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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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\frac{d}{d t} (1 + 2 cos (t))

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\int_{1}^{\infty} \frac{6}{x ^{4}} d x

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