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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + (4) y^{^{'}} + 25 y = 0

\frac{d ^{2} y}{d t ^{2}} + k y = 0

\frac{d ^{2} y}{d x ^{2}} - 9 y = 0, y (0) = 1, y (1) = - 1

y^{^{''}} + 6 y^{^{'}} + 73 y = 0, y (0) = - 5, y^{^{'}} (0) = - 25

y^{^{'''}} = - 3 y^{^{''}} + y^{^{'}}

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