What is the solution for y^{''}-4y^'+4y=2e^{2x} ?

The solution for y^{''}-4y^'+4y=2e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+e^{2x}x^2

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y^{''}-4y^'+4y=2e^{2x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{2 x} x^{2}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = 2 e^{2 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{2 x} x^{2}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + e^{2 x} x^{2}

\frac{d}{d x} (5 e^{- x} + e^{3 x})

\frac{d}{d x} (ln (\frac{x}{1 + x ^{2}}))

\int \frac{2 x}{( 1 + x ^{2} ) ^{2}} d x

y = 14 x, y = x^{2}

4 x^{\frac{2}{3}}

What is the solution for y^{''}-4y^'+4y=2e^{2x} ?
The solution for y^{''}-4y^'+4y=2e^{2x} is y=c_{1}e^{2x}+c_{2}xe^{2x}+e^{2x}x^2