What is the solution for x^2ydx-(x^3+y^3)dy=0 ?

The solution for x^2ydx-(x^3+y^3)dy=0 is y=\sqrt[3]{1/(\W(\frac{x^3){c_{1}})}}x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

x^2ydx-(x^3+y^3)dy=0

x^{2} y d x - (x^{3} + y^{3}) d y = 0

y = 3 \frac{1}{W ( \frac{x ^{3}}{c _{1}} )} x

Solution steps

x^{2} y d x - (x^{3} + y^{3}) d y = 0

Solve using substitution:

y = 3 \frac{1}{W ( \frac{x ^{3}}{c _{1}} )} x

y = 3 \frac{1}{W ( \frac{x ^{3}}{c _{1}} )} x

d er i v a t i v e e^{2 - 5 x}

ma c l a u r in x^{2} e^{- x}

\int cos (e^{x}) d x

\int (1 - cos (x))^{\frac{1}{2}} d x

\int \frac{sin ( 8 x )}{1 + cos ^{2} ( 8 x )} d x

What is the solution for x^2ydx-(x^3+y^3)dy=0 ?
The solution for x^2ydx-(x^3+y^3)dy=0 is y=\sqrt[3]{1/(\W(\frac{x^3){c_{1}})}}x