What is the solution for y^{''}-2y^'+y=3cos(2x) ?

The solution for y^{''}-2y^'+y=3cos(2x) is y=c_{1}e^x+c_{2}xe^x-12/25 sin(2x)-9/25 cos(2x)

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

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

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{12}{25} sin (2 x) - \frac{9}{25} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{12}{25} sin (2 x) - \frac{9}{25} cos (2 x)

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{12}{25} sin (2 x) - \frac{9}{25} cos (2 x)

y^{^{''}} - 16 y = 2 e^{2 x}

y^{^{''}} + 2 y^{^{'}} + 2 y = 4 cos (x) + 2 sin (x)

y^{^{''}} + 5 y^{^{'}} + 6 = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} + 11 y = 5 e^{3 x}

y^{^{''}} + 6 y^{^{'}} + 10 y = 180 sin (2 x)

What is the solution for y^{''}-2y^'+y=3cos(2x) ?
The solution for y^{''}-2y^'+y=3cos(2x) is y=c_{1}e^x+c_{2}xe^x-12/25 sin(2x)-9/25 cos(2x)