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

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

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

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

y = e^{4 t} (a + \frac{5 - 4 a}{6}) - \frac{e ^{- 2 t} ( 5 - 4 a )}{6}

Solution steps

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

Solve linear ODE:

y = e^{4 t} (a + \frac{5 - 4 a}{6}) - \frac{e ^{- 2 t} ( 5 - 4 a )}{6}

y = e^{4 t} (a + \frac{5 - 4 a}{6}) - \frac{e ^{- 2 t} ( 5 - 4 a )}{6}

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