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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 3 y^{^{'}} + 5 y = 2 e^{t}

y^{^{''}} + 4 y^{^{'}} + 6 y = 1^{- t}

y^{^{''}} + y = t - \frac{π}{2}, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 1

y^{^{''}} - y^{^{'}} - 2 y = (x + 1) e^{- x}

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