# [ xy dx-(x^{2}+3y^{2})dy=0

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yunitsiL
Write this equation as

(In the last equality, we divided the numerator and denominator by ${x}^{2}$ .) Therefore, the equation becomes is
$\frac{\text{d}y}{\text{d}x}=\frac{\frac{y}{x}}{1+3\frac{{y}^{2}}{{x}^{2}}}$
(1) which is of the form
$\frac{\text{d}y}{\text{d}x}=h\left(\frac{y}{x}\right)$
Thus, we will use the substitution
$y=ux⇒\frac{\text{d}y}{\text{d}x}=\frac{\text{d}u}{\text{d}x}\cdot x+u$ (2)
Plugging (2) into (1), we get that,
$\frac{\text{d}u}{\text{d}x}\cdot x+u=\frac{u}{1+3{u}^{2}}⇒x\frac{du}{dx}=\frac{u}{1+3{u}^{2}}-u=\frac{u-u\left(1+3{u}^{2}\right)}{1+3{u}^{2}}=-\frac{3{u}^{2}}{1+3{u}^{2}}$ Therefore,
$x\frac{du}{dx}=-\frac{3{u}^{2}}{1+3{u}^{2}}⇒\frac{1+3{u}^{2}}{3{u}^{3}}du=-\frac{dx}{x}$
Integrating both sides:
$\int \frac{1+3{u}^{2}}{3{u}^{3}}du=\int -\frac{dx}{x}$
Now,
$\int -\frac{dx}{x}=-ln|x|+{C}_{2}=ln\frac{1}{|x|}+{C}_{2}$
where ${C}_{2}$ is some constant. (We have also used that $n\cdot ln\cdot y=ln\cdot {y}^{n}$)
On the other hand,
$\int \frac{1+3{u}^{2}}{3{u}^{3}}du=\frac{1}{3}\int \frac{du}{{u}^{3}}+\int \frac{du}{u}$
$=\frac{1}{3}\int {u}^{-3}du+ln|u|$
$=-\frac{1}{6}{u}^{-2}+ln|u|+{C}_{1}$
$=-\frac{1}{6{u}^{2}}+ln|u|+{C}_{1}$
Here ${C}_{1}$ is some constant
Using (3) we now get that
$-\frac{1}{6{u}^{2}}+ln|u|+{C}_{1}=ln\frac{1}{|x|}+{C}_{2}$
Defining $C={C}_{2}-{C}_{1}$, we get
$-\frac{1}{6{u}^{2}}+ln|u|=ln\frac{1}{|x|}+C$
Finally, recall that $u=\frac{y}{x}$ to write the final solution (in the implicit form)
$ln|\frac{y}{x}|-\frac{{x}^{2}}{6{y}^{2}}=ln\frac{1}{|x|}+C$
Answer $ln|\frac{y}{x}|-\frac{{x}^{2}}{6{y}^{2}}=ln\frac{1}{|x|}+C$