What is the solution for y^{''}-4y^'+4y=cos(2x) ?

The solution for y^{''}-4y^'+4y=cos(2x) is y=c_{1}e^{2x}+c_{2}xe^{2x}-1/8 sin(2x)

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - \frac{1}{8} sin (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - \frac{1}{8} sin (2 x)

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - \frac{1}{8} sin (2 x)

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

y^{^{''}} + y = x^{2} + 1, y (0) = 3, y (1) = 0

D (D + 3) y = x (5 + e^{x})

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

y^{^{''}} - y = e^{t} cos (t), y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-4y^'+4y=cos(2x) ?
The solution for y^{''}-4y^'+4y=cos(2x) is y=c_{1}e^{2x}+c_{2}xe^{2x}-1/8 sin(2x)