c=4c = 4, Pmax⁡(p)=p2P_{\max}(p) = p^2, with dPmax⁡dp=2p=f(g(p))\frac{dP_{\max}}{dp} = 2p = f(g(p)).

Line Integrals, Vector Fields, and Conservative Systems

  • Arc Length and Scalar Line Integrals:

    • The arc length of a smooth curve CC parameterized by r(t)=(x(t),y(t),z(t))\mathbf{r}(t) = (x(t), y(t), z(t)) for a≤t≤ba \le t \le b is:     S=∫Cds=∫ab∥r′(t)∥ dt=∫ab(dxdt)2+(dydt)2+(dzdt)2 dtS = \int_C ds = \int_a^b \|\mathbf{r}'(t)\|\,dt = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}\,dt

    • In polar coordinates r=f(θ)r = f(\theta) for α≤θ≤β\alpha \le \theta \le \beta:     S=∫αβ[f(θ)]2+[f′(θ)]2 dθS = \int_\alpha^\beta \sqrt{[f(\theta)]^2 + [f'(\theta)]^2}\,d\theta

    • The scalar line integral of f(x,y)f(x, y) along CC represents the area of a vertical ribbon of height ff above CC:     ∫Cf(x,y) ds=∫abf(x(t),y(t))∥r′(t)∥ dt\int_C f(x, y)\,ds = \int_a^b f(x(t), y(t))\|\mathbf{r}'(t)\|\,dt

  • Vector Line Integrals and Work:

    • The work done by a continuous vector force field F=F1i+F2j+F3k\mathbf{F} = F_1\mathbf{i} + F_2\mathbf{j} + F_3\mathbf{k} moving a particle along curve CC is:     W=∫CF⋅dr=∫CF⋅T ds=∫abF(r(t))⋅r′(t) dt=∫CF1 dx+F2 dy+F3 dzW = \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C \mathbf{F} \cdot \mathbf{T}\,ds = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt = \int_C F_1\,dx + F_2\,dy + F_3\,dz

    • Reversing path orientation inverts the sign: ∫−CF⋅dr=−∫CF⋅dr\int_{-C} \mathbf{F} \cdot d\mathbf{r} = -\int_C \mathbf{F} \cdot d\mathbf{r}.

  • Conservative Vector Fields and Potential Theory:

    • A vector field F\mathbf{F} is conservative if ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r} is independent of path, depending solely on the endpoints AA and BB.

    • Fundamental Theorem for Line Integrals: A vector field F\mathbf{F} is conservative if and only if it is a gradient field, meaning there exists a scalar potential function ff such that F=∇f\mathbf{F} = \nabla f. In this case:     ∫ABF⋅dr=∫AB∇f⋅dr=f(B)−f(A)\int_A^B \mathbf{F} \cdot d\mathbf{r} = \int_A^B \nabla f \cdot d\mathbf{r} = f(B) - f(A)

    • Component Criterion: On a simply connected domain in R2\mathbb{R}^2, F=(F1,F2)\mathbf{F} = (F_1, F_2) is conservative if and only if:     ∂F1∂y=∂F2∂x\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}

  • Conservation of Energy and Central Forces in 3D Space:

    • Under a conservative force F=−∇V(r)\mathbf{F} = -\nabla V(\mathbf{r}), Newton's second law reads mr¨=−∇V(r)m\ddot{\mathbf{r}} = -\nabla V(\mathbf{r}).

  • Taking the inner product with r˙\dot{\mathbf{r}}, the multivariable chain rule yields:   mr¨⋅r˙=−∇V(r)⋅r˙  ⟹  ddt(12m∥r˙∥2)=−dVdt  ⟹  ddt(12m∥r˙∥2+V(r))=0m\ddot{\mathbf{r}} \cdot \dot{\mathbf{r}} = -\nabla V(\mathbf{r}) \cdot \dot{\mathbf{r}} \implies \frac{d}{dt}\left(\frac{1}{2}m\|\dot{\mathbf{r}}\|^2\right) = -\frac{dV}{dt} \implies \frac{d}{dt}\left(\frac{1}{2}m\|\dot{\mathbf{r}}\|^2 + V(\mathbf{r})\right) = 0

  • Hence, total mechanical energy is conserved: E=12m∥r˙∥2+V(r)E = \frac{1}{2}m\|\dot{\mathbf{r}}\|^2 + V(\mathbf{r}).

  • Central Force Dynamics:

    • A force is central if F(r)=F(r)r^\mathbf{F}(\mathbf{r}) = F(r)\hat{\mathbf{r}}, where r=∥r∥r = \|\mathbf{r}\| and r^=rr\hat{\mathbf{r}} = \frac{\mathbf{r}}{r}.

    • Every continuous central force field is conservative, with potential V(r)=−∫F(r) drV(r) = -\int F(r)\,dr.

    • Angular Momentum: L=m(r×v)\mathbf{L} = m(\mathbf{r} \times \mathbf{v}).

    • Differentiating with respect to time yields torque: dLdt=m(r˙×v)+m(r×v˙)=0+r×F\frac{d\mathbf{L}}{dt} = m(\dot{\mathbf{r}} \times \mathbf{v}) + m(\mathbf{r} \times \dot{\mathbf{v}}) = \mathbf{0} + \mathbf{r} \times \mathbf{F}.

    • For central forces, r×F=r×(F(r)r^)=0\mathbf{r} \times \mathbf{F} = \mathbf{r} \times (F(r)\hat{\mathbf{r}}) = \mathbf{0}, proving that angular momentum L\mathbf{L} is strictly conserved.

    • Planar Motion: Since r(t)⋅L=r(t)⋅[m(r(t)×v(t))]=0\mathbf{r}(t) \cdot \mathbf{L} = \mathbf{r}(t) \cdot [m(\mathbf{r}(t) \times \mathbf{v}(t))] = 0, the position vector r(t)\mathbf{r}(t) must remain orthogonal to the constant vector L\mathbf{L}, confining the motion strictly to a two-dimensional plane.

  • Kepler's First Law: In an inverse-square gravitational field F(r)=−GMmr2F(r) = -\frac{GMm}{r^2}, the planar trajectory is a conic section:   r=l1+ecos⁡(θ)r = \frac{l}{1 + e\cos(\theta)}   where ee is the eccentricity and ll is the semi-latus rectum. Bound orbits (planets) have 0≤e<10 \le e < 1 (ellipses/circles with maximum distance apoapsis), whereas unbound trajectories have e≥1e \ge 1 (parabolas/hyperbolas).

First-Order Ordinary Differential Equations: Theory, Geometry, and Numerics

  • Classification and Fundamentals:

    • An Ordinary Differential Equation (ODE) involves derivatives of an unknown function with respect to a single independent variable.

    • A Partial Differential Equation (PDE) involves partial derivatives with respect to two or more independent variables.

    • The order of an ODE is defined as the order of the highest derivative appearing in the equation.

    • A function y=f(x)y = f(x) is a solution to an ODE if direct substitution of yy and its derivatives satisfies the equation identically over an interval II.

    • The general solution contains all possible solutions, incorporating arbitrary constants of integration (nn constants for an nn-th order equation).

    • An Initial Value Problem (IVP) pairs an ODE with specified initial conditions (e.g., y(t0)=y0y(t_0) = y_0), uniquely fixing arbitrary constants.

  • Geometric Interpretation: Slope Fields and Solution Curves:

    • For a first-order ODE dydt=f(t,y)\frac{dy}{dt} = f(t, y), the right-hand side prescribes the slope of the solution curve passing through any point (t,y)(t, y).

    • A slope field (or direction field) is constructed by drawing short line segments with slope f(ti,yj)f(t_i, y_j) at grid points in the tyty-plane, providing a qualitative portrait of solution trajectories without integration.

  • Equilibrium Solutions and Stability Analysis:

    • An equilibrium solution is a constant solution y(t)=cy(t) = c satisfying dydt=0\frac{dy}{dt} = 0 for all tt. It corresponds to roots of f(t,c)=0f(t, c) = 0.

    • In autonomous systems y′=f(y)y' = f(y), equilibrium points partition the phase space.

    • Stability Classification:

    • Stable: An equilibrium y(t)=y0y(t) = y_0 is stable if any solution with initial condition y(0)=y0±ϵy(0) = y_0 \pm \epsilon sufficiently close to y0y_0 satisfies:       lim⁡t→∞y(t)=y0\lim_{t \to \infty} y(t) = y_0       Solution trajectories converge toward y0y_0.

    • Unstable: Solution trajectories diverge away from y0y_0 as t→∞t \to \infty.

    • Semi-stable: Trajectories approach y0y_0 from one side but diverge on the opposite side.

    • Example Analysis:

    • ODE dydt=2y\frac{dy}{dt} = 2y: Equilibrium at y=0y = 0. For y>0y > 0, y′>0y' > 0 (increasing away); for y<0y < 0, y′<0y' < 0 (decreasing away). Hence, y=0y = 0 is unstable.

Slope field of y' = 2y
  • ODE dydt=−3(y−1)\frac{dy}{dt} = -3(y - 1): Equilibrium at y=1y = 1. For y>1y > 1, y′<0y' < 0 (decreasing toward 1); for y<1y < 1, y′>0y' > 0 (increasing toward 1). Hence, y=1y = 1 is stable.

Slope field and solution curves for y' = -3(y - 1)
  • ODE dydt=2t+1\frac{dy}{dt} = 2t + 1: Direct integration yields y(t)=t2+t+Cy(t) = t^2 + t + C. Since y′=0y' = 0 cannot hold for all tt, there are no equilibrium solutions.

Slope field and solution curve for y' = 2t + 1
  • ODE dydt=y(1−y)\frac{dy}{dt} = y(1 - y): Equilibria at y=0y = 0 and y=1y = 1. For y>1y > 1, y′<0y' < 0; for 0<y<10 < y < 1, y′>0y' > 0; for y<0y < 0, y′<0y' < 0. Hence, y=1y = 1 is stable, and y=0y = 0 is unstable.

Slope field and trajectories for logistic equation y' = y(1 - y)
  • ODE dydt=sin⁡(4y)\frac{dy}{dt} = \sin(4y): Setting sin⁡(4y)=0  ⟹  4y=nπ  ⟹  y=nπ4\sin(4y) = 0 \implies 4y = n\pi \implies y = \frac{n\pi}{4} for n∈Zn \in \mathbb{Z}. Infinite equilibrium solutions:

    • For n=2kn = 2k (even integers): y=2kπ4=kπ2y = \frac{2k\pi}{4} = \frac{k\pi}{2} are unstable.

    • For n=2k+1n = 2k + 1 (odd integers): y=(2k+1)π4y = \frac{(2k + 1)\pi}{4} are stable.

Slope field for y' = sin(4y)
* ODE dydt=y2\frac{dy}{dt} = y^2: Equilibrium at y=0y = 0. For y>0y > 0, y′>0y' > 0 (diverges); for y<0y < 0, y′>0y' > 0 (approaches 0). Hence, y=0y = 0 is **semi-stable** (and classified overall as unstable).
  • Picard-Lindelöf Existence and Uniqueness Theorem:

    • Let f(t,y)f(t, y) and ∂f∂y\frac{\partial f}{\partial y} be continuous functions on a closed rectangle R=[t0−h,t0+h]×[y0−k,y0+k]R = [t_0 - h, t_0 + h] \times [y_0 - k, y_0 + k] containing (t0,y0)(t_0, y_0). Then the IVP:     dydt=f(t,y),y(t0)=y0\frac{dy}{dt} = f(t, y), \quad y(t_0) = y_0     has a unique solution y(t)y(t) defined on an open interval containing t0t_0.

    • Crucial Consequence: Solution trajectories cannot cross or intersect one another. If two distinct solution curves intersected, the ODE would have multiple distinct slopes at the same point, violating uniqueness. Therefore, equilibrium solutions act as impermeable barriers partitioning the solution space.

  • Numerical Methods for Initial Value Problems:

    • Euler's Method:

    • Approximates the solution curve by stepping along tangent lines. At step nn, the tangent line at (tn,yn)(t_n, y_n) gives:       yn+1=yn+f(tn,yn)Δt,tn+1=tn+Δty_{n+1} = y_n + f(t_n, y_n)\Delta t, \quad t_{n+1} = t_n + \Delta t

    • Worked Manual Example: dydt=2t\frac{dy}{dt} = 2t, y(0)=0y(0) = 0, Δt=0.2\Delta t = 0.2. Compute y(0.6)y(0.6).

      • Exact analytical solution: y(t)=t2  ⟹  y(0.6)=0.36y(t) = t^2 \implies y(0.6) = 0.36.

      • Step 0: t0=0t_0 = 0, y0=0y_0 = 0.

      • Step 1: t1=0.2t_1 = 0.2, y1=y0+2t0Δt=0+2(0)(0.2)=0y_1 = y_0 + 2t_0\Delta t = 0 + 2(0)(0.2) = 0.

      • Step 2: t2=0.4t_2 = 0.4, y2=y1+2t1Δt=0+2(0.2)(0.2)=0.08y_2 = y_1 + 2t_1\Delta t = 0 + 2(0.2)(0.2) = 0.08.

      • Step 3: t3=0.6t_3 = 0.6, y3=y2+2t2Δt=0.08+2(0.4)(0.2)=0.08+0.16=0.24y_3 = y_2 + 2t_2\Delta t = 0.08 + 2(0.4)(0.2) = 0.08 + 0.16 = 0.24.

      • Numerical error at t=0.6t = 0.6: ∣0.36−0.24∣=0.12|0.36 - 0.24| = 0.12.

    • Implementing with a smaller step size Δt=0.05\Delta t = 0.05 (1212 steps) yields y(0.6)=0.33y(0.6) = 0.33.

    • Python Implementation of Euler's Method:

import matplotlib.pyplot as plt
import numpy as np

t = np.arange(0, 0.6 + 0.05, 0.05)
y = np.zeros(len(t))
y[0] = 0
for i in range(1, len(t)):
  y[i] = y[i - 1] + 2 * t[i - 1] * 0.05
print(y[-1])  # outputs 0.33
plt.plot(t, y, "-")
plt.show()
  • Heun's Method (Improved Euler):

    • Derived by integrating the ODE across a single time step using the Second Fundamental Theorem of Calculus:       y(tn+1)=y(tn)+∫tntn+1f(t,y(t)) dty(t_{n+1}) = y(t_n) + \int_{t_n}^{t_{n+1}} f(t, y(t))\,dt

    • While Euler's method approximates the integral as a left-endpoint rectangle of area f(tn,yn)Δtf(t_n, y_n)\Delta t, Heun's method uses a trapezoid of area 12[f(tn,yn)+f(tn+1,yn+1)]Δt\frac{1}{2}[f(t_n, y_n) + f(t_{n+1}, y_{n+1})]\Delta t.

    • To resolve the implicit dependence on yn+1y_{n+1}, the standard Euler step acts as a predictor:       y~n+1=yn+f(tn,yn)Δt\tilde{y}_{n+1} = y_n + f(t_n, y_n)\Delta t       yn+1=yn+12[f(tn,yn)+f(tn+Δt,y~n+1)]Δty_{n+1} = y_n + \frac{1}{2}\left[f(t_n, y_n) + f(t_n + \Delta t, \tilde{y}_{n+1})\right]\Delta t

    • Separable First-Order ODEs:

  • A first-order ODE is separable if it can be written as:     dydx=f(x)g(y)\frac{dy}{dx} = f(x)g(y)

  • Solution procedure:

    1. Separate variables: 1g(y) dy=f(x) dx\frac{1}{g(y)}\,dy = f(x)\,dx, valid for g(y)≠0g(y) \ne 0.

    2. Integrate both sides: ∫1g(y) dy=∫f(x) dx\int \frac{1}{g(y)}\,dy = \int f(x)\,dx.

    3. Evaluate integrals and solve explicitly for y(x)y(x) when feasible.

  • Singular/Equilibrium Solutions: Points y=ay = a where g(a)=0g(a) = 0 yield constant equilibrium solutions y(x)=ay(x) = a. These solutions may not emerge from the general integral and must be identified separately.

  • Worked Example 1: dydx=xy\frac{dy}{dx} = \frac{x}{y}, y(0)=3y(0) = 3.

    • ∫y dy=∫x dx  ⟹  y22=x22+C1  ⟹  y2=x2+C\int y\,dy = \int x\,dx \implies \frac{y^2}{2} = \frac{x^2}{2} + C_1 \implies y^2 = x^2 + C.

    • Applying y(0)=3y(0) = 3 yields 32=0+C  ⟹  C=93^2 = 0 + C \implies C = 9.

    • Explicit solution: y(x)=x2+9y(x) = \sqrt{x^2 + 9}.

  • Worked Example 2: dydx=sin⁡(x)y\frac{dy}{dx} = \frac{\sin(x)}{y}, y(0)=−1y(0) = -1.

    • ∫y dy=∫sin⁡(x) dx  ⟹  y22=−cos⁡(x)+C1  ⟹  y2=−2cos⁡(x)+C\int y\,dy = \int \sin(x)\,dx \implies \frac{y^2}{2} = -\cos(x) + C_1 \implies y^2 = -2\cos(x) + C.

    • Applying y(0)=−1y(0) = -1 yields (−1)2=−2cos⁡(0)+C  ⟹  1=−2+C  ⟹  C=3(-1)^2 = -2\cos(0) + C \implies 1 = -2 + C \implies C = 3.

    • Selecting the negative branch matching the initial condition: y(x)=−3−2cos⁡(x)y(x) = -\sqrt{3 - 2\cos(x)}.

    • Linear First-Order ODEs and Integrating Factors:

  • A first-order ODE is linear if it can be arranged in standard form:     dydt+p(t)y=q(t)\frac{dy}{dt} + p(t)y = q(t)

  • Linear Operator Property: The differential operator L[y]=dydt+p(t)yL[y] = \frac{dy}{dt} + p(t)y is linear: L[αy1+βy2]=αL[y1]+βL[y2]L[\alpha y_1 + \beta y_2] = \alpha L[y_1] + \beta L[y_2].

  • Integrating Factor Derivation:

    • Multiply by an integrating factor I(t)I(t): I(t)dydt+I(t)p(t)y=I(t)q(t)I(t)\frac{dy}{dt} + I(t)p(t)y = I(t)q(t).

    • Demand that the LHS matches the derivative of a product: ddt[I(t)y]=I(t)dydt+I′(t)y\frac{d}{dt}[I(t)y] = I(t)\frac{dy}{dt} + I'(t)y.

    • Equating terms yields I′(t)=I(t)p(t)  ⟹  dII=p(t) dt  ⟹  I(t)=exp⁡(∫p(t) dt)I'(t) = I(t)p(t) \implies \frac{dI}{I} = p(t)\,dt \implies I(t) = \exp\left(\int p(t)\,dt\right).

    • Any constant multiple of I(t)I(t) divides out, so the integration constant can be taken as zero.

    • The general solution is obtained by direct integration:       I(t)y=∫I(t)q(t) dt+C  ⟹  y(t)=1I(t)(∫I(t)q(t) dt+C)I(t)y = \int I(t)q(t)\,dt + C \implies y(t) = \frac{1}{I(t)}\left(\int I(t)q(t)\,dt + C\right)

  • Worked Example: Solve \frac{dy}{dt} + \left(\frac{2t}{t^2 + 1}\right)y = t^2 + 1$.\n * Here p(t) = \frac{2t}{t^2 + 1},,q(t) = t^2 + 1.\n * Integrating factor: I(t) = \exp\left(\int \frac{2t}{t^2 + 1}\,dt\right) = \exp(\ln(t^2 + 1)) = t^2 + 1.\n * Multiplying through: \frac{d}{dt}[(t^2 + 1)y] = (t^2 + 1)^2 = t^4 + 2t^2 + 1.\n * Integrating: (t^2 + 1)y = \frac{t^5}{5} + \frac{2t^3}{3} + t + C.\n * General solution: y(t) = \frac{\frac{t^5}{5} + \frac{2t^3}{3} + t + C}{t^2 + 1}.\n\n# Applications of First-Order Ordinary Differential Equations\n\n* **Newton's Law of Cooling:**\n * The rate of change of temperature T(t)ofabodyisproportionaltothedifferencebetweenitstemperatureandtheambientmediumof a body is proportional to the difference between its temperature and the ambient mediumT_m:\n    \frac{dT}{dt} = -k(T - T_m), \quad k > 0\n * Standard linear form: \frac{dT}{dt} + kT = kT_m.\n * Integrating factor I(t) = e^{kt}.Generalsolution:. General solution:T(t) = (T_0 - T_m)e^{-kt} + T_m,where, whereT(0) = T_0.\n * **Forensic Post-Mortem Analysis (Carl W. Case):**\n * Body discovered at 11:00 am on 19 April 2010 in a prison cell at T_m = 21.1^\circ\text{C}.\n * Measurements: at 11:00 am (t = 0),),T(0) = 34.8^\circ\text{C};at12:30pm(; at 12:30 pm (t = 1.5\,\text{hours}),),T(1.5) = 34.1^\circ\text{C}.\n * Normal living body temperature at death is assumed to be 37.0^\circ\text{C}.\n * Determine cooling constant k:\n      T(t) = (34.8 - 21.1)e^{-kt} + 21.1 = 13.7e^{-kt} + 21.1\n      T(1.5) = 13.7e^{-1.5k} + 21.1 = 34.1 \implies 13.7e^{-1.5k} = 13.0 \implies e^{-1.5k} = \frac{13.0}{13.7}\n      k = -\frac{1}{1.5}\ln\left(\frac{13.0}{13.7}\right) \approx 0.03496\,\text{hr}^{-1}\n * Determine time of death t_d:\n      37.0 = 13.7e^{-k t_d} + 21.1 \implies 15.9 = 13.7e^{-k t_d} \implies e^{-k t_d} = \frac{15.9}{13.7}\n      t_d = -\frac{1}{k}\ln\left(\frac{15.9}{13.7}\right) = -\frac{\ln(15.9 / 13.7)}{0.03496} \approx -4.26\,\text{hours}\n * Calculating time of death: 11:00\,\text{am} - 4.26\,\text{hours} = 11:00\,\text{am} - 4\,\text{hours } 16\,\text{minutes} = 6:44\,\text{am} \approx 6:45\,\text{am}.\n\n* **Pharmacokinetics: Drug Absorption and Elimination:**\n * Drug amount y(t)(in(in\text{mg})inthebloodstreamchangesviagutabsorption) in the bloodstream changes via gut absorptiona(t)andrenal/hepaticclearanceproportionaltoconcentration(rateand renal/hepatic clearance proportional to concentration (ratek y,,k > 0):\n    \frac{dy}{dt} + ky = a(t)\n * For a single oral dose Dwithabsorptionrateconstantwith absorption rate constantk_a((k_a e k),theabsorptioninputdecaysexponentially:), the absorption input decays exponentially:a(t) = k_a D e^{-k_a t}.\n * Integrating factor I(t) = e^{kt}:\n    \frac{d}{dt}[e^{kt}y] = k_a D e^{(k - k_a)t} \implies e^{kt}y = \frac{k_a D}{k - k_a} e^{(k - k_a)t} + B\n * Initial condition y(0) = 0 \implies B = \frac{k_a D}{k_a - k}. The exact solution is:\n    y(t) = \frac{k_a D}{k_a - k}\left(e^{-kt} - e^{-k_a t}\right)\n * *Numerical Case:* Dose D = 100\,\text{mg},,k_a = 2\,\text{hr}^{-1},,k = 1\,\text{hr}^{-1}.\n    y(t) = \frac{2(100)}{2 - 1}\left(e^{-t} - e^{-2t}\right) = 200\left(e^{-t} - e^{-2t}\right)\n * After 1 hour: y(1) = 200(e^{-1} - e^{-2}) \approx 46.5\,\text{mg}.\n * Peak time: Setting y'(t) = 200(-e^{-t} + 2e^{-2t}) = 0 \implies e^t = 2 \implies t_{\text{peak}} = \ln(2) \approx 0.693\,\text{hours}.\n * Peak amount: y(\ln(2)) = 200(1/2 - 1/4) = 50\,\text{mg}.\n * Continuous IV infusion: If a(t) = R(constantrate),then(constant rate), theny(t) = \frac{R}{k}(1 - e^{-kt}),levellingoffatsteady−stateamount, levelling off at steady-state amounty_{\infty} = \frac{R}{k}.\n\n* **Single-Species Population Dynamics:**\n * **Malthusian Growth Model:** Unlimited growth rate proportional to population P(t)::\frac{dP}{dt} = rP \implies P(t) = P_0 e^{rt}.\n * *Projection of Land Exhaustion:* Current population growth rate r = 0.0125\,\text{yr}^{-1}.Inyear2000,. In year 2000,P(2000) = 6\,\text{billion}.Totalplanetarylandareaisconstantat. Total planetary land area is constant at150 \times 10^6\,\text{km}^2 = 1.5 \times 10^{14}\,\text{m}^2.\n * Carrying capacity constraint: 1 person per square metre yields maximum capacity P_{\text{crit}} = 1.5 \times 10^{14}\,\text{people} = 150000\,\text{billion}.\n * Setting 6 e^{0.0125(t - 2000)} = 150000 \implies e^{0.0125(t - 2000)} = 25000.\n * Solving for t:\n      t = 2000 + \frac{\ln(25000)}{0.0125} = 2000 + 80 \ln(25000) \approx 2000 + 810.13 \approx \text{Year } 2810\n * **Verhulst Logistic Growth Model:** Incorporates environmental carrying capacity \theta:\n    \frac{dP}{dt} = rP\left(1 - \frac{P}{\theta}\right)\n * Nondimensionalisation: Define y = \frac{P}{\theta}andand\tau = rt. The equation simplifies to:\n      \frac{dy}{d\tau} = y(1 - y)\n * Separating variables via partial fractions:\n      \int \left(\frac{1}{y} + \frac{1}{1 - y}\right)\,dy = \int d\tau \implies \ln\left|\frac{y}{1 - y}\right| = \tau + C \implies \frac{y}{1 - y} = A e^\tau\n * Solving for y(\tau)withinitialstatewith initial statey(0) = y_0:\n      y(\tau) = \frac{y_0 e^\tau}{1 - y_0 + y_0 e^\tau}\n * In terms of original dimensional variables with P(0) = P_0:\n      P(t) = \frac{\theta P_0 e^{rt}}{\theta - P_0 + P_0 e^{rt}}\n * **Logistic Model with Harvesting (Pacific Halibut Fishery):**\n * Proportional harvesting rate E P modifies the logistic equation:\n      \frac{dP}{dt} = rP\left(1 - \frac{P}{\theta}\right) - EP = (r - E)P\left(1 - \frac{P}{\theta(1 - E/r)}\right)\n * Equilibria occur at P = 0andandP^* = \theta\left(1 - \frac{E}{r}\right).\n * To prevent population collapse and extinction, the harvesting intensity must satisfy E < r.\n\n* **Coupled Systems of First-Order ODEs:**\n * **The Lotka-Volterra Predator-Prey System:**\n * Let prey population be R(t)(rabbits)andpredatorpopulationbe(rabbits) and predator population beW(t) (wolves):\n      \frac{dR}{dt} = \alpha R - \gamma RW = R(\alpha - \gamma W)\n      \frac{dW}{dt} = -\beta W + \delta RW = W(-\beta + \delta R)\n      where \alpha, \beta, \gamma, \delta > 0.\n * Equilibria occur when \dot{R} = 0andand\dot{W} = 0:\n      (R, W) = (0, 0) \quad \text{and} \quad (R^, W^) = \left(\frac{\beta}{\delta}, \frac{\alpha}{\gamma}\right)\n * Phase plane differential equation:\n      \frac{dW}{dR} = \frac{W(-\beta + \delta R)}{R(\alpha - \gamma W)}\n * *Numerical Parameter Study:* \alpha = 0.08,,\beta = 0.02,,\gamma = 0.001,,\delta = 0.00002.\n * Non-zero equilibrium: R^* = \frac{0.02}{0.00002} = 1000,,W^* = \frac{0.08}{0.001} = 80.\n * Initial condition: R(0) = 1000,,W(0) = 60.\n\n![Predator-prey trajectory on the slope field (t up to 100)](https://assets.knowt.com/pdf-flow-prod/13c9c582-d565-42ce-8982-fc2480ed624e-figures/0.png)\n\n![Populations vs time for predator-prey system (t up to 100)](https://assets.knowt.com/pdf-flow-prod/13c9c582-d565-42ce-8982-fc2480ed624e-figures/1.png)\n\n* Extended integration (t = 400) demonstrates closed periodic orbits in phase space and boom-bust population cycles.\n\n![Trajectory forms an almost-closed loop (t up to 400)](https://assets.knowt.com/pdf-flow-prod/13c9c582-d565-42ce-8982-fc2480ed624e-figures/2.png)\n\n![Boom-bust cycle of rabbit and wolf populations (t up to 400)](https://assets.knowt.com/pdf-flow-prod/13c9c582-d565-42ce-8982-fc2480ed624e-figures/3.png)\n\n![Phase plane trajectories for five initial wolf counts](https://assets.knowt.com/pdf-flow-prod/13c9c582-d565-42ce-8982-fc2480ed624e-figures/4.png)\n\n* **Competing Species Model:**\n * Two species P_1andandP_2 competing for shared resources:\n    \frac{dP_1}{dt} = P_1(\alpha_1 - \beta_1 P_1 - \gamma_{12} P_2)\n    \frac{dP_2}{dt} = P_2(\alpha_2 - \beta_2 P_2 - \gamma_{21} P_1)\n * *Worked Parameter System:* \alpha_1 = \beta_1 = \beta_2 = \gamma_{12} = 1,,\alpha_2 = 0.75,,\gamma_{21} = 0.5.\n    \frac{dP_1}{dt} = P_1(1 - P_1 - P_2)\n    \frac{dP_2}{dt} = P_2(0.75 - P_2 - 0.5P_1)\n * Determining all four equilibrium points (\dot{P}1 = 0andand\dot{P}_2 = 0):\n * If P_1 = 0 \implies P_2(0.75 - P_2) = 0 \implies P_2 = 0ororP_2 = 0.75.\n * If 1 - P_1 - P_2 = 0 \implies P_1 = 1 - P_2.Substitutinginto. Substituting into\dot{P}_2 = 0:\n      P_2(0.75 - P_2 - 0.5(1 - P_2)) = 0 \implies P_2(0.25 - 0.5P_2) = 0\n      Yielding P_2 = 0 \implies P_1 = 1,or, orP_2 = 0.5 \implies P_1 = 0.5.\n * The four equilibria are: (0, 0),,(0, 0.75),,(1, 0),and, and(0.5, 0.5).\n\n![Competing species phase plane trajectories](https://assets.knowt.com/pdf-flow-prod/13c9c582-d565-42ce-8982-fc2480ed624e-figures/5.png)\n\n![Competing species slope field](https://assets.knowt.com/pdf-flow-prod/13c9c582-d565-42ce-8982-fc2480ed624e-figures/6.png)\n\n* Evolution from P_1(0) = 0.6andandP_2(0) = 0.4convergesdirectlytothestableco−existenceequilibriumconverges directly to the stable co-existence equilibrium(0.5, 0.5).\n\n![Evolution toward a stable equilibrium for competing species](https://assets.knowt.com/pdf-flow-prod/13c9c582-d565-42ce-8982-fc2480ed624e-figures/7.png)\n\n# Second-Order Linear Differential Equations\n\n* **Definitions and Structural Foundations:**\n * General second-order linear differential equation:\n    \frac{d^2y}{dt^2} + p(t)\frac{dy}{dt} + q(t)y = r(t)\n * Functions p(t)andandq(t)arethecoefficients;ifare the coefficients; ifr(t) = 0, the equation is **homogeneous**; otherwise it is **inhomogeneous**.\n * **Linear Differential Operator:** Defining L[y] = y'' + p(t)y' + q(t)y, linearity requires:\n    L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2]\n * **Superposition Principle:** If y_1(t)andandy_2(t)aresolutionstoahomogeneouslinearODE(are solutions to a homogeneous linear ODE (L[y] = 0),thenanylinearcombination), then any linear combinationy(t) = c_1 y_1(t) + c_2 y_2(t) is also a solution.\n * **Linear Independence and General Solutions:** Two solutions y_1(t)andandy_2(t)arelinearlyindependentifare linearly independent ify_2(t) e C y_1(t). The general solution of a homogeneous second-order linear ODE is spanned by exactly two linearly independent solutions:\n    y(t) = c_1 y_1(t) + c_2 y_2(t)\n\n* **Reduction of Order:**\n * If one non-zero solution y_1(x)totoy'' + p(x)y' + q(x)y = 0isknown,asecondlinearlyindependentsolutionisfoundbysettingis known, a second linearly independent solution is found by settingy_2(x) = u(x)y_1(x).\n * Derivatives: y_2' = u'y_1 + uy_1',,y_2'' = u''y_1 + 2u'y_1' + uy_1''.\n * Substitution into the ODE gives:\n    u''y_1 + u'(2y_1' + py_1) + u(y_1'' + py_1' + qy_1) = 0\n * Since L[y_1] = 0,thecoefficientof, the coefficient ofu vanishes identically, leaving:\n    u'' + \left(p + \frac{2y_1'}{y_1}\right)u' = 0\n * Setting v = u'yieldsafirst−orderlinearODEinyields a first-order linear ODE inv,solvedusingintegratingfactor, solved using integrating factorI(x) = \exp\left(\int \left(p + \frac{2y_1'}{y_1}\right)dx\right) = y_1^2 \exp\left(\int p\,dx\right).\n * *Worked Example:* Solve x^2 y'' + 2x y' - 6y = 0givengiveny_1(x) = x^2.\n * Set y_2 = u(x)x^2.Derivatives:. Derivatives:y_2' = u'x^2 + 2ux,,y_2'' = u''x^2 + 4u'x + 2u.\n * Substitute: x^2(u''x^2 + 4u'x + 2u) + 2x(u'x^2 + 2ux) - 6ux^2 = 0.\n * Simplification: x^4 u'' + 6x^3 u' = 0 \implies u'' + \frac{6}{x}u' = 0.\n * Set v = u'::v' + \frac{6}{x}v = 0 \implies \frac{dv}{v} = -\frac{6}{x}\,dx \implies v(x) = C_1 x^{-6}.\n * Integrate u' = C_1 x^{-6} \implies u(x) = -\frac{C_1}{5}x^{-5} + C_2 = A x^{-5} + B.\n * Second solution: y_2(x) = x^2(x^{-5}) = x^{-3}.\n * General solution: y(x) = c_1 x^2 + c_2 x^{-3}.\n\n* **Homogeneous Linear Equations with Constant Coefficients:**\n * General form: a y'' + b y' + c y = 0withwitha e 0.\n * Characteristic equation ansatz: y = e^{\lambda t} \implies a\lambda^2 + b\lambda + c = 0.\n * Roots via quadratic formula: \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\n * **Classification of Solution Regimes:**\n * **Case 1: Two Distinct Real Roots (b^2 - 4ac > 0):**\n      y(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}\n * *Example:* y'' - 3y' + 2y = 0 \implies \lambda^2 - 3\lambda + 2 = 0 \implies (\lambda - 1)(\lambda - 2) = 0.\n * Roots: \lambda_1 = 1, \lambda_2 = 2.Generalsolution:. General solution:y(t) = c_1 e^t + c_2 e^{2t}.\n * **Case 2: Repeated Real Root (b^2 - 4ac = 0):**\n      \lambda_1 = \lambda_2 = \mu = -\frac{b}{2a}\n      By reduction of order, the two linearly independent solutions are e^{\mu t}andandt e^{\mu t}, giving:\n      y(t) = (c_1 + c_2 t)e^{\mu t}\n * *Example (IVP):* y'' + 6y' + 9y = 0,,y(0) = 2,,y'(0) = 0.\n * Characteristic equation: \lambda^2 + 6\lambda + 9 = (\lambda + 3)^2 = 0 \implies \lambda = -3.\n * General solution: y(t) = (c_1 + c_2 t)e^{-3t}.\n * Initial conditions: y(0) = c_1 = 2.\n * Derivative: y'(t) = c_2 e^{-3t} - 3(c_1 + c_2 t)e^{-3t} \implies y'(0) = c_2 - 3c_1 = 0 \implies c_2 = 6.\n * Unique solution: y(t) = (2 + 6t)e^{-3t}.\n * **Case 3: Complex Conjugate Roots (b^2 - 4ac < 0):**\n      \lambda = \alpha \pm i\beta, \quad \alpha = -\frac{b}{2a}, \quad \beta = \frac{\sqrt{4ac - b^2}}{2a}\n      Using Euler's formula e^{i\theta} = \cos(\theta) + i\sin(\theta), the real general solution is:\n      y(t) = e^{\alpha t}(b_1 \cos(\beta t) + b_2 \sin(\beta t))\n * *Example (IVP):* y'' - 4y' + 5y = 0,,y(0) = 1,,y'(0) = 0.\n * Characteristic roots: \lambda = \frac{4 \pm \sqrt{16 - 20}}{2} = 2 \pm i \implies \alpha = 2, \beta = 1.\n * General solution: y(t) = e^{2t}(b_1 \cos(t) + b_2 \sin(t)).\n * y(0) = b_1 = 1.\n * Derivative: y'(t) = 2e^{2t}(b_1 \cos(t) + b_2 \sin(t)) + e^{2t}(-b_1 \sin(t) + b_2 \cos(t)).\n * y'(0) = 2b_1 + b_2 = 0 \implies b_2 = -2.\n * Unique solution: y(t) = e^{2t}(\cos(t) - 2\sin(t)).\n\n* **Inhomogeneous Linear Second-Order Equations:**\n * General equation: a y'' + b y' + c y = r(x).\n * Structure of General Solution: y(x) = y_H(x) + y_P(x),where, wherey_H(x) = c_1 y_1(x) + c_2 y_2(x)isthecomplementaryfunctionsolvingthehomogeneousequation,andis the complementary function solving the homogeneous equation, andy_P(x) is any particular solution to the inhomogeneous equation.\n * **Method of Undetermined Coefficients:**\n * *Case 1: Polynomial Forcing:* For r(x)apolynomialofdegreea polynomial of degreen,choose, choosey_P(x) = A_n x^n + \dots + A_1 x + A_0.\n * Example: y'' + y' - 2y = x^2 - 2x + 3.\n * Homogeneous roots: \lambda^2 + \lambda - 2 = 0 \implies \lambda = 1, -2,so, soy_H = c_1 e^x + c_2 e^{-2x}.\n * Particular trial: y_P = A x^2 + B x + C.Derivatives:. Derivatives:y_P' = 2Ax + B,,y_P'' = 2A.\n * Substitute: 2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2 - 2x + 3.\n * Match coefficients:\n * x^2: -2A = 1 \implies A = -1/2\n * x^1: 2A - 2B = -2 \implies -1 - 2B = -2 \implies B = 1/2\n * x^0: 2A + B - 2C = 3 \implies -1 + 1/2 - 2C = 3 \implies -2C = 7/2 \implies C = -7/4\n * Particular solution: y_P(x) = -\frac{1}{2}x^2 + \frac{1}{2}x - \frac{7}{4} = \frac{1}{4}(-2x^2 + 2x - 7).\n * *Case 2: Exponential-Polynomial Forcing:* For r(x) = P(x)e^{kx},set, sety_P(x) = u(x)e^{kx}.\n * *Case 3: Trigonometric Forcing:* For r(x) = P(x)\cos(\omega x)ororP(x)\sin(\omega x),set, sety_P(x) = A(x)\cos(\omega x) + B(x)\sin(\omega x).\n * *Case 4: Superposition of Forcing Terms:* If r(x) = r_1(x) + r_2(x),and, andy{P1}, y_{P2}solvetheODEforsolve the ODE forr_1, r_2respectively,thenrespectively, theny_P = y_{P1} + y_{P2}.\n\n# Physical Oscillations: Harmonic, Damped, Forced, and Resonant Systems\n\n* **The Undamped Mass-Spring Oscillator:**\n * A mass mattachedtoalinearspringofconstantattached to a linear spring of constantk > 0hangingundergravityhasnaturalequilibriumatstretchhanging under gravity has natural equilibrium at stretchssatisfyingsatisfyingks = mg.\n * Measuring displacement x(t)fromequilibrium(positivedownwards),Hooke′srestoringforceisfrom equilibrium (positive downwards), Hooke's restoring force isF_s = -k(s + x).\n * Total force: F = -k(s + x) + mg = -kx.\n * By Newton's second law: m\ddot{x} + kx = 0.\n * Defining natural angular frequency \omega = \sqrt{\frac{k}{m}}, the equation of motion is:\n    \ddot{x} + \omega^2 x = 0\n * Characteristic roots: \lambda = \pm i\omega.\n * General solution:\n    x(t) = c_1 \cos(\omega t) + c_2 \sin(\omega t) = A \cos(\omega t - \phi)\n    where amplitude A = \sqrt{c_1^2 + c_2^2}andphaseshiftand phase shift\phi = \tan^{-1}\left(\frac{c_2}{c_1}\right).\n\n* **The Damped Oscillator:**\n * Adding a viscous damping force proportional to velocity F_d = -\gamma \dot{x}((\gamma > 0):\n    m\ddot{x} + \gamma \dot{x} + kx = 0 \implies \ddot{x} + 2p\dot{x} + \omega^2 x = 0\n    where \omega^2 = \frac{k}{m} > 0andand2p = \frac{\gamma}{m} > 0.\n * Characteristic equation: \lambda^2 + 2p\lambda + \omega^2 = 0,withdiscriminant, with discriminantD = 4(p^2 - \omega^2).\n * **Classification of Damping Regimes:**\n * **Underdamping (p < \omega):**\n * Roots: \lambda = -p \pm i\omega_p,wherequasi−frequency, where quasi-frequency\omega_p = \sqrt{\omega^2 - p^2} = \omega \sqrt{1 - (p/\omega)^2}.\n * General solution:\n        x(t) = e^{-pt}\left(c_1 \cos(\omega_p t) + c_2 \sin(\omega_p t)\right) = A e^{-pt}\cos(\omega_p t - \phi)\n * Exhibits oscillatory decay under an exponentially decaying envelope A(t) = A e^{-pt}.\n * Because \omega_p < \omega, damping induces a frequency red-shift and increases the period.\n * **Critical Damping (p = \omega):**\n * Discriminant D = 0.Repeatedroot. Repeated root\lambda = -p.\n * General solution:\n        x(t) = (c_1 + c_2 t)e^{-pt} = e^{-pt}[x(0) + (x'(0) + px(0))t]\n * Provides the most rapid return to equilibrium without sustained oscillations.\n * Solution crosses equilibrium at most once if x'(0) < -p x(0).\n * **Overdamping (p > \omega):**\n * Discriminant D > 0.Twodistinctnegativerealroots. Two distinct negative real roots\lambda_1 = -p - \sqrt{p^2 - \omega^2},,\lambda_2 = -p + \sqrt{p^2 - \omega^2}.\n * General solution:\n        x(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}\n * Strictly non-oscillatory. Governed at large tbytheslower−decayingexponentby the slower-decaying exponente^{\lambda_2 t}, making the return to equilibrium slower than in critical damping.\n\n* **Forced Oscillations and Mechanical Resonance:**\n * Applying periodic external forcing: \ddot{x} + 2p\dot{x} + \omega^2 x = \alpha \cos(\Omega t).\n * **Damped Forced Response (p < \omega):**\n * Solution: x(t) = x_H(t) + x_P(t) = e^{-pt}(c_1 \cos(\omega_p t) + c_2 \sin(\omega_p t)) + A \cos(\Omega t - \phi).\n * The homogeneous part x_H(t) \to 0asast \to \infty (transient solution).\n * The long-time asymptotic response (steady-state solution) oscillates at the forcing frequency \Omega:\n      x_{\text{steady}}(t) = A(\Omega)\cos(\Omega t - \phi)\n      A(\Omega) = \frac{\alpha}{\sqrt{(\omega^2 - \Omega^2)^2 + 4p^2 \Omega^2}}, \quad \phi = \tan^{-1}\left(\frac{2p\Omega}{\omega^2 - \Omega^2}\right)\n * **Undamped Forced Oscillations and Pure Resonance (p = 0):**\n * Equation: \ddot{x} + \omega^2 x = \alpha \cos(\Omega t).\n * For \Omega e \omegawithinitialconditionswith initial conditionsx(0) = x_0, \dot{x}(0) = v_0:\n      x(t) = x_0 \cos(\omega t) + \frac{v_0}{\omega}\sin(\omega t) + \frac{\alpha}{\omega^2 - \Omega^2}(\cos(\Omega t) - \cos(\omega t))\n* Taking the resonant limit as \Omega \to \omega using L'Hôpital's rule on the forcing term:\n  \lim_{\Omega \to \omega} \frac{\alpha(\cos(\Omega t) - \cos(\omega t))}{\omega^2 - \Omega^2} = \lim_{\Omega \to \omega} \frac{-\alpha t \sin(\Omega t)}{-2\Omega} = \frac{\alpha t}{2\omega}\sin(\omega t)\n* **Resonant Solution:**\n  x(t) = x_0 \cos(\omega t) + \frac{v_0}{\omega}\sin(\omega t) + \frac{\alpha t}{2\omega}\sin(\omega t)\n* The amplitude of oscillation grows linearly without bound (\propto t), causing catastrophic structural failure in physical systems lacking adequate damping.\n\n# The Energy Principle in One-Dimensional Dynamical Systems\n\n* **Conservation of Mechanical Energy in 1D:**\n * Consider a mass mmovingalongacoordinatelinemoving along a coordinate lineyunderaposition−dependentforceunder a position-dependent forceF(y)::m\ddot{y} = F(y).\n * Potential energy V(y)isdefinedasthenegativeworkoftheforce:is defined as the negative work of the force:F(y) = -V'(y) \implies V(y) = -\int F(y)\,dy.\n * Multiplying the equation of motion by \dot{y} and using the chain rule:\n    m\ddot{y}\dot{y} = F(y)\dot{y} \implies \frac{d}{dt}\left(\frac{1}{2}m\dot{y}^2\right) = -\frac{dV}{dy}\frac{dy}{dt} = -\frac{dV}{dt}\n * Integrating with respect to time yields the total mechanical energy conservation law:\n    E = \frac{1}{2}m\dot{y}^2 + V(y) = \text{constant}\n * Kinetic energy T = \frac{1}{2}m\dot{y}^2isnon−negative(is non-negative (T \ge 0), which imposes the fundamental physical constraint:\n    V(y) \le E\n\n* **Potential Energy Diagrams and Phase Space Topology:**\n * Plotting V(y)againstagainstyonanenergydiagramalongsideafixedtotalenergylevelon an energy diagram alongside a fixed total energy levelE dictates system dynamics:\n * **Allowed Regions:** Intervals of ywherewhereV(y) \le E. Motion is physically constrained within these regions.\n * **Forbidden Regions:** Intervals where V(y) > E,whichwouldrequirenegativekineticenergy(, which would require negative kinetic energy (\dot{y}^2 < 0), which is impossible in classical mechanics.\n * **Turning Points:** Points where V(y) = E.Attheseboundaries,kineticenergyvanishes(. At these boundaries, kinetic energy vanishes (T = 0),instantaneousvelocityiszero(), instantaneous velocity is zero (\dot{y} = 0),andtheparticlereversesdirectionunderforce), and the particle reverses direction under forceF = -V'(y).\n * **Equilibria and Stability:** Equilibrium points occur at critical points of the potential energy: F(y) = 0 \implies V'(y) = 0.\n * **Stable Equilibrium:** Local minimum of V(y)((V''(y) > 0). Small perturbations produce restoring forces pushing the particle back toward the minimum.\n * **Unstable Equilibrium:** Local maximum of V(y)((V''(y) < 0). Perturbations cause the particle to accelerate away from the crest.\n\n* **Worked Dynamical Applications:**\n * **Constant Gravitational Acceleration:**\n * Equation: m\ddot{y} = -mg.\n * Potential energy: V(y) = -\int (-mg)\,dy = mgy(takinggroundreference(taking ground referenceV(0) = 0).\n * Energy equation: \frac{1}{2}m\dot{y}^2 + mgy = E.\n * Maximum height y_{\max}occursattheturningpointwhereoccurs at the turning point where\dot{y} = 0:\n      mgy_{\max} = E = \frac{1}{2}m v_0^2 + mgy_0 \implies y_{\max} = y_0 + \frac{v_0^2}{2g}\n * For a projectile launched from ground level (y_0 = 0)atspeed) at speedv_0 = 98\,m/sunderunderg = 9.8\,m/s^2:\n      y_{\max} = \frac{98^2}{2(9.8)} = \frac{9604}{19.6} = 490\,m\n * **Harmonic Oscillator Energy:**\n * Restoring force: F(x) = -kx.\n * Elastic potential energy: V(x) = -\int (-kx)\,dx = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2.\n * Harmonic trajectory: x(t) = A\cos(\omega t - \phi),,\dot{x}(t) = -A\omega\sin(\omega t - \phi).\n * Total energy:\n      E = \frac{1}{2}m(-A\omega\sin(\omega t - \phi))^2 + \frac{1}{2}m\omega^2(A\cos(\omega t - \phi))^2 = \frac{1}{2}mA^2\omega^2(\sin^2(\dots) + \cos^2(\dots)) = \frac{1}{2}mA^2\omega^2\n * Total energy is strictly proportional to both the square of the amplitude and the square of the natural frequency.\n * *Example Calculation:* Mass m = 9\,kg,springconstant, spring constantk = 4\,N/m.Pulleddown. Pulled downx(0) = 1\,m,upwardkick, upward kick\dot{x}(0) = -0.5\,m/s.\n * Natural frequency: \omega = \sqrt{4/9} = \frac{2}{3}\,rad/s.\n * Amplitude: A = \sqrt{x(0)^2 + \left(\frac{\dot{x}(0)}{\omega}\right)^2} = \sqrt{1^2 + \left(\frac{-0.5}{2/3}\right)^2} = \sqrt{1 + \frac{9}{16}} = \frac{5}{4} = 1.25\,m.\n * Total energy:\n        E = \frac{1}{2}(9)\left(\frac{5}{4}\right)^2\left(\frac{2}{3}\right)^2 = \frac{1}{2}(9)\left(\frac{25}{16}\right)\left(\frac{4}{9}\right) = \frac{25}{8} = 3.125\,J\n * Speed at equilibrium (x = 0):\n        \frac{1}{2}m v_{\text{eq}}^2 = E \implies \frac{9}{2}v_{\text{eq}}^2 = \frac{25}{8} \implies v_{\text{eq}}^2 = \frac{25}{36} \implies v_{\text{eq}} = \pm \frac{5}{6} \approx \pm 0.833\,m/s$$