I want to find the general solution of the following:

$$\frac{dy}{dx} = \frac{y}{x+y+2}$$

I substitute $u = x + y+2$ and got the following $du/dx = 1 + dy/dx$.

Then I got an expression as such: $$\frac{du}{dx} = 2 - \frac{x+2}{u}$$

But from this expression, I cannot seem to separate it further. Using an integrating factor also doesn't seem to work. It also doesn't have the form of a bernoulli's equation, due to the $+2$.

How should I further proceed?

  • $\begingroup$ it is an equation of Abel-type $\endgroup$ – Dr. Sonnhard Graubner Feb 12 '18 at 15:17
  • $\begingroup$ I did a check online to see what an Abel equation is. Unfortunately I've not learnt this type of equation before, so I couldn't follow. Could this problem be solved with a simpler set of tools? $\endgroup$ – mathworker123 Feb 12 '18 at 15:19
  • $\begingroup$ see here encyclopediaofmath.org/index.php/Abel_differential_equation $\endgroup$ – Dr. Sonnhard Graubner Feb 12 '18 at 15:19
  • $\begingroup$ and how can i help you? $\endgroup$ – Dr. Sonnhard Graubner Feb 12 '18 at 15:24
  • $\begingroup$ Thanks for the reference, this does look like an interesting form of equation to learn. What does it mean that I have to normalize the Abel equation according to the reference you gave? Also, according to the form for such an equation, I have $f(x,y)y$ instead of $f(x)y$ on the right hand side. Is this still alright? $\endgroup$ – mathworker123 Feb 12 '18 at 15:27

Hint...you can write the differential equation upside down i.e. $$\frac {dx}{dy}=\frac xy+1+\frac 2y$$

Now use an integrating factor since you have $$\frac{dx}{dy}-\frac 1yx=1+\frac 2y$$

  • $\begingroup$ That's a very interesting way to look at things. I've learned a new way of looking at ODEs -- flipping the equations might make it so much easier to solve at times. $\endgroup$ – mathworker123 Feb 12 '18 at 15:52
  • $\begingroup$ Glad you like it! $\endgroup$ – David Quinn Feb 12 '18 at 16:27

You can work it through as transforming the given ODE to an exact equation :

$$\frac{dy}{dx} = \frac{y}{x+y+2} \Rightarrow y'(x)(x+y+2) -y = 0$$

If you let $R(x,y) = -y$ and $S(x,y) = x+y+2$, then you can see that this is not an exact equation, since :

$$\frac{\partial R}{\partial y} \neq \frac{\partial S}{\partial x}$$

Let's find an integrating factor $μ(y)$, such that the ODE :

$$\frac{\partial μ R}{\partial y} + \frac{\partial μS}{\partial x} = 0$$

is exact.

This means, that :

$$\frac{\partial μR}{\partial y} = \frac{\partial μS}{\partial x}\Rightarrow-\bigg(y\frac{dμ(y)}{dy}\bigg)-μ(y) =μ(y) \Rightarrow \frac{\frac{\partial μ(y)}{\partial y}}{μ(y)} = -\frac{2}{y} \Rightarrow \ln(μ(y))=-2\ln(y) \Rightarrow μ(y) = \frac{1}{y^2}$$

So, now multiply both sides of the initial expression by $μ(y)$ :

$$-\frac{1}{y(x)} + \frac{(x+y(x)+2)\frac{dy(x)}{dx}}{y^2(x)}=0 \Rightarrow \frac{dg(y)}{dy} = \frac{y+2}{y^2} \Rightarrow g(y) = \int \frac{y+2}{y^2}dy = -\frac{2}{y} +\ln(y)$$

And finally, solving for $y$, will yield a not that common result, as the Product-log function will take part :

$$y(x) = \frac{x+2}{W\bigg(\frac{x+2}{e^{c_1}}\bigg)} \Rightarrow y(x) = \frac{x+2}{W(c_1(x+2))}$$

Of course, if you are not familiar with the $W$ function, you may as well stop at the integration step, which is still considered a solution to the given ODE and is of course 100% correct.

Sampling a bit over initial values $y(0)$ for some trajectories, will yield you the following trajectory spectre for the family of the solutions :

$\quad \quad \quad \quad \quad \quad \quad \quad$enter image description here

  • $\begingroup$ That's a very detailed post. Learned tremendously. $\endgroup$ – mathworker123 Feb 12 '18 at 15:50
  • $\begingroup$ yes very nice and complete answer Rebellos +1 $\endgroup$ – MtGlasser Feb 12 '18 at 16:30
  • $\begingroup$ Thank you both ! Glad I helped ! $\endgroup$ – Rebellos Feb 12 '18 at 16:55

To solve $$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{y}{x+y+2} $$ we can substitute $u=x+2$: $$ \frac{\mathrm{d}y}{\mathrm{d}u}=\frac{y}{u+y} $$ Then simple manipulation gives $y\,\mathrm{d}u-u\,\mathrm{d}y=y\,\mathrm{d}y$; therefore, $$ \begin{align} \mathrm{d}\frac uy &=\frac{y\,\mathrm{d}u-u\,\mathrm{d}y}{y^2}\\ &=\frac{\mathrm{d}y}y\\[6pt] &=\mathrm{d}\log(y) \end{align} $$ Solving yields $$ x+2=y(c+\log(y)) $$


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