What is the solution for y^{''}-4y^'-5y=0,y(0)=-5,y^'(0)=31 ?

The solution for y^{''}-4y^'-5y=0,y(0)=-5,y^'(0)=31 is y=(-5+28/3)e^{5t}-28/3 e^{-t}

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y^{''}-4y^'-5y=0,y(0)=-5,y^'(0)=31

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = - 5, y^{^{'}} (0) = 31

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} - 2 y = 0

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

4 x^{^{''}} + x^{^{'}} + 7 x = 0

y^{^{''}} + 0.025 \cdot y + 0.1 y^{^{'}} = 0

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

What is the solution for y^{''}-4y^'-5y=0,y(0)=-5,y^'(0)=31 ?
The solution for y^{''}-4y^'-5y=0,y(0)=-5,y^'(0)=31 is y=(-5+28/3)e^{5t}-28/3 e^{-t}