What is the solution for x^'x=-1/2 (x^2-25) ?

The solution for x^'x=-1/2 (x^2-25) is x=sqrt(e^{-t-c_{1)}+25},x=-sqrt(e^{-t-c_{1)}+25}

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x^'x=-1/2 (x^2-25)

x^{^{'}} x = - \frac{1}{2} (x^{2} - 25)

x = e^{- t - c_{1}} + 25, x = - e^{- t - c_{1}} + 25

Solution steps

x^{'} (t) x = - \frac{1}{2} (x^{2} - 25)

Solve separable ODE:

x = e^{- t - c_{1}} + 25, x = - e^{- t - c_{1}} + 25

x = e^{- t - c_{1}} + 25, x = - e^{- t - c_{1}} + 25

x \frac{d y}{d x} = \frac{1 - 5 y ^{2}}{3 y}

x^{2} e^{y} \frac{d y}{d x} + 2 x e^{y} = x

\frac{y d y}{d x} + 4 x^{2} + 6 x + 1 = 0

y^{^{'}} = e^{3 x - 4 y}, y (0) = 0

\frac{d y}{d x} + 9 x^{8} y = 0, y = 6 e^{- x^{9}}

What is the solution for x^'x=-1/2 (x^2-25) ?
The solution for x^'x=-1/2 (x^2-25) is x=sqrt(e^{-t-c_{1)}+25},x=-sqrt(e^{-t-c_{1)}+25}