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

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

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

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

y = - 2 e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = - 2 e^{- 2 t}

y = - 2 e^{- 2 t}

cos^{2} (8 t)

y = cos (9 x)

\frac{d y}{d x}, y = (sec (x) + tan (x)) (sec (x) - tan (x))

\int sin (x) tan (x) d x

x \to 2 lim (2 x^{2} - 10 x + 1)

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