What is the solution for x^{''}-2x-5=0 ?

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

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x^{''}-2x-5=0

x^{^{''}} - 2 x - 5 = 0

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

Solution steps

x^{''} (t) - 2 x - 5 = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (e^{k x})

1 0^{2 x}

\int 6 + 6 tan^{2} (x) d x

\int - 7 d x

\int_{0}^{1} \frac{x ^{2}}{x ^{3} - 5} d x

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