What is the solution for y^{''}-4y^'-96y=0,y(0)=13,y^'(0)=36 ?

The solution for y^{''}-4y^'-96y=0,y(0)=13,y^'(0)=36 is y=7e^{12t}+6e^{-8t}

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y^{''}-4y^'-96y=0,y(0)=13,y^'(0)=36

y^{^{''}} - 4 y^{^{'}} - 96 y = 0, y (0) = 13, y^{^{'}} (0) = 36

y = 7 e^{12 t} + 6 e^{- 8 t}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 96 y = 0

Solve linear ODE:

y = 7 e^{12 t} + 6 e^{- 8 t}

y = 7 e^{12 t} + 6 e^{- 8 t}

y^{^{''}} + 25 y^{^{'}} + \frac{1}{0.046} y = 0

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} = 0

y^{^{''}} - 3 y^{^{'}} = 0, y (0) = 2 - 2 e^{3}, y (1) = 0

y^{^{''}} - 19 y^{^{'}} + 90 y = 0

What is the solution for y^{''}-4y^'-96y=0,y(0)=13,y^'(0)=36 ?
The solution for y^{''}-4y^'-96y=0,y(0)=13,y^'(0)=36 is y=7e^{12t}+6e^{-8t}