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

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

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{1}{2} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{1}{2} cos (x)

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{1}{2} cos (x)

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

y^{^{''}} - y = tan (x)

y^{^{''}} + y = 6 x sin (x)

y^{^{''}} + 6 y^{^{'}} + 13 y = 4 x^{2} - 5 x + 2 + 3 e^{x}

y^{^{''}} - 2 y^{^{'}} - 35 y = 13 sin (x) - e^{3 x} + 1

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