What is the solution for y^{''}+4y^'=3*sin(2x) ?

The solution for y^{''}+4y^'=3*sin(2x) is y=c_{1}+c_{2}e^{-4x}-3/20 sin(2x)-3/10 cos(2x)

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y^{''}+4y^'=3*sin(2x)

y^{^{''}} + 4 y^{^{'}} = 3 \cdot sin (2 x)

y = c_{1} + c_{2} e^{- 4 x} - \frac{3}{20} sin (2 x) - \frac{3}{10} cos (2 x)

Solution steps

y^{''} (x) + 4 y^{'} (x) = 3 \cdot sin (2 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} - \frac{3}{20} sin (2 x) - \frac{3}{10} cos (2 x)

y = c_{1} + c_{2} e^{- 4 x} - \frac{3}{20} sin (2 x) - \frac{3}{10} cos (2 x)

y^{^{''}} - 6 y^{^{'}} + 8 y = x e^{3 x} cos (4 x)

y^{^{''}} - 8 y^{^{'}} - 20 y = 5 e^{- 2 x}

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{4} x

y^{^{''}} + 9 y^{^{'}} = sin (t), y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + 9 y = 5 sin (3 x), y (0) = 6, y^{^{'}} (0) = 0

What is the solution for y^{''}+4y^'=3*sin(2x) ?
The solution for y^{''}+4y^'=3*sin(2x) is y=c_{1}+c_{2}e^{-4x}-3/20 sin(2x)-3/10 cos(2x)