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

The solution for y^{''}-8y^'+16y=x is y=c_{1}e^{4x}+c_{2}xe^{4x}+x/(16)+1/32

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 6 t^{2} e^{5 t}

y^{^{''}} - 12 y^{^{'}} + 36 y = 18 x + 2

y^{^{''}} - 5 y = 2 x^{3} - 4 x^{2} - x + 6

y^{^{''}} + 4 y^{^{'}} + 5 y = 10 e^{- 2 x} cos (x)

What is the solution for y^{''}-8y^'+16y=x ?
The solution for y^{''}-8y^'+16y=x is y=c_{1}e^{4x}+c_{2}xe^{4x}+x/(16)+1/32