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

The solution for yy^'=x(1+y^2) is y=sqrt(e^{x^2+c_{1)}-1},y=-sqrt(e^{x^2+c_{1)}-1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

yy^'=x(1+y^2)

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

y = e^{x^{2} + c_{1}} - 1, y = - e^{x^{2} + c_{1}} - 1

Solution steps

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

Solve separable ODE:

y = e^{x^{2} + c_{1}} - 1, y = - e^{x^{2} + c_{1}} - 1

y = e^{x^{2} + c_{1}} - 1, y = - e^{x^{2} + c_{1}} - 1

\frac{d}{d x} (e^{e^{2 x}})

\int \frac{2 x + 1}{x ^{2} + 2 x + 65} d x

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

\int sin^{3} (b) cos^{3} (b)

x \to - 2 lim (x^{4} + 5^{x} - 2)

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