What is the solution for y^{''}+y^'-20y=0 ?

The solution for y^{''}+y^'-20y=0 is y=c_{1}e^{4t}+c_{2}e^{-5t}

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y^{''}+y^'-20y=0

y^{^{''}} + y^{^{'}} - 20 y = 0

y = c_{1} e^{4 t} + c_{2} e^{- 5 t}

Solution steps

y^{''} (t) + y^{'} (t) - 20 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- 5 t}

y = c_{1} e^{4 t} + c_{2} e^{- 5 t}

\int_{0}^{x} sin (t^{2}) d t

\frac{d}{d x} (- 2 x^{2} - \frac{1}{2} x + 10)

y^{^{'}} + (1 + \frac{1}{t}) y = 1

y = 6 + 5 x

x \to 8 lim (x^{2})

What is the solution for y^{''}+y^'-20y=0 ?
The solution for y^{''}+y^'-20y=0 is y=c_{1}e^{4t}+c_{2}e^{-5t}