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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

y y^{^{'}} = \frac{x}{x ^{2} + 1}

y = ln (x^{2} + 1) + c_{1}, y = - ln (x^{2} + 1) + c_{1}

Solution steps

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

Solve separable ODE:

y = ln (x^{2} + 1) + c_{1}, y = - ln (x^{2} + 1) + c_{1}

y = ln (x^{2} + 1) + c_{1}, y = - ln (x^{2} + 1) + c_{1}

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