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

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

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

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

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{4}{25} sin (2 x) - \frac{3}{25} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{4}{25} sin (2 x) - \frac{3}{25} cos (2 x)

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{4}{25} sin (2 x) - \frac{3}{25} cos (2 x)

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

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

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

y^{^{''}} + 2 y^{^{'}} - 15 y = (2 x + 1) e^{3 x} + 10

y^{^{''}} - y^{^{'}} - 2 y = 2 t^{2} - 5

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