What is the solution for y^{''}-4y^'-5y=0,y(0)=1,y^'(0)=2 ?

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

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y^{''}-4y^'-5y=0,y(0)=1,y^'(0)=2

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = (1 - \frac{1}{2}) e^{5 t} + \frac{1}{2} e^{- t}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{1}{2}) e^{5 t} + \frac{1}{2} e^{- t}

y = (1 - \frac{1}{2}) e^{5 t} + \frac{1}{2} e^{- t}

y^{^{'''}} + 5 y^{^{''}} - 13 y^{^{'}} + 7 y = 0

0.05 y^{^{''}} + 2 y^{^{'}} + 100 y = 0, y (0) = 5, y^{^{'}} (0) = 0

(D^{4} - 81) y = 0

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

y^{^{''}} - 12 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = 16

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