The general second-order linear ODE with constant coefficients:
$$a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$$
For the homogeneous equation ($f(x) = 0$), the general solution depends on the characteristic equation.
Substitute $y = e^{mx}$: $am^2 + bm + c = 0$.
Discriminant $\Delta = b^2 - 4ac$.
| Case | Roots $m$ | General solution |
|---|---|---|
| $\Delta > 0$ | Two real distinct: $m_1, m_2$ | $y = Ae^{m_1 x} + Be^{m_2 x}$ |
| $\Delta = 0$ | Repeated real: $m = -b/(2a)$ | $y = (A + Bx)e^{mx}$ |
| $\Delta < 0$ | Complex: $m = \alpha \pm \beta i$ | $y = e^{\alpha x}(A\cos\beta x + B\sin\beta x)$ |
where $A$ and $B$ are arbitrary constants.
Example 1: $y’’ - 5y’ + 6y = 0$.
Characteristic: $m^2 - 5m + 6 = 0 \Rightarrow (m-2)(m-3) = 0 \Rightarrow m = 2, 3$.
$$y = Ae^{2x} + Be^{3x}$$
Example 2: $y’’ + 4y’ + 4y = 0$.
Characteristic: $m^2+4m+4 = (m+2)^2 = 0 \Rightarrow m = -2$ (repeated).
$$y = (A+Bx)e^{-2x}$$
Example 3: $y’’ + 2y’ + 5y = 0$.
Characteristic: $m^2+2m+5 = 0 \Rightarrow m = \dfrac{-2\pm\sqrt{4-20}}{2} = -1\pm2i$.
$$y = e^{-x}(A\cos2x + B\sin2x)$$
Apply initial conditions to find $A$ and $B$.
Example 4: $y’’ - 3y’ + 2y = 0$, $y(0) = 1$, $y’(0) = 0$.
Roots: $m=1, 2$. General solution: $y = Ae^x + Be^{2x}$.
$y(0) = A + B = 1$.
$y’ = Ae^x + 2Be^{2x}$, so $y’(0) = A + 2B = 0$.
Solving: $B = -1$, $A = 2$. Particular solution: $y = 2e^x - e^{2x}$.
For $ay’’ + by’ + cy = f(x)$, the general solution is $y = y_h + y_p$ where:
- $y_h$ is the general solution of the homogeneous equation
- $y_p$ is a particular solution (trial function method)
| $f(x)$ | Trial $y_p$ |
|---|---|
| $ke^{\lambda x}$ (not a root) | $Ce^{\lambda x}$ |
| $k\cos(\omega x)$ or $k\sin(\omega x)$ | $C\cos(\omega x)+D\sin(\omega x)$ |
| Polynomial degree $n$ | Polynomial degree $n$ |
Example 5: $y’’ - y’ - 2y = 3e^x$.
Homogeneous: $m^2-m-2=(m-2)(m+1)=0$, so $y_h = Ae^{2x}+Be^{-x}$.
Try $y_p = Ce^x$: $C e^x - Ce^x - 2Ce^x = 3e^x \Rightarrow -2C = 3 \Rightarrow C = -3/2$.
General solution: $y = Ae^{2x}+Be^{-x} - \dfrac{3}{2}e^x$.
KEY TAKEAWAY: The characteristic equation determines the structure of the homogeneous solution. The three cases (distinct real, repeated, complex) give three different solution forms.
EXAM TIP: Write the characteristic equation and its roots clearly before writing the solution. Identify which case applies and write the correct solution form.
COMMON MISTAKE: In the repeated root case, writing $y = Ae^{mx} + Be^{mx} = (A+B)e^{mx}$ — this gives only one constant. The correct form is $(A+Bx)e^{mx}$.
