Revision of Differential Equations: Variable Separable and Linear DEs Linear Differential Equations

Revision of Variable Separable Differential Equations

  • Welcome to second-year engineering calculus. The lecture begins with a revision of first-year engineering calculus, specifically focusing on differential equations (DEs).

  • The first type of differential equation to be revised is the variable separable DE.

  • Core Methodology: To solve a variable separable DE, the goal is to isolate all $x$ variables on one side of the equation and all $y$ variables on the other side, then integrate both sides to find the solution.

Example 1: Variable Separable DE

  • The initial problem provided is:   x×dydx=1−y2x \times \frac{dy}{dx} = 1 - y^2

  • Step-by-Step Separation Process:

    • Multiply through by $dx$:     x×dy=(1−y2)×dxx \times dy = (1 - y^2) \times dx

    • Isolate variables by dividing both sides by $x$ and $(1 - y^2)$:     dy1−y2=dxx\frac{dy}{1 - y^2} = \frac{dx}{x}

  • Integration and Factorization:

    • integrate both sides:     ∫11−y2 dy=∫1x dx\int \frac{1}{1 - y^2} \, dy = \int \frac{1}{x} \, dx

    • Factorize the denominator on the left-hand side into $(1 - y)(1 + y)$:     ∫1(1−y)(1+y) dy=∫1x dx\int \frac{1}{(1 - y)(1 + y)} \, dy = \int \frac{1}{x} \, dx

  • Partial Fraction Decomposition:

    • The left-hand side requires partial fraction decomposition (students should revise this independently).

    • The resulting decomposition is:     ∫[12×11−y+12×11+y] dy=∫1x dx\int [\frac{1}{2} \times \frac{1}{1 - y} + \frac{1}{2} \times \frac{1}{1 + y}] \, dy = \int \frac{1}{x} \, dx

  • Performing the Integration:

    • Integrating the first term:     ∫12×11−y dy=12×ln⁡(1−y)×(−1)\int \frac{1}{2} \times \frac{1}{1 - y} \, dy = \frac{1}{2} \times \ln(1 - y) \times (-1)     (Dividing by the derivative of $1 - y$, which is $-1$, is equivalent to multiplying by $-1$).

    • Integrating the second term:     ∫12×11+y dy=12×ln⁡(1+y)\int \frac{1}{2} \times \frac{1}{1 + y} \, dy = \frac{1}{2} \times \ln(1 + y)

    • Integrating the right-hand side:     ∫1x dx=ln⁡(x)+C\int \frac{1}{x} \, dx = \ln(x) + C

  • Simplification using Log Laws:

    • The equation currently stands as:     −12ln⁡(1−y)+12ln⁡(1+y)=ln⁡(x)+C-\frac{1}{2} \ln(1 - y) + \frac{1}{2} \ln(1 + y) = \ln(x) + C

    • Take out a common factor of $\frac{1}{2}$ and rearrange terms:     12(ln⁡(1+y)−ln⁡(1−y))=ln⁡(x)+C\frac{1}{2} (\ln(1 + y) - \ln(1 - y)) = \ln(x) + C

    • Define the constant $C$ as $\ln(b)$ because the constant can be anything, and this makes simplification easier:     12ln⁡(1+y1−y)=ln⁡(x)+ln⁡(b)\frac{1}{2} \ln\left(\frac{1 + y}{1 - y}\right) = \ln(x) + \ln(b)

    • Multiply the entire equation by $2$:     ln⁡(1+y1−y)=2×ln⁡(x)+2×ln⁡(b)\ln\left(\frac{1 + y}{1 - y}\right) = 2 \times \ln(x) + 2 \times \ln(b)

  • Solving for the Final Solution:

    • Raise both sides to the base $e$ to remove the natural logarithms:     1+y1−y=e2ln⁡(x)+2ln⁡(b)\frac{1 + y}{1 - y} = e^{2 \ln(x) + 2 \ln(b)}

    • Simplify the right-hand side using exponent properties:     1+y1−y=e2ln⁡(x)×e2ln⁡(b)\frac{1 + y}{1 - y} = e^{2 \ln(x)} \times e^{2 \ln(b)}     1+y1−y=eln⁡(x2)×eln⁡(b2)\frac{1 + y}{1 - y} = e^{\ln(x^2)} \times e^{\ln(b^2)}

    • Since $e$ and $\ln$ are inverse functions, they cancel out:     1+y1−y=x2×a\frac{1 + y}{1 - y} = x^2 \times a     (Where the constant $a$ represents $e^{2 \ln(b)}$ or $b^2$).

    • This expression is the final solution for the differential equation.

Example 2: Variable Separable DE

  • The second problem is:   ex×dy=(1+y2)×dxe^x \times dy = (1 + y^2) \times dx

  • Variables Separation:

    • Rearranging the equation gives:     dy1+y2=dxex\frac{dy}{1 + y^2} = \frac{dx}{e^x}     dy1+y2=e−x dx\frac{dy}{1 + y^2} = e^{-x} \, dx

  • Integration:

    • Integrate both sides:     ∫11+y2 dy=∫e−x dx\int \frac{1}{1 + y^2} \, dy = \int e^{-x} \, dx

    • The integral of $\frac{1}{1 + y^2}$ is $\arctan(y)$.

    • The integral of $e^{-x}$ is $e^{-x}$ divided by the derivative of $-x$ (which is $-1$):     arctan⁡(y)=−e−x+C\arctan(y) = -e^{-x} + C

  • This concludes the solution for the second example.

Linear Differential Equations and the Integrating Factor

  • Standard Form: A linear differential equation has the form:   dydt+y×P(t)=Q(t)\frac{dy}{dt} + y \times P(t) = Q(t)   (Where $P(t)$ and $Q(t)$ are functions of $t$).

  • Objective: To solve a linear DE, one must first find the integrating factor, denoted as $\mu$.

  • Definition of Integrating Factor ($\mu$):   μ=e∫P(t) dt\mu = e^{\int P(t) \, dt}

    • The term $P(t)$ is always identified as the function multiplier in front of the $y$ term.

Example 3: Linear Differential Equation

  • The Problem:   dydt+2ycot⁡(t)=cos⁡(t)\frac{dy}{dt} + 2y \cot(t) = \cos(t)

  • Finding the Integrating Factor ($\mu$):

    • Here, $P(t) = 2 \cot(t)$.

    • $\mu = e^{\int 2 \cot(t) \, dt}$

    • Rewrite $\cot(t)$ as $\frac{\cos(t)}{\sin(t)}$:     μ=e2∫cos⁡(t)sin⁡(t) dt\mu = e^{2 \int \frac{\cos(t)}{\sin(t)} \, dt}

    • Use a u-substitution to integrate $\frac{\cos(t)}{\sin(t)}$:

    • Let $u = \sin(t)$

    • Then $\frac{du}{dt} = \cos(t) \Rightarrow du = \cos(t) \, dt$

    • Substituted expression: $\mu = e^{2 \int \frac{1}{u} \, du} = e^{2 \ln(u)}$

    • Simplify $\mu$:     μ=eln⁡(u2)=u2\mu = e^{\ln(u^2)} = u^2     (Since $e$ and $\ln$ are inverse functions).

    • Substitute back for $u$:     μ=sin⁡2(t)\mu = \sin^2(t)

  • Applying the Integrating Factor:

    • Multiply the entire differential equation by $\sin^2(t)$:     sin⁡2(t)×dydt+2ycot⁡(t)×sin⁡2(t)=cos⁡(t)×sin⁡2(t)\sin^2(t) \times \frac{dy}{dt} + 2y \cot(t) \times \sin^2(t) = \cos(t) \times \sin^2(t)

  • Simplification Rule:

    • If the integrating factor is found correctly, the left-hand side will always simplify to the derivative of ($y \times \mu$):     ddt[ysin⁡2(t)]=cos⁡(t)sin⁡2(t)\frac{d}{dt} [y \sin^2(t)] = \cos(t) \sin^2(t)

  • Integration to Solve:

    • Integrate both sides with respect to $dt$:     ∫ddt[ysin⁡2(t)] dt=∫sin⁡2(t)cos⁡(t) dt\int \frac{d}{dt} [y \sin^2(t)] \, dt = \int \sin^2(t) \cos(t) \, dt

    • The left-hand side cancels (integral vs differential), leaving:     ysin⁡2(t)=∫sin⁡2(t)cos⁡(t) dty \sin^2(t) = \int \sin^2(t) \cos(t) \, dt

    • Use the same u-substitution ($u = \sin(t)$, $du = \cos(t) \, dt$) for the right-hand side:     ∫u2 du=u33+C\int u^2 \, du = \frac{u^3}{3} + C

    • Substituting back:     ysin⁡2(t)=sin⁡3(t)3+Cy \sin^2(t) = \frac{\sin^3(t)}{3} + C

  • Final Form:

    • Divide through by $\sin^2(t)$ to solve for $y$:     y=13sin⁡(t)+Csin⁡2(t)y = \frac{1}{3} \sin(t) + \frac{C}{\sin^2(t)}

    • Rewrite the second term using $\csc^2(t)$:     y=13sin⁡(t)+Ccsc⁡2(t)y = \frac{1}{3} \sin(t) + C \csc^2(t)

Example 4: Linear Differential Equation

  • The Problem:   A linear differential equation in the form:   dydt−yt+1=t\frac{dy}{dt} - \frac{y}{t + 1} = t

  • Finding the Integrating Factor ($\mu$):

    • Here, $P(t) = -\frac{1}{t + 1}$.

    • $\mu = e^{\int -\frac{1}{t + 1} \, dt} = e^{-\ln(t + 1)}$

    • Apply log laws to move the negative sign: $e^{\ln((t + 1)^{-1})}$

    • $\mu = (t + 1)^{-1} = \frac{1}{t + 1}$

  • Applying the Integrating Factor:

    • Multiply the whole equation by $\frac{1}{t + 1}$:     1t+1×dydt−y(t+1)2=tt+1\frac{1}{t + 1} \times \frac{dy}{dt} - \frac{y}{(t + 1)^2} = \frac{t}{t + 1}

    • The left-hand side always simplifies to the derivative of ($y \times \mu$):     ddt[yt+1]=tt+1\frac{d}{dt} [\frac{y}{t + 1}] = \frac{t}{t + 1}

  • Integration and Algebraic Trick:

    • Integrate both sides:     yt+1=∫tt+1 dt\frac{y}{t + 1} = \int \frac{t}{t + 1} \, dt

    • To integrate the right-hand side, use the trick of adding and subtracting 1 in the numerator:     ∫t+1−1t+1 dt=∫[t+1t+1−1t+1] dt\int \frac{t + 1 - 1}{t + 1} \, dt = \int [\frac{t + 1}{t + 1} - \frac{1}{t + 1}] \, dt     ∫[1−1t+1] dt=t−ln⁡(t+1)+C\int [1 - \frac{1}{t + 1}] \, dt = t - \ln(t + 1) + C

  • Final Solution:   yt+1=t−ln⁡(t+1)+C\frac{y}{t + 1} = t - \ln(t + 1) + C

    • Multiply through by $(t + 1)$ to isolate $y$:     y=(t+1)[t−ln⁡(t+1)+C]y = (t + 1) [t - \ln(t + 1) + C]

Proof of the Linear DE Simplification

  • The Question: Why does $\mu(t) \times (y' + P(t) y)$ equal the derivative of $(\mu(t) \times y)$?

  • Method: Work from the right-hand side (RHS) of the identity back to the left-hand side.

  • Derivation:

    • Let RHS be $\frac{d}{dt} [y \times \mu(t)]$.

    • Apply the Product Rule:     Derivative=y′μ(t)+yμ′(t)\text{Derivative} = y' \mu(t) + y \mu'(t)

  • Finding an expression for $\mu'(t)$:

    • Recall the definition $\mu = e^{\int P(t) \, dt}$.

    • Take the natural log of both sides:     ln⁡(μ)=∫P(t) dt\ln(\mu) = \int P(t) \, dt

    • Differentiate both sides with respect to $t$:     ddt[ln⁡(μ)]=ddt[∫P(t) dt]\frac{d}{dt} [\ln(\mu)] = \frac{d}{dt} [\int P(t) \, dt]

    • The derivative of $\ln(\mu)$ (using the chain rule) is $\frac{1}{\mu} \times \mu'$.

    • The derivative of the integral of $P(t)$ is simply $P(t)$.

    • Therefore: $\frac{\mu'}{\mu} = P(t) \Rightarrow \mu' = \mu P(t)$.

  • Substitution:

    • Substitute $\mu' = \mu P(t)$ back into the product rule expansion:     y′μ(t)+y[μ(t)P(t)]y' \mu(t) + y [\mu(t) P(t)]

    • Take out the common factor of $\mu(t)$:     μ(t)[y′+yP(t)]\mu(t) [y' + y P(t)]

  • This proves that the left-hand side is indeed equal to the derivative of the product of $y$ and the integrating factor, concluding the lecture.