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Second-Order Linear Equations

For constant-coefficient equations, form the characteristic equation and solve for its roots.

$$ay''+by'+cy=0 \quad \Rightarrow \quad ar^2+br+c=0$$

Constant-Coefficient Initial Value

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

Characteristic equation:

$$r^2-5r+6=0=(r-2)(r-3)$$
$$y=C_1e^{2x}+C_2e^{3x}$$

Apply initial conditions: $C_1+C_2=2$ and $2C_1+3C_2=5$.

Thus $C_1=1$ and $C_2=1$.

Final answer: $y=e^{2x}+e^{3x}$.

Problem: Distinct Real Roots

Solve y'' - 5y' + 6y = 0.

$$r^2-5r+6=0 \Rightarrow (r-2)(r-3)=0$$

Answer: $y=C_1e^{2x}+C_2e^{3x}$.

Problem: Repeated Root

Solve y'' - 4y' + 4y = 0.

$$r^2-4r+4=(r-2)^2=0$$

Answer: $y=(C_1+C_2x)e^{2x}$.

Problem: Complex Roots

Solve y'' + 9y = 0.

$$r^2+9=0 \Rightarrow r=\pm 3i$$

Answer: $y=C_1\cos 3x+C_2\sin 3x$.

Problem: Undetermined Coefficients Trial Form

For y'' + y = 4x, state a suitable trial particular solution.

The forcing term is a first-degree polynomial, so use a first-degree polynomial trial.

$$y_p=Ax+B$$

Answer: A suitable trial form is $y_p=Ax+B$.

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