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

The solution for y^{''}+3y^'+2y=e^t,y(0)=1,y^'(0)=-1 is y=(-1/3+5/6)e^{-t}+1/3 e^{-2t}+1/6 e^t

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y^{''}+3y^'+2y=e^t,y(0)=1,y^'(0)=-1

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}, y (0) = 1, y^{^{'}} (0) = - 1

y = (- \frac{1}{3} + \frac{5}{6}) e^{- t} + \frac{1}{3} e^{- 2 t} + \frac{1}{6} e^{t}

Solution steps

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

Solve linear ODE:

y = (- \frac{1}{3} + \frac{5}{6}) e^{- t} + \frac{1}{3} e^{- 2 t} + \frac{1}{6} e^{t}

y = (- \frac{1}{3} + \frac{5}{6}) e^{- t} + \frac{1}{3} e^{- 2 t} + \frac{1}{6} e^{t}

What is the solution for y^{''}+3y^'+2y=e^t,y(0)=1,y^'(0)=-1 ?
The solution for y^{''}+3y^'+2y=e^t,y(0)=1,y^'(0)=-1 is y=(-1/3+5/6)e^{-t}+1/3 e^{-2t}+1/6 e^t