What is the solution for y^{''}-8y^'+16y=e^{4x} ?

The solution for y^{''}-8y^'+16y=e^{4x} is y=c_{1}e^{4x}+c_{2}xe^{4x}+(e^{4x}x^2)/2

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y^{''}-8y^'+16y=e^{4x}

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x}

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

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = e^{4 x}

Solve linear ODE:

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

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

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What is the solution for y^{''}-8y^'+16y=e^{4x} ?
The solution for y^{''}-8y^'+16y=e^{4x} is y=c_{1}e^{4x}+c_{2}xe^{4x}+(e^{4x}x^2)/2