What is the solution for y^{''}+2y^'+2y=t^2,y(0)= 1/2 ,y^'(0)=0 ?

The solution for y^{''}+2y^'+2y=t^2,y(0)= 1/2 ,y^'(0)=0 is y=e^{-t}sin(t)+(t^2)/2-t+1/2

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y^{''}+2y^'+2y=t^2,y(0)= 1/2 ,y^'(0)=0

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

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

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

y^{^{'''}} + 265 y^{^{'}} = tan (16 x)

y^{^{''}} - 10 y^{^{'}} + 9 y = e^{2 x}

y^{^{''}} + y^{^{'}} - 6 y = e^{2 t} + t

y^{^{'''}} - 3 y^{^{''}} - y^{^{'}} + 3 y = 1 + e^{x}

y^{^{''}} - y = 2 e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+2y^'+2y=t^2,y(0)= 1/2 ,y^'(0)=0 ?
The solution for y^{''}+2y^'+2y=t^2,y(0)= 1/2 ,y^'(0)=0 is y=e^{-t}sin(t)+(t^2)/2-t+1/2