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

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

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x^{''}-5x=0

x^{^{''}} - 5 x = 0

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

Solution steps

x^{''} (t) - 5 x = 0

Solve linear ODE:

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

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

2 5^{^{'}}

\frac{\partial}{\partial x} (3 (x - y) e^{3 y + 3 x^{2}})

\frac{d}{d x} (\frac{5 x + 4}{5 x - 4})

\frac{d}{d t} (cosh (t))

cot^{2} (cos (θ))

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