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

The solution for y^{''}+2y^'=1,y(0)=1,y^'(0)=0 is y=1-1/4+1/4 e^{-2t}+t/2

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

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

y = 1 - \frac{1}{4} + \frac{1}{4} e^{- 2 t} + \frac{t}{2}

Solution steps

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

Solve linear ODE:

y = 1 - \frac{1}{4} + \frac{1}{4} e^{- 2 t} + \frac{t}{2}

y = 1 - \frac{1}{4} + \frac{1}{4} e^{- 2 t} + \frac{t}{2}

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What is the solution for y^{''}+2y^'=1,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+2y^'=1,y(0)=1,y^'(0)=0 is y=1-1/4+1/4 e^{-2t}+t/2