What is the solution for x^'=(x^2-1)t ?

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

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x^'=(x^2-1)t

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

- 2 \cdot e^{- l n (2) \cdot s}

\frac{\partial}{\partial y} (4 x y)

\frac{d}{d x} (\frac{1}{3 + x ^{2}})

\int \frac{1}{x + 1 + x - 1} d x

\int \frac{4}{( 4 + x ^{2} ) ^{2}} d x

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