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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = \frac{t}{e ^{t}} + \frac{e ^{t}}{t}

4 y^{^{''}} - 7 y^{^{'}} + 3 y = t e^{2 t} sin (- 2 t)

(D^{2} - 3 D + 2) y = e^{5 x}

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

y^{^{''}} + 3 y^{^{'}} + 6 = 0

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