Solve the differential equation y'' - 3y' + 2y = e^x.

Step-by-step solution

1. 1

   Solve the homogeneous equation first. Substituting $y = e^{rx}$ into $y'' - 3y' + 2y = 0$ gives the characteristic equation

   $r^2 - 3r + 2 = (r-1)(r-2) = 0 \quad\Longrightarrow\quad r = 1,\ 2$

   so $y_h = C_1e^{x} + C_2e^{2x}$. This has to be found before the particular solution, because it tells you whether the forcing term is resonant.

2. 2

   Detect the resonance. The forcing term $e^{x}$ has exponent $1$, which is one of the characteristic roots — $e^{x}$ is already a homogeneous solution. Trying $y_p = Ae^{x}$ would substitute to $0$, never to $e^{x}$, so that trial can never work.

3. 3

   Multiply the trial by x. The standard fix for a simple root is one extra factor of $x$: take $y_p = Axe^{x}$. Then

   $y_p' = A(1 + x)e^{x}, \qquad y_p'' = A(2 + x)e^{x}$

4. 4

   Substitute and match coefficients.

   $A(2+x)e^{x} - 3A(1+x)e^{x} + 2Axe^{x} = A e^{x}\left[(2+x) - 3(1+x) + 2x\right] = -Ae^{x}$

   The $x$ terms cancel exactly — the signature of a correctly chosen resonant trial. Setting $-Ae^x = e^x$ gives $A = -1$, so $y_p = -xe^{x}$.

5. 5

   Assemble and verify the general solution.

   $y = C_1e^{x} + C_2e^{2x} - xe^{x}$

   Check: with $y_p = -xe^x$, $y_p' = -(1+x)e^x$ and $y_p'' = -(2+x)e^x$, so $y_p'' - 3y_p' + 2y_p = e^x\left[-(2+x) + 3(1+x) - 2x\right] = e^x$, as required.