What is the solution for (12x^2-20y^2)dx+(-20xy+36x^{-1}y^3)dy=0 ?

The solution for (12x^2-20y^2)dx+(-20xy+36x^{-1}y^3)dy=0 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(12x^2-20y^2)dx+(-20xy+36x^{-1}y^3)dy=0

(12 x^{2} - 20 y^{2}) d x + (- 20 x y + 36 x^{- 1} y^{3}) d y = 0

y = - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = - - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}, y = - \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}

Solution steps

(12 x^{2} - 20 y^{2}) d x + (- 20 x y + 36 x^{- 1} y^{3}) d y = 0

Solve using substitution:

y = - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = - - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}, y = - \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}

y = - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = - - \frac{- 5 x ^{2} + - 2 x ^{4} - c _{1}}{9 x ^{2}} x, y = \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}, y = - \frac{5 x ^{2} + - 2 x ^{4} - c _{1}}{3}

\int e^{x} d x

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

y = 5 x^{3} + 4 x^{2} - 6, (1, 3)

\int e^{- 2 x} \cdot sin (x) d x

x \to 2 lim ((- 2)^{3})

What is the solution for (12x^2-20y^2)dx+(-20xy+36x^{-1}y^3)dy=0 ?
The solution for (12x^2-20y^2)dx+(-20xy+36x^{-1}y^3)dy=0 is