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

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

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y^{''}-4y^'-5y=0

y^{^{''}} - 4 y^{^{'}} - 5 y = 0

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

Solution steps

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

Solve linear ODE:

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

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

a re a x + y^{2} = 42, x = y

\int_{0}^{\frac{π}{8}} x (sec (2 x))^{2} d x

\int arctan (9 t) d t

y^{^{'}} = - 2 y

\int (2 x - 4) (x^{2} - 4 x + 3)^{5} d x

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