Math 242 Notes — Section 1.1 & 1.2: DEs, ODEs vs PDEs, Separation of Variables, Direct Integration, and IVPs

Separation of Variables (Key Example)

• Goal: Solve a separable DE by placing all x-terms and dx on one side and all y-terms and dy on the other, then integrate.
• The given DE (from the transcript) is separable:$\frac{dy}{dx}=\frac{2y}{x}.$ This can be rewritten by separating variables as:
  $\frac{dy}{y}=\frac{2}{x}\,dx.$
• Integrate both sides:
  $\int \frac{dy}{y}=\int \frac{2}{x}\,dx\quad\Rightarrow\quad \ln|y|=2\ln|x|+C.$
• Exponentiate to solve for y:
  $|y|=e^{C}|x|^{2}.$
  Let $K=\pm e^{C}$ (K may be any real constant, including 0 for the zero-solution).
  Thus the general solution is
  $y=Kx^{2}.$
  Note: for a specific K, the origin can cause issues with the tangent-intercept argument; the derivation of the tangency property assumes the base point x_0 ≠ 0.
• Tangent-line check (geometric property of the family): if the solution is y=Kx^2 and the tangent is taken at (x0, y0) with y0=Kx0^2, then the tangent line has slope$y'(x<em>0)=2Kx</em>0.$
  The tangent line is
  $y-y<em>0=2Kx</em>0\,(x-x<em>0).$ Its x-intercept occurs when y=0: $-y</em>0=2Kx<em>0\,(x</em>{int}-x<em>0)\Rightarrow x</em>{int}=x<em>0-\frac{y</em>0}{2Kx<em>0}=x</em>0-\frac{Kx<em>0^2}{2Kx</em>0}=\frac{x<em>0}{2}.$ Hence the tangent line intersects the x-axis at $(x</em>0/2,\,0)$, as stated in the problem.

1.1: WORDS (Definitions, Concepts, and Examples)

• Differential Equation (DE): an equation that involves a function, its independent variables, and its derivatives.
  • Examples:
  • (a) $y''+y=0\quad (\text{where } y=y(x))$
  • (b) $y'=x\quad (\text{where } y=y(x))$
  • (c) $u<em>{xx}+u</em>{yy}=0\quad (\text{where } u=u(x,y))$
• Solution of a DE: a function that satisfies the differential equation.
  • Examples of solutions:
  • (a) For $y''+y=0$, all functions of the form$y=a\sin x+b\cos x$are solutions, since
    $y' = a\cos x - b\sin x,\quad y'' = -a\sin x - b\cos x,$
    and $y''+y=(-a\sin x-b\cos x)+(a\sin x+b\cos x)=0.$
  • (b) For $y'=x$, all functions of the form$y=\frac{x^{2}}{2}+C,$where C is a constant.
  • (c) For $u<em>{xx}+u</em>{yy}=0$, functions such as$u=e^{x}\cos(a y)$(with constant a) satisfy the PDE since
    $u<em>x = e^{x}\cos(a y),\ u</em>{xx}=e^{x}\cos(a y),\ u<em>y=-a e^{x}\sin(a y),\ u</em>{yy}=-a^{2}e^{x}\cos(a y),$
    so that $u<em>{xx}+u</em>{yy}=(1-a^{2})e^{x}\cos(a y)=0$ when a is chosen appropriately (here a acts as a constant parameter; in the particular calculation a is used to illustrate the process).
• The order of a DE: the largest order of derivative that appears in the equation.
  • Examples: $y''+y=0$ is second-order; $y'=x$ is first-order.
• Ordinary vs Partial Differential Equations:
  • ODE: involves a function of a single variable. PDE: involves a function of multiple variables.
  • In Math 242 we study ODEs; PDEs are studied in Math 521.
• Example family: Find all r such that $y=e^{rx}$ is a solution of $y''-2y'-3y=0.$
  • Substitute: $y'=re^{rx}$, $y''=r^2e^{rx}$, so
    $r^{2}e^{rx}-2r e^{rx}-3e^{rx}=0\Rightarrow e^{rx}(r^{2}-2r-3)=0.$
  • Since $e^{rx}\neq 0$, solve $r^{2}-2r-3=0=(r-3)(r+1)$, giving
    $r=3,\; r=-1.$
  • Therefore the general solution is a linear combination of the two independent solutions:
    $y=C<em>{1}e^{3x}+C</em>{2}e^{-x}.$
• Example (IVP technique hint): often we use initial conditions to determine constants in the general solution.
• Additional context: The chapter emphasizes understanding what constitutes a solution, how to verify it, and how to interpret DEs conceptually (order, independence, and types of equations).

1.1: Section on Exponential Solutions and Characteristic Equations

• If we try $y=e^{rx}$ in a linear constant-coefficient DE, we obtain a characteristic equation for $r$.
• Example: y''-2y'-3y=0
  • Characteristic equation: $r^{2}-2r-3=0$ with roots $r=3, -1.$
  • Corresponding exponential solutions: $e^{3x}, e^{-x}.$
  • General solution: $y=C<em>{1}e^{3x}+C</em>{2}e^{-x}.$

1.1.1: Example 2.1 (DE from a Tangent Intercept Condition)

• Problem: Find and solve a DE describing a function $y=g(x)$ such that the tangent to the curve at $(x<em>0,y</em>0)$ intersects the x-axis at $(2,0)$ for all points on the curve.
• Geometric setup:
  • Slope of tangent at $(x<em>0,y</em>0)$ is $y'(x_0)$.
  • Slope of the line through $(x<em>0,y</em>0)$ and $(2,0)$ is
    $m=\frac{y<em>0-0}{x</em>0-2}=\frac{y<em>0}{x</em>0-2}.$
  • Since the tangent line has slope $y'(x<em>0)$, we set $y'(x</em>0)=\frac{y<em>0}{x</em>0-2} = \frac{g(x<em>0)}{x</em>0-2}.$
  • Thus the DE describing the curve is
    $\frac{dy}{dx} = \frac{y}{x-2}.$
• Solve by separation:
  $\frac{dy}{y}=\frac{dx}{x-2}.$
  Integrate:
  $\ln|y|=\ln|x-2|+C.$
  Exponentiate:
  $|y|=C|x-2|.$
  Hence
  $y=K(x-2)$ with an arbitrary constant $K$ (including $K=0$ giving the zero-solution).
• Verification (qualitative): Each member $y=K(x-2)$ is a straight line through $(2,0)$. The tangent to the curve at every point coincides with the line itself, so the tangent line at any point passes through $(2,0)$ as required.

1.2: DIRECT INTEGRATION

• Direct integration method for first-order DEs of the form $y' = f(x)$:
  • Integrate both sides: $y = \int f(x) \, dx + C.$
• Example: Solve the IVP $y' = \ln x$, with $y(1)=5$.
  • Integrate: $y = \int \ln x \, dx = x\ln x - x + C.$
  • Apply initial condition: $y(1)=5$ gives
    $5 = y(1) = 1\cdot\ln 1 - 1 + C = -1 + C \Rightarrow C=6.$
  • Solution: $y = x\ln x - x + 6.$
• Integration by Parts (reformulation of the product rule):
  • Product rule: $d(uv) = u\,dv + v\,du.$
  • Integration by Parts formula:
    $\int u \, dv = uv - \int v \, du.$
• Example: Use integration by parts to compute $\int \ln x \, dx.$
  • Take $u=\ln x$, $dv=dx$; then $du=dx/x$, $v=x$:
    $\int \ln x \, dx = x\ln x - \int x\cdot \frac{1}{x} dx = x\ln x - x + C.$
• Observations:
  • Integration by parts is suited for products of unrelated functions, inverse functions, or tricky trigonometric integrals.

1.2: Example Verifications and IVPs

• Example: Verify that $y = x\cos(\ln x)$ satisfies the DE
  $x^{2}y'' - x y' + 2y = 0.$

  • Compute derivatives:
  • $y = x\cos(\ln x)$.
  • $y' = \cos(\ln x) - \sin(\ln x)$ (one can derive this via product rule and chain rule; note that a common transcription in the notes shows a variant but the correct derivative is this).
  • $y'' = -\frac{\sin(\ln x) + \cos(\ln x)}{x}$.
  • Substitute into the left-hand side:
    $x^{2}y'' - x y' + 2y = x^{2}\left(-\frac{\sin(\ln x) + \cos(\ln x)}{x}\right) - x(\cos(\ln x) - \sin(\ln x)) + 2x\cos(\ln x)$
    $= -x\sin(\ln x) - x\cos(\ln x) - x\cos(\ln x) + x\sin(\ln x) + 2x\cos(\ln x) = 0.$
  • Conclusion: $y = x\cos(\ln x)$ is a solution.

• Example: Solve the IVP for $y'' + y = 0$ with $y(0)=3$, $y'(0)=4$.

  • General solution of $y'' + y = 0$ is
    $y(x) = a\sin x + b\cos x.$
  • Apply initial conditions:
  • $y(0) = a\sin 0 + b\cos 0 = b = 3$ ⇒ $b=3$.
  • $y'(x) = a\cos x - b\sin x$; thus $y'(0) = a\cos 0 - b\sin 0 = a = 4$ ⇒ $a=4$.
  • Solution: $y(x) = 4\sin x + 3\cos x.$
  • Verification: $y'' = -4\sin x - 3\cos x,$ so $y'' + y = (-4\sin x - 3\cos x) + (4\sin x + 3\cos x) = 0.$
  • Initial conditions check: $y(0) = 3$, $y'(0) = 4$.

• Note on IVPs: An Initial Value Problem (IVP) is a DE together with initial conditions; a well-posed IVP has a unique solution (this is a central concern in Math 242).

Quick Reference (Formulas)

• Separation of Variables: $\frac{dy}{dx}=\frac{f(y)}{g(x)}\Rightarrow \frac{dy}{f(y)}=\frac{dx}{g(x)}$ and then integrate.
• Direct Integration: $y=\int f(x)\,dx + C.$
• Integration by Parts: $\int u\,dv = uv - \int v\,du.$
• Tangent-line intercept check (geometry of DEs) and general solution forms depend on the specific problem context.
• Common families:
  • Linear constant-coefficient DEs: use characteristic equations.
  • Separable DEs: reduce to two single-variable integrals.
  • IVPs yield constants determined by initial conditions.