What is the solution for y^{''}-3y^'-40y=0,y(0)=-3,y^'(0)=4 ?

The solution for y^{''}-3y^'-40y=0,y(0)=-3,y^'(0)=4 is y=(-3+28/13)e^{8t}-28/13 e^{-5t}

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y^{''}-3y^'-40y=0,y(0)=-3,y^'(0)=4

y^{^{''}} - 3 y^{^{'}} - 40 y = 0, y (0) = - 3, y^{^{'}} (0) = 4

y = (- 3 + \frac{28}{13}) e^{8 t} - \frac{28}{13} e^{- 5 t}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 40 y = 0

Solve linear ODE:

y = (- 3 + \frac{28}{13}) e^{8 t} - \frac{28}{13} e^{- 5 t}

y = (- 3 + \frac{28}{13}) e^{8 t} - \frac{28}{13} e^{- 5 t}

y^{^{''}} + 33 y = 0

y^{^{''}} - 8 y + 16 y = 0

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

y^{^{''}} - 13 y^{^{'}} + 36 y = 0

y^{^{''}} - 5 y^{^{'}} - 24 y = 0

What is the solution for y^{''}-3y^'-40y=0,y(0)=-3,y^'(0)=4 ?
The solution for y^{''}-3y^'-40y=0,y(0)=-3,y^'(0)=4 is y=(-3+28/13)e^{8t}-28/13 e^{-5t}