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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2}}{d x ^{2}} (3 x - 4 x^{\frac{9}{10}})

\frac{d y}{d x}, e^{y} = x + y

0.4 x - 1.295, - 1.4206, - 1.486

n = 2 \sum \infty (- 0.75)^{n}

\frac{1}{x ^{2}} 0

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