What is the solution for y^{''}-2y^'-8y=0,y(0)=2,y^'(0)=-1 ?

The solution for y^{''}-2y^'-8y=0,y(0)=2,y^'(0)=-1 is y=(2-3/2)e^{4t}+3/2 e^{-2t}

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y^{''}-2y^'-8y=0,y(0)=2,y^'(0)=-1

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 22 y^{^{'}} + α y = 0

x^{^{''}} + a \cdot x^{^{'}} + b \cdot x = 0

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

5 y^{^{''}} - 30 y^{^{'}} + 170 y = 0

y^{^{''}} + 14 y^{^{'}} + 49 y = 0, y (- 4) = 1, y^{^{'}} (- 4) = 5

What is the solution for y^{''}-2y^'-8y=0,y(0)=2,y^'(0)=-1 ?
The solution for y^{''}-2y^'-8y=0,y(0)=2,y^'(0)=-1 is y=(2-3/2)e^{4t}+3/2 e^{-2t}