What is the solution for y^{''}+2y^'-8y=0,y(0)=8,y^'(0)=-14 ?

The solution for y^{''}+2y^'-8y=0,y(0)=8,y^'(0)=-14 is y=3e^{2t}+5e^{-4t}

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y^{''}+2y^'-8y=0,y(0)=8,y^'(0)=-14

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = 8, y^{^{'}} (0) = - 14

y = 3 e^{2 t} + 5 e^{- 4 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 8 y = 0

Solve linear ODE:

y = 3 e^{2 t} + 5 e^{- 4 t}

y = 3 e^{2 t} + 5 e^{- 4 t}

y^{^{''}} + 5 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 25 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y^{^{''}} + 6 y^{^{'}} + 34 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

9 y^{^{''}} + 16 y = 0

a \frac{d ^{2} y}{d x ^{2}} + b \frac{d y}{d x} + cy = 0

What is the solution for y^{''}+2y^'-8y=0,y(0)=8,y^'(0)=-14 ?
The solution for y^{''}+2y^'-8y=0,y(0)=8,y^'(0)=-14 is y=3e^{2t}+5e^{-4t}