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

The solution for y^'=(y^2-1)x is y=-(e^{c_{1}})/(e^{x^2)-e^{c_{1}}}-(e^{x^2})/(e^{x^2)-e^{c_{1}}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(y^2-1)x

y^{^{'}} = (y^{2} - 1) x

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

Solution steps

y^{'} (x) = (y^{2} - 1) x

Solve Bernoulli ODE:

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

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

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

\frac{d}{d x} (\frac{ln ( 2 x )}{2 x})

\frac{\partial}{\partial u} (u \cdot sin (v))

\int_{- 2}^{1} \frac{1}{2 - x} d x

\frac{d}{d x} (x e^{- 6 x})

What is the solution for y^'=(y^2-1)x ?
The solution for y^'=(y^2-1)x is y=-(e^{c_{1}})/(e^{x^2)-e^{c_{1}}}-(e^{x^2})/(e^{x^2)-e^{c_{1}}}