What is the solution for y^{''}-2y^'-3y=sin(x) ?

The solution for y^{''}-2y^'-3y=sin(x) is y=c_{1}e^{3x}+c_{2}e^{-x}-1/5 sin(x)+1/10 cos(x)

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y^{''}-2y^'-3y=sin(x)

y^{^{''}} - 2 y^{^{'}} - 3 y = sin (x)

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{1}{5} sin (x) + \frac{1}{10} cos (x)

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = sin (x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{1}{5} sin (x) + \frac{1}{10} cos (x)

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{1}{5} sin (x) + \frac{1}{10} cos (x)

y^{^{''}} + 2 y^{^{'}} + 2 y = cos (t)

y^{^{''}} - 6 y^{^{'}} + 9 y = sin (3 t)

y^{^{''}} - 2 y^{^{'}} - 8 y = 6 + 3 x

2 y^{^{''}} + 4 y = 6 t

y^{^{''}} + 6 y^{^{'}} + 8 y = 8 t + 22, y (0) = - 7, y^{^{'}} (0) = 20

What is the solution for y^{''}-2y^'-3y=sin(x) ?
The solution for y^{''}-2y^'-3y=sin(x) is y=c_{1}e^{3x}+c_{2}e^{-x}-1/5 sin(x)+1/10 cos(x)