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)/(16)+(e^4)/(32)

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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}{16} + \frac{e ^{4}}{32}

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}{16} + \frac{e ^{4}}{32}

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{e ^{4} x}{16} + \frac{e ^{4}}{32}

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)/(16)+(e^4)/(32)