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

The solution for y^{''}-8y^'+16y=e^x is y=c_{1}e^{4x}+c_{2}xe^{4x}+1/9 e^x

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

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

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{1}{9} e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{1}{9} e^{x}

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{1}{9} e^{x}

d^{2} y = (2 x^{2} + 4) (x^{2} + 3 x)^{4}

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} - 4 y = x^{2}

y^{^{''}} + 3 y^{^{'}} + 2 y = - 6 - 4 x - 7 x^{2} - 2 x^{3}

(D^{2} + 1) y = 12 cos^{2} (x)

(D^{2} + 3 D + 2) y = 1 + 3 x + x^{2}

What is the solution for y^{''}-8y^'+16y=e^x ?
The solution for y^{''}-8y^'+16y=e^x is y=c_{1}e^{4x}+c_{2}xe^{4x}+1/9 e^x