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

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

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: Particular Solution

The right-hand side $f(x)=e^{3x}$ has exponent $\alpha=3$. Since the roots of the characteristic equation are $r=1$ and $r=2$, $r=\alpha=3$ is not a root, so the multiplicity is $s=0$ and we try $y^*=Ae^{3x}$:

\[ y^{*\prime}=3Ae^{3x}, \qquad y^{*\prime\prime}=9Ae^{3x}. \]

Substitute into the equation: \[9Ae^{3x}-3(3Ae^{3x})+2Ae^{3x}=e^{3x} \implies (9A-9A+2A)e^{3x}=e^{3x} \implies 2A=1 \implies A=\frac{1}{2}.\]

\[ y^* = \frac{e^{3x}}{2}. \]

Step 3: Verification

\[ y^{*\prime\prime}-3y^{*\prime}+2y^* = \frac{9}{2}e^{3x}-\frac{9}{2}e^{3x}+e^{3x} = e^{3x}.\ \checkmark \]

Step 4: General Solution