What is the solution for y^2+x^2y^'=xyy^' ?

The solution for y^2+x^2y^'=xyy^' is y=-\W(-(c_{1})/x)x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^2+x^2y^'=xyy^'

y^{2} + x^{2} y^{^{'}} = x y y^{^{'}}

y = - W (- \frac{c _{1}}{x}) x

Solution steps

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

Solve using substitution:

y = - W (- \frac{c _{1}}{x}) x

y = - W (- \frac{c _{1}}{x}) x

\int 8 sin^{2} (t - \frac{π}{12}) d t

2 sin^{2} (t)

f (x) = \frac{x}{e ^{x} - 1}

\int_{9}^{16} \frac{1}{8 - x} d x

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

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