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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 49 y = t^{2} cos (7 t) + t e^{- 7 t}

y^{^{'''}} - y^{^{'}} = 11 t

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{x}

y^{^{''}} - 4 y^{^{'}} = 1, y (0) = 1, y^{^{'}} (0) = 4

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

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