What is the solution for y^{''}+2y^'+2y=sin(x) ?

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

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y^{''}+2y^'+2y=sin(x)

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (x)

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{5} sin (x) - \frac{2}{5} cos (x)

Solution steps

y^{''} (x) + 2 y^{'} (x) + 2 y = sin (x)

Solve linear ODE:

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{5} sin (x) - \frac{2}{5} cos (x)

y = e^{- x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{5} sin (x) - \frac{2}{5} cos (x)

\int \frac{1}{2 y - 1} d y

n = 0 \sum \infty 0. 9^{n}

\int tan^{7} (x) sec^{2} (x) d x

\int_{1}^{e} \frac{( x ^{2} + 1 )}{x} d x

x \to 4 + lim (ln (x^{2} - 16))

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