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

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

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y^{''}+3y+2y=t

y^{^{''}} + 3 y + 2 y = t

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

Solution steps

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

Solve linear ODE:

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

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

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