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

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

Step 1: Homogeneous Solution

\[ r^2-2r+1=0 \implies (r-1)^2=0 \implies r=1 \ \ (\text{double root}). \]

\[ Y_h = C_1e^{x} + C_2xe^{x}. \]

Step 2: Particular Solution

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

\[ y^{*\prime}=2Ae^{2x}, \qquad y^{*\prime\prime}=4Ae^{2x}. \]

Substitute into the equation: \[4Ae^{2x}-2(2Ae^{2x})+Ae^{2x}=e^{2x} \implies (4A-4A+A)e^{2x}=e^{2x} \implies A=1.\]

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

Step 3: Verification

\[ y^{*\prime\prime}-2y^{*\prime}+y^* = 4e^{2x}-4e^{2x}+e^{2x} = e^{2x}.\ \checkmark \]

Step 4: General Solution