What is the solution for y^{''}+y^'-2y=-4,y(0)=2,y^'(0)=3 ?

The solution for y^{''}+y^'-2y=-4,y(0)=2,y^'(0)=3 is y=e^t-e^{-2t}+2

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y^{''}+y^'-2y=-4,y(0)=2,y^'(0)=3

y^{^{''}} + y^{^{'}} - 2 y = - 4, y (0) = 2, y^{^{'}} (0) = 3

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

Solution steps

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

Solve linear ODE:

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

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

- y^{^{''}} + y^{^{'}} + 2 y = e^{t}

y^{^{''}} - 4 y^{^{'}} - 21 y = 800 e^{- t}

y^{^{''}} + 5 y^{^{'}} + 4 y = 13 t e^{4 t}

y^{^{''}} - 2 y^{^{'}} - 8 y = (x + 1) e^{x}

y^{^{''}} - y^{^{'}} - 6 y = 56 cos (3 x)

What is the solution for y^{''}+y^'-2y=-4,y(0)=2,y^'(0)=3 ?
The solution for y^{''}+y^'-2y=-4,y(0)=2,y^'(0)=3 is y=e^t-e^{-2t}+2