What is the solution for y^{''}+4y^'+5y=sin(x) ?

The solution for y^{''}+4y^'+5y=sin(x) is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/8 sin(x)-1/8 cos(x)

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y^{''}+4y^'+5y=sin(x)

y^{^{''}} + 4 y^{^{'}} + 5 y = sin (x)

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = sin (x)

Solve linear ODE:

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

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

y^{^{''}} = 2 y + 20 tan^{3} (x), y (0) = 10 tan (x)

\frac{1}{6} y^{^{''}} + 6 y = 2 tan (6 t) - \frac{1}{4} e^{3 t}

y^{^{''}} + 4 y = 11 t^{3} sin (2 t)

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

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

What is the solution for y^{''}+4y^'+5y=sin(x) ?
The solution for y^{''}+4y^'+5y=sin(x) is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/8 sin(x)-1/8 cos(x)