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

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

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y^{''}-8y^'+16=(1-x)e^{4x}

y^{^{''}} - 8 y^{^{'}} + 16 = (1 - x) e^{4 x}

y = c_{1} e^{8 x} + c_{2} - \frac{1}{16} e^{4 x} + \frac{e ^{4 x} x}{16} + 2 x

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 = (1 - x) e^{4 x}

Solve linear ODE:

y = c_{1} e^{8 x} + c_{2} - \frac{1}{16} e^{4 x} + \frac{e ^{4 x} x}{16} + 2 x

y = c_{1} e^{8 x} + c_{2} - \frac{1}{16} e^{4 x} + \frac{e ^{4 x} x}{16} + 2 x

y^{^{'''}} + 4 y^{^{''}} - 5 y = x e^{x} + 14 x

y^{^{''}} - 2 y^{^{'}} + y = - 5 e^{x} + 8 x^{2}

y^{^{''}} + 9 y = 4 x

y^{^{''}} + 2 y^{^{'}} + y = e^{- x}, y^{^{'}} (0) = 0

(D^{4} - 4 D^{2}) y = 5 x^{2} - e^{2 x}

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