What is the solution for y^{''}+2y^'+2y=3,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+2y^'+2y=3,y(0)=1,y^'(0)=0 is y=e^{-t}(-1/2 cos(t)-1/2 sin(t))+3/2

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y^{''}+2y^'+2y=3,y(0)=1,y^'(0)=0

y^{^{''}} + 2 y^{^{'}} + 2 y = 3, y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

What is the solution for y^{''}+2y^'+2y=3,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+2y^'+2y=3,y(0)=1,y^'(0)=0 is y=e^{-t}(-1/2 cos(t)-1/2 sin(t))+3/2