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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (sec^{2} (3 x) \cdot 3)

x \to - 10 - lim (4 x + 40)

f (x) = 26 - x^{2}, (- 5, 1)

f (x) = 2 x e^{- x}

\int \frac{1}{2 x + 1} d x

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