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

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

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

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}

2 y^{^{''''}} - 4 y^{^{''}} + 2 y = 0

\frac{d ^{2} y}{d x ^{2}} + 44 y = 0

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

(0.1) x^{^{''}} + (0.4) x^{^{'}} + 2 x = 0

10 x^{^{''}} + 40 x^{^{'}} + \frac{1}{50} x = 0

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