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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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