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

The solution for y^{''}+4y^'=sin(t) is y=c_{1}+c_{2}e^{-4t}-1/17 sin(t)-4/17 cos(t)

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

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

y = c_{1} + c_{2} e^{- 4 t} - \frac{1}{17} sin (t) - \frac{4}{17} cos (t)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} - \frac{1}{17} sin (t) - \frac{4}{17} cos (t)

y = c_{1} + c_{2} e^{- 4 t} - \frac{1}{17} sin (t) - \frac{4}{17} cos (t)

y^{^{''}} - 5 y^{^{'}} + 6 y = - 5 e^{3 x}

y^{^{''}} - 2 y^{^{'}} + y = t^{2} + 1

y^{^{''}} + 8 y^{^{'}} + 16 y = 6 cos (4 t)

x^{^{''}} + 4 x = - 6 sin (2 t) + 3 cos (2 t)

y^{^{''}} - 7 y^{^{'}} + 12 = x

What is the solution for y^{''}+4y^'=sin(t) ?
The solution for y^{''}+4y^'=sin(t) is y=c_{1}+c_{2}e^{-4t}-1/17 sin(t)-4/17 cos(t)