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Problem: Find all solutions of the differential equation $$ x^2\left( \dfrac{dy}{dx}\right)^2 + 2xy \dfrac{dy}{dx} + y^2 - 1 =0. $$

What are two solutions that pass through the origin?

My Solution: By the perfect square identity, we can write $$ \left( x\dfrac{dy}{dx} + y \right)^2 = 1. $$ Then, $x\dfrac{dy}{dx} + y = \mp 1 \implies \dfrac{dy}{y \mp 1} = -\dfrac{dx}{x} $ and we find the general solutions $$\boxed{y = \mp 1 + \dfrac{C}{x}}$$

But this method doesn't give me the solution at the origin. Because $y = \mp 1 + \dfrac{C}{x}$ is undefined for $x=0$. If $y'(0)< \infty$ then, we can easily see that the equation has no solution for $y(0)=0$. Probably the solutions with $y (0) = 0$ have a tangent perpendicular to the $x$-axis at the origin. That is $y'(0) \to \infty$. Like a function $y=\sqrt{|x|}$ ... Therefore, I tried looking for a solution in the form $y =ax^n$ ($a,n\in \mathbb R$). Unfortunately, I couldn't get the suitable values for $a$ and $n$. There is no solution in the form $y =ax^n$.

How can we find the solutions with $y(0)=0$? Thanks.

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  • $\begingroup$ Actually the solutions are $| y \pm 1| = C/|x|$ for $C > 0$. $\endgroup$
    – Kaind
    Mar 31, 2021 at 21:45
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    $\begingroup$ Are you sure you wrote the problem down correctly? Substituting $0$ for $x$ in the original ODE gives $[y(0)]^2=1$, so any solution defined around $x=0$ must satisfy $y(0)=\pm 1$. $\endgroup$ Mar 31, 2021 at 21:46
  • $\begingroup$ Yes @AlannRosas , I wrote the problem correctly. May be $y' \to \infty $ for $x \to 0$. $\endgroup$
    – scarface
    Mar 31, 2021 at 21:51
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    $\begingroup$ You have that $(yx)'=\pm1 \implies xy=\pm x+c \implies c=0$ from initial condition $y(0)=0$ so that $y=\pm1$ $\endgroup$ Mar 31, 2021 at 22:24
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    $\begingroup$ Hi @Aryadeva , by your solution $(yx)'=\pm1 \implies xy=\pm x+c \implies c=0$, we can find that $y= \mp \text{sgn} (x)$. These functions satisfy the O.P. Also $y= \mp \text{sgn} (0) = 0$. (On the other hand, is there a contradiction to finding a discontinuous solution for the equation, I am not sure.) I think that the solutions with $y(0)=0$ are $y= \mp \text{sgn} (x)$ Thanks for your effort. $\endgroup$
    – scarface
    Mar 31, 2021 at 22:55

2 Answers 2

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This initial value problem has no solution. If $y$ was any solution defined on an interval containing $0$, then the differential equation would imply that $[y(0)]^2=1$, showing that $y(0)$ is necessarily nonzero.

You mentioned in the comments that a solution with $y(0)=0$ might have an unbounded derivative near $0$. Such a function isn't really a "solution" to the ODE in the ordinary sense of the word, as a solution (by definition) must satisfy

$$x^2[y'(x)]^2+2xy(x)y'(x)+[y(x)]^2-1=0, y(0)=0$$

for all $x$ in some open interval containing $0$. In particular, the solution must be differentiable at $0$, so if $y'(x)\to\infty$ as $x\to0^{\pm}$, it clearly can't be a solution.

Thus, it seems like what you're really after is a continuous function satisfying $y(0)=0$ and $x^2[y'(x)]^2+2xy(x)y'(x)+[y(x)]^2-1=0$ for all nonzero $x$; such a function doesn't exist either. You found that the original ODE can be rewritten as $(xy'+y)^2=1$ (kudos to you for that!), so $xy'+y=\pm 1$. This further simplifies to $\frac{d}{dx}(xy)=\pm 1$, leaving us with $xy=c\pm x$ for some constant $c$. For $y(0)=0$, $c$ is necessarily $0$, so any solution to the ODE satisfying the initial condition must satisfy $y(x)=\pm 1$ on any interval not containing $0$. This immediately implies that $\lim_{x\to 0^{\pm}}y(x)\neq 0$, so $y$ can't be continuous.

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  • $\begingroup$ Your explanation is very clear to me. Thanks for sharing your time. I wish you a good day. $\endgroup$
    – scarface
    Mar 31, 2021 at 23:29
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Notice that the left-hand-side of $xy'+y=\pm1$ is of the form

$$ xy'(x)+y(x)=(xy(x))' $$ and so, integrating both sides of the differential equation over an interval, say $[0,x]$, gives

$$ \int^x_0 (ty(t))'\,dt=xy(x)-0\cdot y(0)=\pm\int^x_0\,ds=\pm x $$

This means that $y(x)=\pm1$ are two (and the only ones) solutions to the differential equation, none of which satisfy the initial condition $y(0)=0$. So there is no solution to the initial value problem

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