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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + a y^{^{'}} - b = 0

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

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

y^{^{''}} + 4 y^{^{'}} + 5 y = 50 x + 13 e^{3 x}

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

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