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

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

Step 1: Homogeneous solution

\[ r^2-4r+4=0 \implies (r-2)^2=0 \implies r_1=r_2=2 \ \ (\text{double root}). \]

\[ Y_h = C_1e^{2x} + C_2xe^{2x}. \]

Step 2: Diagnose resonance

The right-hand side is $f(x)=e^{\alpha x}$ with $\alpha=2$, which coincides with the characteristic root $r=2$, here of multiplicity $s=2$. Both the plain trial $Ae^{2x}$ and the once-multiplied trial $Axe^{2x}$ already lie in the homogeneous solution space, so the trial must be multiplied by $x^s=x^2$:

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

Step 3: Substitute

\[ y^{*\prime}=2Ae^{2x}(x^2+x), \qquad y^{*\prime\prime}=2Ae^{2x}(2x^2+4x+1). \]

Insert into $y''-4y'+4y=e^{2x}$ and divide by $e^{2x}$: $2A(2x^2+4x+1) - 8A(x^2+x) + 4Ax^2 = 1.$

Step 4: Collect and solve

Collecting powers of $x$: \[(4A-8A+4A)x^2 + (8A-8A)x + 2A = 1 \implies 0\cdot x^2 + 0\cdot x + 2A = 1.\]

The $x^2$ and $x^1$ terms cancel automatically — as they must, since $s=2$ was chosen precisely to remove them — leaving: \[2A=1 \implies A=\frac{1}{2}, \qquad y^* = \frac{x^2e^{2x}}{2}.\]

Step 5: Verify

\[ y^{*\prime}=e^{2x}(x+x^2), \qquad y^{*\prime\prime}=e^{2x}(2x^2+4x+1). \]

\[ y^{*\prime\prime}-4y^{*\prime}+4y^* = e^{2x}(2x^2+4x+1)-4e^{2x}(x+x^2)+2x^2e^{2x} = e^{2x}\bigl[(2-4+2)x^2+(4-4)x+1\bigr] = e^{2x}.\ \checkmark \]

Step 6: General solution

The solution is recorded in the result below.