What is the solution for y^{''}-8y^'+16y=8x+7 ?

The solution for y^{''}-8y^'+16y=8x+7 is y=c_{1}e^{4x}+c_{2}xe^{4x}+x/2+11/16

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y^{''}-8y^'+16y=8x+7

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 x + 7

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{x}{2} + \frac{11}{16}

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = 8 x + 7

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{x}{2} + \frac{11}{16}

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{x}{2} + \frac{11}{16}

y^{^{''}} - 6 y^{^{'}} - 7 y = - 9 e^{- 2 x}

y^{^{''}} - 2 y^{^{'}} = - 2 e^{x}

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

y^{^{''}} + 9 y = 13 e^{2 x} + 10 sin (2 x)

y^{^{''}} - 3 y^{^{'}} = sin (2 x)

What is the solution for y^{''}-8y^'+16y=8x+7 ?
The solution for y^{''}-8y^'+16y=8x+7 is y=c_{1}e^{4x}+c_{2}xe^{4x}+x/2+11/16