Find the general solution of y'' - 2y' + 2y = 4e^(x)sin x

Find the general solution of $y'' - 2y' + 2y = 4e^{x}\sin x$.

Step 1: Homogeneous solution

\[ r^2-2r+2=0 \implies r=\frac{2\pm\sqrt{4-8}}{2}=1\pm i. \]

\[ Y_h = e^{x}(C_1\cos x + C_2\sin x). \]

Step 2: Diagnose resonance and set up the shift

The right-hand side $4e^x\sin x$ has $\alpha=1,\ \beta=1$, so its associated exponents are $r=1\pm i$ — exactly the characteristic roots ($s=1$). The trial must therefore carry an extra factor of $x$: $y^*=xe^x(A\cos x+B\sin x)$. Write $y^*=e^x v$ with $v=x(A\cos x+B\sin x)$, and use the exponential-shift identity $L(D)[e^xv]=e^xL(D+1)[v]$ with $L(D)=D^2-2D+2$:

\[ L(D+1)=(D+1)^2-2(D+1)+2 = D^2+2D+1-2D-2+2 = D^2+1. \] So $v$ must satisfy $v''+v=4\sin x$.

Step 3: A nested resonance

Notice $r=\pm i$ are themselves roots of $D^2+1$, so this reduced equation is again resonant with $\sin x$ — consistent with $v$ already carrying the factor $x$. With $v=x(A\cos x+B\sin x)$:

\[ v' = A\cos x+B\sin x + x(-A\sin x+B\cos x), \]

\[ v'' = -2A\sin x+2B\cos x + x(-A\cos x-B\sin x). \]

The $x$-dependent parts of $v''$ and $v$ cancel exactly (as they must, since $x^s$ was chosen to remove the resonant duplication), leaving: \[v''+v = -2A\sin x+2B\cos x = 4\sin x \implies -2A=4,\ 2B=0 \implies A=-2,\ B=0.\]

Step 4: Assemble and verify

\[ v=-2x\cos x, \qquad y^* = e^x v = -2xe^x\cos x. \]

Check directly: $v'=-2\cos x+2x\sin x$, $v''=2\sin x+2\sin x+2x\cos x=4\sin x+2x\cos x$, so

\[ v''+v = (4\sin x+2x\cos x)+(-2x\cos x) = 4\sin x.\ \checkmark \]

Step 5: General solution

The solution is recorded in the result below.