What is the solution for y^{''}+3y^'+2y=12e^t,y(0)=2,y^'(0)=3 ?

The solution for y^{''}+3y^'+2y=12e^t,y(0)=2,y^'(0)=3 is y=e^{-t}-e^{-2t}+2e^t

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y^{''}+3y^'+2y=12e^t,y(0)=2,y^'(0)=3

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 e^{t}, y (0) = 2, y^{^{'}} (0) = 3

y = e^{- t} - e^{- 2 t} + 2 e^{t}

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 12 e^{t}

Solve linear ODE:

y = e^{- t} - e^{- 2 t} + 2 e^{t}

y = e^{- t} - e^{- 2 t} + 2 e^{t}

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 x^{2} + 5

y^{^{''}} - y^{^{'}} - 6 y = 73 cos (3 x)

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 8 t

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 7 t

y^{^{''}} - 2 y^{^{'}} + y = 6 t - 2, y (- 1) = 3, y^{^{'}} (- 1) = 7

What is the solution for y^{''}+3y^'+2y=12e^t,y(0)=2,y^'(0)=3 ?
The solution for y^{''}+3y^'+2y=12e^t,y(0)=2,y^'(0)=3 is y=e^{-t}-e^{-2t}+2e^t