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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} - 32 y = 108 e^{2 t}

y^{^{''}} + 16 y = 32 sec (2 x)

y^{^{''''}} - y^{^{'''}} - y^{^{''}} - y^{^{'}} - 2 y = 8 x^{5}

(d^{2} - 2 d + 5) y = e^{x}

(d^{3} - d) y = 2 cosh (x)

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