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

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

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

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

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

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

y^{^{''}} + y = cos (a x)

y^{^{''}} + 5 y^{^{'}} - 36 y = - 128 t + 72 t^{2}

y^{^{''}} + 4 y^{^{'}} - 32 y = - 112 t + 64 t^{2}

y^{^{''}} - 36.6 y^{^{'}} + 334.89 y = 18.5 u

4 y^{^{''}} + y = 2 cos (x)

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