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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y = e^{t}

y^{^{''}} - 2 y^{^{'}} + 2 y = \frac{e ^{x}}{cos ( x )}

y^{^{'''}} + 9 y^{^{'}} = - 18 t

y^{^{''}} + 16 y = 5 sec (4 x)

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

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=(-c_{2}+5/6)e^{-t}+c_{2}e^{-2t}+1/6 e^t