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⬅ Back to Differential Equations

📘 Key Concepts: Second-Order Linear Equations

Standard form: $ay'' + by' + cy = f(x)$. The characteristic equation is $ar^2 + br + c = 0$.

Cases for the homogeneous solution $y_h$:

Two distinct real roots $r_1, r_2$: $y_h = C_1 e^{r_1 x} + C_2 e^{r_2 x}$
Repeated real root $r$: $y_h = (C_1 + C_2 x)\,e^{rx}$
Complex roots $r = \alpha \pm \beta i$: $y_h = e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)$

Non-homogeneous ($f(x) \neq 0$): General solution = $y_h + y_p$, where $y_p$ is a particular solution found by the method of undetermined coefficients.

Undetermined coefficients guide:

$f(x) = e^{ax}$ → $y_p = Ae^{ax}$ (or $Axe^{ax}$ if $a$ is a root)
$f(x) = \sin bx$ or $\cos bx$ → $y_p = A\sin bx + B\cos bx$ (or multiply by $x$ if resonance)
$f(x) = $ polynomial of degree $n$ → $y_p = $ polynomial of degree $n$

Problem: Distinct Real Roots (IVP)

Solve $y'' - 5y' + 6y = 0$, $y(0) = 2$, $y'(0) = 5$.

DE – 2nd Order – Problem 1 – Setup

DE – 2nd Order – Problem 1 – Diagram

DE – 2nd Order – Problem 1 – Diagram

Solution

Solution

Solution

Solution

Problem: Repeated Real Root (IVP)

Solve $y'' - 4y' + 4y = 0$, $y(0) = 1$, $y'(0) = 2$.

DE – 2nd Order – Problem 2 – Setup

DE – 2nd Order – Problem 2 – Diagram

DE – 2nd Order – Problem 2 – Diagram

Solution

Solution

Solution

Solution

Problem: Complex Conjugate Roots (IVP)

Solve $y'' + 2y' + 5y = 0$, $y(0) = 1$, $y'(0) = 1$.

DE – 2nd Order – Problem 3 – Setup

DE – 2nd Order – Problem 3 – Diagram

DE – 2nd Order – Problem 3 – Diagram

Solution

Solution

Solution

Solution

Problem: Non-Homogeneous — Undetermined Coefficients

Solve $y'' + y = \cos x$ (resonance case). Find the general solution.

DE – 2nd Order – Problem 4 – Setup

DE – 2nd Order – Problem 4 – Diagram

DE – 2nd Order – Problem 4 – Diagram

Solution

Solution

Solution

Solution

Problem: Particular Trial for Sine Forcing

For y'' + 4y = 3 sin x, choose and solve for a trial particular solution.

Use $y_p=A\sin x+B\cos x$.

$$y_p''+4y_p=3A\sin x+3B\cos x=3\sin x$$

Answer: $A=1$, $B=0$, so $y_p=\sin x$.