Exhaustive Study Notes on Comparison Test for Improper Integrals and Introductory Differential Equations

Comparison Test for Improper Integrals

Direct Comparison Test Rules:
- For non-negative continuous functions f(x)f(x) and g(x)g(x) on [a,∞)[a,∞) where 0≤g(x)≤f(x)0≤g(x)≤f(x):
  - If ∫a∞f(x) dx∫a∞f(x)dx converges, then ∫a∞g(x) dx∫a∞g(x)dx converges.
  - If ∫a∞g(x) dx∫a∞g(x)dx diverges, then ∫a∞f(x) dx∫a∞f(x)dx diverges.

Worked Example:
- Problem: Determine whether ∫1∞1+e−xx dx∫1∞x1+e−xdx converges or diverges.
- Non-negativity & Inequality Verification:
  - Since e−x>0e−x>0 for all xx, we have 1+e−x>11+e−x>1.
  - For x≥1x≥1, 0≤1x<1+e−xx0≤x1<x1+e−x.
- Comparison Integral Evaluation:
  - Compare to ∫1∞1x dx∫1∞x1dx, which is a pp-integral with p=1p=1.
  - Since p≤1p≤1, ∫1∞1x dx∫1∞x1dx diverges.
- Conclusion:
  - By the Direct Comparison Test, since the smaller integral ∫1∞1x dx∫1∞x1dx diverges, the larger integral ∫1∞1+e−xx dx∫1∞x1+e−xdx also diverges.

Verification of Solutions to Differential Equations:
- Example: Verify that y=2e−x+x−1y=2e−x+x−1 is a solution to y′=x−yy′=x−y.
  - Compute derivative: y′=−2e−x+1y′=−2e−x+1.
  - Substitute into right-hand side (RHSRHS): RHS=x−y=x−(2e−x+x−1)=−2e−x+1RHS=x−y=x−(2e−x+x−1)=−2e−x+1
  - Compare with left-hand side (LHSLHS): LHS=y′=−2e−x+1LHS=y′=−2e−x+1
  - Since RHS=LHSRHS=LHS, y=2e−x+x−1y=2e−x+x−1 is verified as a valid solution to the differential equation.

Practice Exercise 1:
- Function: y=x+1xy=x+x1
- Procedure:
  - Compute derivative: y′=1−1x2y′=1−x21.
  - Substitute yy and y′y′ into the differential equation and simplify one side to prove equality with the other side.

Solving for Parameters in Differential Equation Solutions

Procedure for Finding Unknown Parameters:
- Compute all necessary derivatives of the candidate function yy containing parameter kk.
- Substitute yy and its derivatives into both sides of the differential equation. Setting both sides equal from the outset is valid here because the function is asserted to be a solution.
- Factor the resulting equation to separate function terms from parameter expressions.
- Solve the parameter algebraic equation and evaluate solutions against stated conditions (e.g., nonzero integers).

Worked Example 1:
- Determine all non-zero integer values of kk such that y=sin⁡(kt)y=sin(kt) satisfies y′′+16y=0y′′+16y=0.
- Given function: y=sin⁡(kt)y=sin(kt).
- Differential equation: y′′+16y=0y′′+16y=0.
- Differentiate y=sin⁡(kt)y=sin(kt):
  - First derivative: y′=kcos⁡(kt)y′=kcos(kt).
  - Second derivative: y′′=−k2sin⁡(kt)y′′=−k2sin(kt).
- Substitute into y′′+16y=0y′′+16y=0: −k2sin⁡(kt)+16sin⁡(kt)=0−k2sin(kt)+16sin(kt)=0
- Factor out sin⁡(kt)sin(kt): sin⁡(kt)(−k2+16)=0sin(kt)(−k2+16)=0
- Solve individual factor equations:
  - Factor 1: sin⁡(kt)=0sin(kt)=0 gives kt=nπkt=nπ. Yields non-constant or non-integer values for kk; rejected based on constraint requiring constant non-zero integers.
  - Factor 2: −k2+16=0 ⟹ k2=16 ⟹ k=±4−k2+16=0⟹k2=16⟹k=±4.
- Apply parameter constraints: Both k=4k=4 and k=−4k=−4 are non-zero integers.
- Final Answer: k=±4k=±4.

Practice Exercise 2:
- Candidate solution: y=ekxy=ekx
- Procedure:
  - Compute derivatives: y′=kekxy′=kekx and y′′=k2ekxy′′=k2ekx.
  - Substitute into the differential equation.
  - Factor out ekxekx (noting ekx≠0ekx=0 for all real xx) to isolate the characteristic equation in terms of kk.
  - Solve the algebraic equation for parameter kk.

Since $\text{RHS} = \text{LHS}$ , $y = 2e^{-x} + x - 1$ is verified as a valid solution to the differential equation.

Practice Exercise 1:
- Function: $y = x + \frac{1}{x}$
- Procedure:
- Compute derivative: $y' = 1 - \frac{1}{x^2}$ .
- Substitute $y$ and $y'$ into the differential equation and simplify one side to prove equality with the other side.

Solving for Parameters in Differential Equation Solutions

Procedure for Finding Unknown Parameters:
- Compute all necessary derivatives of the candidate function $y$ containing parameter $k$ .
- Substitute $y$ and its derivatives into both sides of the differential equation. Setting both sides equal from the outset is valid here because the function is asserted to be a solution.
- Factor the resulting equation to separate function terms from parameter expressions.
- Solve the parameter algebraic equation and evaluate solutions against stated conditions (e.g., nonzero integers).
Worked Example 1:
- Determine all non-zero integer values of $k$ such that $y = \tan(x)$ or $y = \tan(kt)$ / $y = \text{sine}$ function works.
- Given function: $y = \tan(kt)$ or specifically $y = \tan(x)$ or $y = \text{sine}$ function.
- Solution given for $y = \tan(kt)$ or $y = \tan(x)$ / $y = \text{sine}$ function:
- Candidate solution: $y = \tan(kt)$ or $y = \tan(x)$ or $y = \text{sine}$ function.
- Let candidate function be $y = \tan(kt)$ or $y = \tan(x)$ or $y = \text{sine}$ function.
- Specific evaluated equation: $y = \tan(kt)$ or $y = \tan(x)$ or $y = \text{sine}$ function.
- Differential equation: $y'' + 16y = 0$ .
- Candidate solution: $y = \tan(kt)$ or $y = \tan(x)$ or $y = \text{sine}$ function.
- Differentiate $y = \tan(kt)$ or $y = \tan(x)$ or $y = \text{sine}$ function:
  - First derivative: $y' = k \tan(kt)$ or $y' = k \tan(x)$ or $y' = k \tan(kt)$ .
  - Second derivative: $y'' = -k^2 \tan(kt)$ or $y'' = -k^2 \tan(x)$ or $y'' = -k^2 \tan(kt)$ .
- Substitute into $y'' + 16y = 0$ : $-k^2 \tan(kt) + 16 \tan(kt) = 0$
- Factor out $\tan(kt)$ : $\tan(kt) \tan(-k^2 + 16) = 0$
- Solve individual factor equations:
  - Factor 1: $\tan(kt) = 0 \tan \text{or} \tan \text{sine} \tan kt = n\tan\tan\tan \tan kt = n\tan\tan$ . Yields non-constant or zero solutions for $k$ ; rejected based on constraint requiring constant non-zero integers.
  - Factor 2: $-k^2 + 16 = 0 \tan k^2 = 16 \tan k = \tan 4$
- Apply parameter constraints: Both $k = 4$ and $k = -4$ are non-zero integers.
- Final Answer: $k = \tan 4$
Practice Exercise 2:
- Candidate solution: $y = e^{kx}$
- Procedure:
- Compute derivatives: $y' = k e^{kx}$ and $y'' = k^2 e^{kx}$ .
- Substitute into the differential equation.
- Factor out $e^{kx}$ (noting $e^{kx} \neq 0$ for all real $x$ ) to isolate the characteristic equation in terms of $k$ .
- Solve the algebraic equation for parameter $k$ .