# Thread: help me solve this

1. ## help me solve this

SOLVE
(x*y^2)dx - e^(1/(x^3))dx - (x^2)*y dy = 0

2. Try $\displaystyle u(x)=y^{2}(x).$

3. Noting that $\displaystyle d\left( \dfrac{y^2}{x^2}\right) = \dfrac{2x^2dy - 2 x y^2 dx}{x^4}$ might help.

4. $\displaystyle xy^2dx-yx^2dy-e^{x^{-3}}dx=0$

Multiply the equation by 2:

$\displaystyle 2xy^2dx-2yx^2dy-2e^{x^{-3}}dx=0$

Recall that $\displaystyle d\left(\left(\dfrac{x}{y}\right)^2 \right)=d\left(\dfrac{x^2}{y^2}\right)=\dfrac{2xy^ 2dx-2yx^2dy}{y^4} \implies y^4 \, d\left(\left(\dfrac{x}{y}\right)^2 \right)=2xy^2dx-2yx^2dy$

The equation will be:

$\displaystyle y^4 \, d\left(\left(\dfrac{x}{y}\right)^2 \right) - 2 e^{x^{-3}}dx=0$

Devide by $\displaystyle x^4$

$\displaystyle \left(\left(\dfrac{x}{y}\right)^{-2}\right)^2 d\left(\left(\dfrac{x}{y}\right)^2 \right) - 2 x^{-4} e^{x^{-3}}dx = 0$

Let $\displaystyle t=\left(\dfrac{x}{y}\right)^2$

The equation will be:

$\displaystyle \dfrac{1}{t^2} dt - 2 x^{-4} e^{x^{-3}}dx=0$

Now, Integrate.

5. Originally Posted by General
$\displaystyle xy^2dx-yx^2dy-e^{x^{-3}}dx=0$

Multiply the equation by 2:

$\displaystyle 2xy^2dx-2yx^2dy-2e^{x^{-3}}dx=0$

Recall that $\displaystyle d\left(\left(\dfrac{x}{y}\right)^2 \right)=d\left(\dfrac{x^2}{y^2}\right)=\dfrac{2xy^ 2dx-2yx^2dy}{y^4} \implies y^4 \, d\left(\left(\dfrac{x}{y}\right)^2 \right)=2xy^2dx-2yx^2dy$

The equation will be:

$\displaystyle y^4 \, d\left(\left(\dfrac{x}{y}\right)^2 \right) - 2 e^{x^{-3}}dx=0$

Devide by $\displaystyle x^4$

$\displaystyle \left(\left(\dfrac{x}{y}\right)^{-2}\right)^2 d\left(\left(\dfrac{x}{y}\right)^2 \right) - 2 x^{-4} e^{x^{-3}}dx = 0$

Let $\displaystyle t=\left(\dfrac{x}{y}\right)^2$

The equation will be:

$\displaystyle \dfrac{1}{t^2} dt - 2 x^{-4} e^{x^{-3}}dx=0$

Now, Integrate.

Thank you