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

The solution for y^{''}+2y^'=3+sin(t) is y=c_{1}+c_{2}e^{-2t}+(3t)/2-1/5 sin(t)-2/5 cos(t)

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

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

y = c_{1} + c_{2} e^{- 2 t} + \frac{3 t}{2} - \frac{1}{5} sin (t) - \frac{2}{5} cos (t)

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 t} + \frac{3 t}{2} - \frac{1}{5} sin (t) - \frac{2}{5} cos (t)

y = c_{1} + c_{2} e^{- 2 t} + \frac{3 t}{2} - \frac{1}{5} sin (t) - \frac{2}{5} cos (t)

What is the solution for y^{''}+2y^'=3+sin(t) ?
The solution for y^{''}+2y^'=3+sin(t) is y=c_{1}+c_{2}e^{-2t}+(3t)/2-1/5 sin(t)-2/5 cos(t)