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

The solution for y^{''}+3y^'+2y=sin(2x) is y=c_{1}e^{-x}+c_{2}e^{-2x}-1/20 sin(2x)-3/20 cos(2x)

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

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

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - \frac{1}{20} sin (2 x) - \frac{3}{20} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - \frac{1}{20} sin (2 x) - \frac{3}{20} cos (2 x)

y = c_{1} e^{- x} + c_{2} e^{- 2 x} - \frac{1}{20} sin (2 x) - \frac{3}{20} cos (2 x)

What is the solution for y^{''}+3y^'+2y=sin(2x) ?
The solution for y^{''}+3y^'+2y=sin(2x) is y=c_{1}e^{-x}+c_{2}e^{-2x}-1/20 sin(2x)-3/20 cos(2x)