What is the solution for y^{''}+4y^'=3sin(2t) ?

The solution for y^{''}+4y^'=3sin(2t) is y=c_{1}+c_{2}e^{-4t}-3/20 sin(2t)-3/10 cos(2t)

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y^{''}+4y^'=3sin(2t)

y^{^{''}} + 4 y^{^{'}} = 3 sin (2 t)

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

Solution steps

y^{''} (t) + 4 y^{'} (t) = 3 sin (2 t)

Solve linear ODE:

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

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

y^{^{'''}} + y^{^{'}} = sec (t), y^{^{'}} (0) = 3

(d^{2} - 2 d) y = e^{x} sin (x)

x^{^{''}} + 6 x^{^{'}} - 6 x = 7 t^{2}

\frac{1}{2} y^{^{''}} + 2 y = 3 tan (2 t) - \frac{1}{3} e^{4 t}

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

What is the solution for y^{''}+4y^'=3sin(2t) ?
The solution for y^{''}+4y^'=3sin(2t) is y=c_{1}+c_{2}e^{-4t}-3/20 sin(2t)-3/10 cos(2t)