How to find the initial value of a solution to a differential equation in order to comply with certain limits

For the IVP, $$y'+\frac{2x^2-4xy-y^2}{3x^2}=0, x>0, y(1)=y_0$$

I got this, once all the calculations have been done: $$y'=\frac{y_0^2(-2x^2+4x+1)+4y_0(x^2+x+1)-2(x^2+4x-2)}{(2+y_0-x(y_0-1))^2}$$

I am confident with this calculation, and I also verified with Wolfram Alpha. Then the question asked there is exactly one value of $$y_0$$ such that the IVP satisfied $$\lim_{x\to 0}y'(x)\neq 1$$, while $$\lim_ {x\to0}y'(x)=1$$ for all other values of $$y_0$$. What is this value of $$y_0$$ corresponding to the different limits?

So I took the limit $$\lim_{x\to0}y'(x)=\frac{y_0^2+4y_0+4}{(2+y_0)^2}=1$$

So no matter what value of $$y_0$$ I have, the limit will always be 1. Where did I misunderstand or did wrong? Any help will be great, stuck about 2 days...

Observe that for $$y_0=-2$$ the solution $$y$$ of the IVP has the property that $$y'$$ is constant:

$$y'(x)=-2.$$

• @Lutz: thanks for your comment. I edited my solution. – Fred Sep 29 at 8:23

This is, among other things, a Riccati equation. Set thus $$y=-3x^2\frac{u'}{u}$$ to find $$0=y'+\frac{2x^2-4xy-y^2}{3x^2} =\left[-3x^2\frac{u''}{u}+3x^2\frac{u'^2}{u^2}-6x\frac{u'}{u}\right] +\left[\frac23+4x\frac{u'}{u}-3x^2\frac{u'^2}{u^2}\right] \\~\\ 0=3x^2u''+2xu'-\frac23u$$ This now is an Euler-Cauchy DE with characteristic polynomial $$0=3m(m-1)+2m-\tfrac23=\tfrac13((3m-\tfrac12)^2-2-\tfrac14)=\tfrac13(3m-2)(3m+1)$$ So \begin{align} u&=Ax^{-1/3}+Bx^{2/3},~~~|A|+|B|\ne 0 \\~\\ \implies y(x)&=-3x^2\frac{-\frac13Ax^{-4/2}+\frac23Bx^{-1/3}}{Ax^{-1/3}+Bx^{2/3}} \\ &=x\frac{A-2Bx}{A+Bx}\\&=x-\frac{3Bx^2}{A+Bx} \\ y'(x)&=1-\frac{6ABx+3B^2x^2}{(A+Bx)^2} \end{align} For $$A\ne 0$$ the solution has slope 1 in $$x=0$$, only for $$A=0$$ one gets a deviating result with $$y(x)=-2x$$, $$y'(0)=-2$$. Then $$y_0=y(1)=-2$$.

You excluded this case from consideration in the limit $$\lim_{x\to 0}y'(x)$$, but did not treat this case separately. The quotient law in this limit can only be applied for $$y_0+2\ne 0$$.

• This DE could as well be solved as a homogeneous DE (this is probably what is implied by your "Among other things"). For the OP: mathworld.wolfram.com/… – Jean Marie Sep 29 at 6:46
• Yes, I guessed that this was the way OP got to his solution, as the formulation of the DE is very suggestive in that direction. Guessing $y(x)=x$ as a solution and transforming via the Bernoulli equation for $u(x)=y(x)-x$ seems less likely. – Lutz Lehmann Sep 29 at 6:55
• Thank you very much for your time. – Will Sep 29 at 7:50