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.

  • 2
    $\begingroup$ What have you try so far? $\endgroup$ – Chee Han Sep 9 '17 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 Sep 9 '17 at 18:02
  • 1
    $\begingroup$ 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 Sep 9 '17 at 18:26

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 $$I=e^{\int p(z)dz}=\dfrac{1}{z^3}=\dfrac{1}{(u+x)^3}=\color{blue}{\dfrac{1}{(y^2+x)^3}}$$

| cite | improve this answer | |
  • $\begingroup$ Wow! Never used that strategy in a question. Thanks for this solution. $\endgroup$ – user1611542 Sep 9 '17 at 18:32
  • $\begingroup$ Your welcome. Accept answer when you are sure about the correctness of it! $\endgroup$ – Nosrati Sep 9 '17 at 18:33

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:

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

| cite | improve this answer | |
  • $\begingroup$ 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 Sep 9 '17 at 18:11
  • $\begingroup$ Yes, I do agree. I will bear that in mind when answering future questions. $\endgroup$ – Student 1 Sep 9 '17 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 Sep 9 '17 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.