What is the solution for y^{''}-4y^'-5y=0,y(-1)=3 ?

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

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y^{''}-4y^'-5y=0,y(-1)=3

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (- 1) = 3

y = (3 e^{5} - e^{6} c_{2}) e^{5 t} + c_{2} e^{- t}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 5 y = 0

Solve linear ODE:

y = (3 e^{5} - e^{6} c_{2}) e^{5 t} + c_{2} e^{- t}

y = (3 e^{5} - e^{6} c_{2}) e^{5 t} + c_{2} e^{- t}

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

x^{^{''}} + 4 x^{^{'}} + 16 x = 0

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

y^{^{''}} - 2 y^{^{'}} + \frac{1}{2} y = 0

y^{^{''}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = - 6

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