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Show that following differential equation admits an integrating factor which is a function of $(x+y^2)$.

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

Approach : Write $y^{'}$ as $dy/dx$. Multiply the equation by $dx$ and $f(x+y^2)$. Now equation is of the form $Mdx+Ndy=0$. $f$ is integrating factor if

$dM/dy = dN/dx$

However, this isn't the case in this equation. I feel that original equation need to be modified in some manner to get the desired result. However, I am unable to get it.

PS : It is not a homework. This is an exam question. And I was trying it for practice.

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    $\begingroup$ What have you try so far? $\endgroup$
    – Chee Han
    Commented Sep 9, 2017 at 17:41
  • $\begingroup$ Now you've intrigued my curiosity as to any other questions you've asked prior to this one, and their quality or lack there of. This is not a do-my-homework for me site. Please show your workings of your attempts, provide the source wrt where you encountered the problem, explain where, exactly, you are stuck... or some of the above. That's what we mean showing context when you ask a question (and it's why we have a justification for putting on hold questions which lack such context. $\endgroup$
    – amWhy
    Commented Sep 9, 2017 at 18:02
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    $\begingroup$ I am sorry. I have edited the question. I very well know this is not do-my-homework for me site. In my previous questions too I have clearly mentioned the approach I have been using. Sometimes I skip it If I find it completely wrong and not heading in any direction. But will take care from next time. $\endgroup$
    – user1611542
    Commented Sep 9, 2017 at 18:26

2 Answers 2

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At first we see $$(3y^2-x)+2y(y^2-3x)y'=0$$ may write of the form $(3y^2-x)+(y^2-3x)(y^2)'=0$ or $(3u-x)dx+(u-3x)du=0$. Now with $M=3u-x$ and $N=u-3x$ we have $$\dfrac{\partial M}{\partial u}=3~~~,~~~\dfrac{\partial N}{\partial z}=-3$$ which shows not exact.

We know if $I(x,u)$ be an integral factor for non-exact equation $Mdx+Ndu=0$ then for $IMdx+INdu=0$ to be exact \begin{eqnarray*} \frac{\partial(IM)}{\partial u} &=& \frac{\partial(IN)}{\partial x} \\ \frac{\partial I}{\partial u}M+I\frac{\partial M}{\partial u} &=& \frac{\partial I}{\partial x}N+I\frac{\partial N}{\partial x} \\ I(\frac{\partial M}{\partial u}-\frac{\partial N}{\partial x}) &=&\frac{\partial I}{\partial x}N-\frac{\partial I}{\partial u}M \end{eqnarray*} in our equation, we see for artificial variable $z=x+u$, $$p(z)=\dfrac{\frac{\partial M}{\partial u}-\frac{\partial N}{\partial x}}{N-M}=\dfrac{6}{-2(u+x)}=\dfrac{-3}{z}$$ is a variable of $z$, then our integral factor is $$I=e^{\int p(z)dz}=\dfrac{1}{z^3}=\dfrac{1}{(u+x)^3}=\color{blue}{\dfrac{1}{(y^2+x)^3}}$$

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  • $\begingroup$ Wow! Never used that strategy in a question. Thanks for this solution. $\endgroup$
    – user1611542
    Commented Sep 9, 2017 at 18:32
  • $\begingroup$ Your welcome. Accept answer when you are sure about the correctness of it! $\endgroup$
    – Nosrati
    Commented Sep 9, 2017 at 18:33
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I am unsure of your integrating factor function but by $(x+y^2)$ do you mean if $(x+y^2) = u$ then $y^2 = -x + u$ ?

Then if $y^2=-x$ satisfies as a solution to the differential equation, the substitution of $y^2 = -x+u$ will lead you to a general solution to the problem.

Hence for $y^2=-x$ , $2y\frac{dy}{dx}=-1$

Substitute these results back into the differential equation:

$$(3(-x)-x)+2y((-x)-3x)\frac{-1}{2y}=0$$ $$(-4x) - (-4x) = 0 $$

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    $\begingroup$ Tip: When you are unclear about what a question is actually suggesting, as in a comment. You're entire first sentence should have been written in a comment; and only after the asker has clarified the question, should you post an answer. Otherwise, you're answering what makes sense to you, but not necessarily answering the question asked. $\endgroup$
    – amWhy
    Commented Sep 9, 2017 at 18:11
  • $\begingroup$ Yes, I do agree. I will bear that in mind when answering future questions. $\endgroup$
    – Student 1
    Commented Sep 9, 2017 at 18:13
  • $\begingroup$ It means that if equation is multiplied by some function of $x +y^2$ i.e $f(x+y^2)$ then it becomes a proper differential. $\endgroup$
    – user1611542
    Commented Sep 9, 2017 at 18:29

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