What is the solution for y^{''}+10y^'+24y=0,y(0)=0,y^'(0)=-5 ?

The solution for y^{''}+10y^'+24y=0,y(0)=0,y^'(0)=-5 is y=-5/2 e^{-4t}+5/2 e^{-6t}

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y^{''}+10y^'+24y=0,y(0)=0,y^'(0)=-5

y^{^{''}} + 10 y^{^{'}} + 24 y = 0, y (0) = 0, y^{^{'}} (0) = - 5

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

Solution steps

y^{''} (t) + 10 y^{'} (t) + 24 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 8 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'''}} + 9 y^{^{''}} + 8 y^{^{'}} = 0

y^{^{''}} - y^{^{'}} + cy = 0

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

y^{^{''}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 6

What is the solution for y^{''}+10y^'+24y=0,y(0)=0,y^'(0)=-5 ?
The solution for y^{''}+10y^'+24y=0,y(0)=0,y^'(0)=-5 is y=-5/2 e^{-4t}+5/2 e^{-6t}