What is the solution for y^{''}+2y^'-8y=0,y(0)=-5,y^'(0)=26 ?

The solution for y^{''}+2y^'-8y=0,y(0)=-5,y^'(0)=26 is y=e^{2t}-6e^{-4t}

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y^{''}+2y^'-8y=0,y(0)=-5,y^'(0)=26

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = - 5, y^{^{'}} (0) = 26

y = e^{2 t} - 6 e^{- 4 t}

Solution steps

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

Solve linear ODE:

y = e^{2 t} - 6 e^{- 4 t}

y = e^{2 t} - 6 e^{- 4 t}

What is the solution for y^{''}+2y^'-8y=0,y(0)=-5,y^'(0)=26 ?
The solution for y^{''}+2y^'-8y=0,y(0)=-5,y^'(0)=26 is y=e^{2t}-6e^{-4t}