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

Question

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...

Answer 1

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

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

Answer 2

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 $y_0=y(1)=-2$.

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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 $y_0+2\ne 0$.