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

The solution for (1+x^2)y^{''}=2x(y^')^2 is y=integral of-1/(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

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

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

y = \int - \frac{1}{ln ( x ^{2} + 1 ) + c _{1}} d x

Solution steps

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

Solve using substitution:

y = \int - \frac{1}{ln ( x ^{2} + 1 ) + c _{1}} d x

y = \int - \frac{1}{ln ( x ^{2} + 1 ) + c _{1}} d x

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