What is the solution for y^{''}-2y^'+y=25sin(2x) ?

The solution for y^{''}-2y^'+y=25sin(2x) is y=c_{1}e^x+c_{2}xe^x-3sin(2x)+4cos(2x)

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y^{''}-2y^'+y=25sin(2x)

y^{^{''}} - 2 y^{^{'}} + y = 25 sin (2 x)

y = c_{1} e^{x} + c_{2} x e^{x} - 3 sin (2 x) + 4 cos (2 x)

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 25 sin (2 x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - 3 sin (2 x) + 4 cos (2 x)

y = c_{1} e^{x} + c_{2} x e^{x} - 3 sin (2 x) + 4 cos (2 x)

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 65 x e^{x}

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

y^{^{'''}} - y^{^{''}} - 8 y^{^{'}} + 12 y = 2 e^{t}

y^{^{''}} + 4 y = 3 x^{3}

x^{^{''}} + 2 x^{^{'}} + x = e^{- t}

What is the solution for y^{''}-2y^'+y=25sin(2x) ?
The solution for y^{''}-2y^'+y=25sin(2x) is y=c_{1}e^x+c_{2}xe^x-3sin(2x)+4cos(2x)