What is the solution for y^{''}-y^'-30y=0,y(0)=0,y^'(0)=5 ?

The solution for y^{''}-y^'-30y=0,y(0)=0,y^'(0)=5 is y= 5/11 e^{6t}-5/11 e^{-5t}

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y^{''}-y^'-30y=0,y(0)=0,y^'(0)=5

y^{^{''}} - y^{^{'}} - 30 y = 0, y (0) = 0, y^{^{'}} (0) = 5

y = \frac{5}{11} e^{6 t} - \frac{5}{11} e^{- 5 t}

Solution steps

y^{''} (t) - y^{'} (t) - 30 y = 0

Solve linear ODE:

y = \frac{5}{11} e^{6 t} - \frac{5}{11} e^{- 5 t}

y = \frac{5}{11} e^{6 t} - \frac{5}{11} e^{- 5 t}

49 y^{^{'''}} + 28 y^{^{''}} + 4 y^{^{'}} = 0

y^{^{'''}} - 5 y^{^{''}} - 36 y^{^{'}} = 0

\frac{3}{2} y^{^{''}} + 3 y^{^{'}} + 5 y = 0

56 y^{^{''}} + 33 y^{^{'}} - 14 y = 0

(d^{3} - 2 d^{2} - d + 2) y = 0

What is the solution for y^{''}-y^'-30y=0,y(0)=0,y^'(0)=5 ?
The solution for y^{''}-y^'-30y=0,y(0)=0,y^'(0)=5 is y= 5/11 e^{6t}-5/11 e^{-5t}