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

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

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + e^{x} sin (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + e^{x} sin (x)

y = c_{1} e^{x} + c_{2} x e^{x} + e^{x} sin (x)

y^{^{''}} - 4 y = 4

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = x^{2} - 1

y^{^{''}} - 4 y^{^{'}} + 3 y = 6

2 y^{^{''}} - 5 y^{^{'}} - 3 y = 3 - x

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 t + e^{3 t}

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