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

The solution for y^{''}-4y^'+4=sin(x) is y=c_{1}e^{4x}+c_{2}-1/17 sin(x)+4/17 cos(x)+x

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

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

y = c_{1} e^{4 x} + c_{2} - \frac{1}{17} sin (x) + \frac{4}{17} cos (x) + x

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} - \frac{1}{17} sin (x) + \frac{4}{17} cos (x) + x

y = c_{1} e^{4 x} + c_{2} - \frac{1}{17} sin (x) + \frac{4}{17} cos (x) + x

y^{^{''}} + 2 y^{^{'}} = 4, y (0) = 3, y^{^{'}} (0) = 5

y^{^{''}} + y = t sin (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = x^{2} - 1

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 e^{2} x

y^{^{''}} - 4 y^{^{'}} + 5 y = \frac{2 e ^{2 x}}{sin ( x )}

What is the solution for y^{''}-4y^'+4=sin(x) ?
The solution for y^{''}-4y^'+4=sin(x) is y=c_{1}e^{4x}+c_{2}-1/17 sin(x)+4/17 cos(x)+x