What is the solution for linear y^{''}+y^'-2y=0,y(0)=4,y^'(0)=5 ?

The solution for linear y^{''}+y^'-2y=0,y(0)=4,y^'(0)=5 is y=(4+1/3)e^t-1/3 e^{-2t}

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linear y^{''}+y^'-2y=0,y(0)=4,y^'(0)=5

linear

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 4, y^{^{'}} (0) = 5

y = (4 + \frac{1}{3}) e^{t} - \frac{1}{3} e^{- 2 t}

Solution steps

y^{''} (t) + y^{'} (t) - 2 y = 0

Solve linear ODE:

y = (4 + \frac{1}{3}) e^{t} - \frac{1}{3} e^{- 2 t}

y = (4 + \frac{1}{3}) e^{t} - \frac{1}{3} e^{- 2 t}

y^{^{''}} + 3 y^{^{'}} + 12 y = 0

\frac{4}{9} y^{^{''}} + 36 y = 0

2 y^{^{''}} + 5 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = - 1.5

4 y^{^{''}} + y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 13 y^{^{'}} = 0

What is the solution for linear y^{''}+y^'-2y=0,y(0)=4,y^'(0)=5 ?
The solution for linear y^{''}+y^'-2y=0,y(0)=4,y^'(0)=5 is y=(4+1/3)e^t-1/3 e^{-2t}