I am stuck with a question of differential equation.
My book solves it in a peculiar way by multiplying it by $~\dfrac{y}{x}~$ and forming perfect derivatives of $~x^3y^6~$ and $~\ln\left(\dfrac{y}{x}\right)~$.
I could not understand the intuition behind this rearrangement of the terms.
I tried other methods like forming homogeneous equation by substitution or trigonometric substitutions with no success.
So what is the exact logic behind forming these perfect differentials in the question?

Also is there any other method to solve the question?


The task now actually is $$ y^4dx+2xy^3dy=\dfrac{ydx-xdy}{x^3y^3} $$ then the same transformations as below for the former reproduction of the equation apply, only the residual factors will have to account for the moved factor $y^3$. This gives $$ d(xy^2)=-\frac{d\left(\frac yx\right)}{xy^5} \implies (xy^2)^2\,d(xy^2)=-\frac{d\left(\frac yx\right)}{\left(\frac yx\right)} $$ which integrates to $$ \frac13(xy^2)^3=C-\ln\left(\frac yx\right) $$ Isolating $x$ or $y$ is now more difficult, but not impossible.

The original reproduction of the equation was $$ y^4dx+2xy^3dy=\dfrac{ydx-xdy}{x^3}y^3. $$ Divide by $y^2$ to get $$ d(xy^2)=-\frac{y}{x}\,d\left(\frac yx\right) $$ This can be integrated to $$ 2xy^2=C-\frac{y^2}{x^2} $$ which can be solved for $y$ resp. $y^2$ as $$ y^2=\frac{Cx^2}{2x^3+1} $$ I do not see were exponentials or logarithms enter the equation chain.

  • $\begingroup$ I am really sorry,😥 I entered the question wrong(the y^3 term is in the denominator). Thanks for the effort,but could you look into the question again? $\endgroup$ Sep 12 '19 at 11:59
  • $\begingroup$ See second section. The whole approach is more reverse engineering how the task was constructed than finding integrating factors. $\endgroup$ Sep 12 '19 at 12:15
  • $\begingroup$ Thanks a lot. That really helped. $\endgroup$ Sep 12 '19 at 12:17

It is probably done by inspection rather than intuition.

See that the RHS is $\frac{y^3}{x}$ of $-d(\frac{y}{x})$. Dividing both sides by that will leave us with: $$-d\frac{y}{x} = yxdx + 2x^2dy$$

Now notice the appearance of $2dy$ and $ydx$. This would push us to believe the path of multiplying and dividing the expression by $y$ will lead us somewhere, which is the sum rule for $(y^2)$ as by the chain/product rule $d(y^2) = 2ydy$

Now we have $$-d\frac{y}{x} = \frac{xy^2dx+x*x*2ydy}{y}$$

Or, after a rearrangement, $$-d\frac{y}{x} =\frac{y^2dx + x*2ydy}{\frac{y}{x}}$$

You may recognize the numerator of the RHS as $d(xy^2)$. So, after a rearrangement we get $$-\frac{y}{x}d\frac{y}{x} = d(xy^2)$$

And after a simple integration we have the general solution as $$-\frac{y^2}{2x^2} + k = xy^2$$

Edit: It seems OP had entered the incorrect question (and that someone else has also posted this solution). I leave it up as my comments on the thought process may serve useful to the OP (and other readers).


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