What is the solution for y^{''}+y^'-2y=0,y(0)=1,y(0)=1 ?

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

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

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

y = (1 - c_{2}) e^{t} + c_{2} e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = (1 - c_{2}) e^{t} + c_{2} e^{- 2 t}

y = (1 - c_{2}) e^{t} + c_{2} e^{- 2 t}

2 \frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} - y = 0

y^{^{''}} + 10 y^{^{'}} + \frac{8}{3} y = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (\frac{π}{2}) = 0, y (π) = - 1

15 y^{^{''}} - 4 y^{^{'}} - 3 y = 0

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

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