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

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

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y^{''}+y^'-2y=4e^{2t},y(0)=1,y^'(0)=-1

y^{^{''}} + y^{^{'}} - 2 y = 4 e^{2 t}, y (0) = 1, y^{^{'}} (0) = - 1

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

Solution steps

y^{''} (t) + y^{'} (t) - 2 y = 4 e^{2 t}

Solve linear ODE:

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

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

y^{^{''}} + 3 y + 2 = 0

y^{^{''}} - 14 y^{^{'}} + 49 y = 21 x + 6

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

D (t) = 30 e^{- 0.1 (3)}

2 y^{^{''}} - y^{^{'}} - 6 y = - 7 e^{2 x}

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