What is the solution for y^{''}-2y^'-8y=0,y(0)=-4,y^'(0)=14 ?

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

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y^{''}-2y^'-8y=0,y(0)=-4,y^'(0)=14

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = - 4, y^{^{'}} (0) = 14

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

Solution steps

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

Solve linear ODE:

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

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

9 \frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + y = 0

u^{''} - u^{'} - 2 H (t) = 0

y^{^{''}} + 0.25 y^{^{'}} + y = 0

2 y^{^{'''}} + 5 y^{^{''}} - 28 y^{^{'}} - 15 y = 0

4 y^{^{''}} - 20 y^{^{'}} + 169 y = 0

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