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

The solution for y^{''}-4y^'+4y=sin(4x) is y=c_{1}e^{2x}+c_{2}xe^{2x}-3/100 sin(4x)+1/25 cos(4x)

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - \frac{3}{100} sin (4 x) + \frac{1}{25} cos (4 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - \frac{3}{100} sin (4 x) + \frac{1}{25} cos (4 x)

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - \frac{3}{100} sin (4 x) + \frac{1}{25} cos (4 x)

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{3 x} ln (x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 6 e^{x} - 2 x e^{2 x} + 2

y^{^{''}} + 3 y^{^{'}} - 6 y = x e^{2 x}

y^{^{''}} + 2 y^{^{'}} = x + 1

x^{^{''}} + x = 8 cos (6 t)

What is the solution for y^{''}-4y^'+4y=sin(4x) ?
The solution for y^{''}-4y^'+4y=sin(4x) is y=c_{1}e^{2x}+c_{2}xe^{2x}-3/100 sin(4x)+1/25 cos(4x)