What is the solution for x^{''}+2x^'-8x=0,x^'(0)=22,x(0)=-1 ?

The solution for x^{''}+2x^'-8x=0,x^'(0)=22,x(0)=-1 is x=3e^{2t}-4e^{-4t}

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x^{''}+2x^'-8x=0,x^'(0)=22,x(0)=-1

x^{^{''}} + 2 x^{^{'}} - 8 x = 0, x^{^{'}} (0) = 22, x (0) = - 1

x = 3 e^{2 t} - 4 e^{- 4 t}

Solution steps

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

Solve linear ODE:

x = 3 e^{2 t} - 4 e^{- 4 t}

x = 3 e^{2 t} - 4 e^{- 4 t}

\frac{d ^{2} x}{d t ^{2}} - 4 \frac{d x}{d t} + 5 x = 0

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y^{1} (0) = e^{x} cos (2 x)

4 y^{^{''}} + 24 y^{^{'}} + 37 y = 0, y (π) = 1

y^{^{'''}} - 3 y^{^{''}} - 5 y^{^{'}} + 39 y = 0

y^{^{''}} + 29.4 y = 0

What is the solution for x^{''}+2x^'-8x=0,x^'(0)=22,x(0)=-1 ?
The solution for x^{''}+2x^'-8x=0,x^'(0)=22,x(0)=-1 is x=3e^{2t}-4e^{-4t}