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

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

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{1}{4} e^{4 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{1}{4} e^{4 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{1}{4} e^{4 x}

y^{^{'''}} + 7 y^{^{''}} - 8 y = x e^{x} + 12

y^{^{''}} + 4000 y^{^{'}} + 4000000 y = 36

\frac{d ^{2} y}{d x ^{2}} + 14 \frac{d y}{d x} + 49 y = 4 e^{5 x}

y^{^{''}} + 5 y^{^{'}} - y = 11

100 y^{^{''}} + 13 y^{^{'}} + \frac{6 y}{25} = 600

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