What is the solution for y^{''}+2y^'-24y=0,y(0)=1,y^'(0)=-2 ?

The solution for y^{''}+2y^'-24y=0,y(0)=1,y^'(0)=-2 is y=(1-3/5)e^{4t}+3/5 e^{-6t}

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y^{''}+2y^'-24y=0,y(0)=1,y^'(0)=-2

y^{^{''}} + 2 y^{^{'}} - 24 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y = (1 - \frac{3}{5}) e^{4 t} + \frac{3}{5} e^{- 6 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 24 y = 0

Solve linear ODE:

y = (1 - \frac{3}{5}) e^{4 t} + \frac{3}{5} e^{- 6 t}

y = (1 - \frac{3}{5}) e^{4 t} + \frac{3}{5} e^{- 6 t}

y^{^{'''}} - 13 y^{^{''}} + 46 y^{^{'}} - 48 y = 0

y^{^{''}} + a y = 0, y^{^{'}} (0) = 0

y^{^{''}} + 5 y + 4 y = 0

\frac{1}{4} \frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} + x = 0

9 y^{^{''}} - 18 y^{^{'}} + 34 y = 0

What is the solution for y^{''}+2y^'-24y=0,y(0)=1,y^'(0)=-2 ?
The solution for y^{''}+2y^'-24y=0,y(0)=1,y^'(0)=-2 is y=(1-3/5)e^{4t}+3/5 e^{-6t}