Find the general solution of y'' - 3y' + 2y = e^x

Find the general solution of $y'' - 3y' + 2y = e^x$.

Step 1: Homogeneous solution

\[ r^2-3r+2=0 \implies (r-1)(r-2)=0 \implies r_1=1,\ r_2=2. \]

\[ Y_h = C_1e^{x} + C_2e^{2x}. \]

Step 2: Diagnose resonance

The right-hand side is $f(x)=e^{\alpha x}$ with $\alpha=1$. Comparing $\alpha$ to the characteristic roots $\{1,2\}$, we see $\alpha=1$ is a root, and it occurs with multiplicity $s=1$ (a simple root). This is the resonance case: the naive trial $y^*=Ae^{x}$ would solve the homogeneous equation and could never balance the right-hand side, so the trial must be multiplied by $x^s=x^1=x$:

\[ y^* = Axe^{x}. \]

Step 3: Substitute

\[ y^{*\prime}=Ae^x(1+x), \qquad y^{*\prime\prime}=Ae^x(2+x). \]

Insert into $y''-3y'+2y=e^x$: $Ae^x(2+x) - 3Ae^x(1+x) + 2Axe^x = e^x.$

Factor out $Ae^x$ and collect the bracket: $Ae^x\bigl[(2+x)-3(1+x)+2x\bigr] = Ae^x(-1) = e^x.$

Step 4: Solve for the coefficient

\[ -A = 1 \implies A=-1, \qquad y^* = -xe^{x}. \]

Step 5: Verify

\[ y^{*\prime}=-e^x(1+x), \qquad y^{*\prime\prime}=-e^x(2+x). \]

\[ y^{*\prime\prime}-3y^{*\prime}+2y^* = -e^x(2+x)+3e^x(1+x)-2xe^x = e^x\bigl[-(2+x)+3(1+x)-2x\bigr] = e^x.\ \checkmark \]

Step 6: General solution

The solution is recorded in the result below.