What is the solution for y^{''}-2y^'+y=8e^x-5 ?

The solution for y^{''}-2y^'+y=8e^x-5 is y=c_{1}e^x+c_{2}xe^x+4x^2e^x-5

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} e^{x} - 5

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} e^{x} - 5

y = c_{1} e^{x} + c_{2} x e^{x} + 4 x^{2} e^{x} - 5

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 16 y = 16 sin (x)

y^{^{''}} - 2 y^{^{'}} = 7 cos (3 t)

y^{^{''}} - 16 y^{^{'}} + 64 y = 504 sin (x) - 128 cos (x)

y^{^{''''}} - 3 y^{^{'''}} + 2 y^{^{''}} = e^{3 x}

y^{^{''}} + y^{^{'}} = - cos (x)

What is the solution for y^{''}-2y^'+y=8e^x-5 ?
The solution for y^{''}-2y^'+y=8e^x-5 is y=c_{1}e^x+c_{2}xe^x+4x^2e^x-5