I am taking an online course on IIT's MOOC site https://onlinecourses.nptel.ac.in/noc17_ma11/ on Ordinary Differential equations. One of the questions in the assignment is the following :

Solve the differential equation


I am having some trouble attempting this question. Here are my quick thoughts:

  1. This is not in the separable form.
  2. Substituting $y=ux$ or $x=vy$ does not yield a homogenous function.
  3. The coefficient of $dx$ is not linear.
  4. This is not an exact differential equation.




Obviously, this must have an integrating factor then. Are my initial thoughts correct? Is that the correct way to proceed? I tried googling, but didn't find much luck.

Any hints in the right direction would be great.


Hint: set $v=x+2y\implies\frac{dv}{dx}=1+2\frac{dy}{dx}$

  • $\begingroup$ This did cross my mind, let me see, if I can find a solution. $\endgroup$ – Quasar Aug 6 '17 at 7:21
  • $\begingroup$ @Quasar note, the following separable equation does involve making the substitution $t=\tan\left(\frac{v}{2}\right)$ $\endgroup$ – Teh Rod Aug 6 '17 at 7:22
  • $\begingroup$ I posted an answer, does the solution look right? $\endgroup$ – Quasar Aug 6 '17 at 8:31

I am getting the following general solution. Could you help me verify, if it looks right?

Let $x+2y=u$.

Then, $1+2\frac{dy}{dx}=\frac{du}{dx}$.

We obtain,

$$\begin{align} 1+2(\sin{u}+\cos{u})=\frac{du}{dx}\\ dx=\frac{du}{1+2(\sin{u}+\cos{u})} \end{align}$$

The second term is a rational function of $\sin{u}$ and $\cos{u}$.

Let us subsititute : $$\tan\frac{u}{2}=t$$. $$\sin u=\frac{2t}{1+t^{2}}$$ $$\cos u=\frac{1-t^{2}}{1+t^{2}}$$ $$du=\frac{2dt}{1+t^{2}}$$

We obtain

$$\begin{align} \frac{\frac{2dt}{1+t^{2}}}{1+\frac{4t}{1+t^{2}}+\frac{2-2t^{2}}{1+t^{2}}}&=dx\\ \frac{2dt}{(1+t^{2})+4t+(2-2t^{2})}&=dx\\ \frac{2dt}{3+4t-t^{2}}&=dx\\ dx-2\frac{dt}{t^{2}-4t-3}&=0\\ dx-2\frac{dt}{(t-2)^{2}-7}&=0\\ \int dx-2\int\frac{dt}{(t-2)^{2}-7}&=c\\ x-\frac{1}{\sqrt{7}}\log\left(\frac{(t-2)-\sqrt{7}}{(t-2)+\sqrt{7}}\right)&=c\\ x-\frac{1}{\sqrt{7}}\log\left(\frac{\tan(u/2)-2-\sqrt{7}}{\tan(u/2)-2+\sqrt{7}}\right)&=c\\ x-\frac{1}{\sqrt{7}}\log\left(\frac{\tan((x+2y)/2)-2-\sqrt{7}}{\tan((x+2y)/2))-2+\sqrt{7}}\right) &=c \end{align}$$

  • 1
    $\begingroup$ You are missing a $\frac{1}{\sqrt{7}}$ in front of your log $\endgroup$ – Teh Rod Aug 6 '17 at 8:36
  • $\begingroup$ Thanks so much, edited the answer! :) $\endgroup$ – Quasar Aug 6 '17 at 8:39

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