# Solutions of $x^2\left( \dfrac{dy}{dx}\right)^2 + 2xy \dfrac{dy}{dx} + y^2 - 1 =0$, $y(0)=0$

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.

• Actually the solutions are $| y \pm 1| = C/|x|$ for $C > 0$. Mar 31, 2021 at 21:45
• 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 $y(0)=\pm 1$. Mar 31, 2021 at 21:46
• Yes @AlannRosas , I wrote the problem correctly. May be $y' \to \infty$ for $x \to 0$. Mar 31, 2021 at 21:51
• You have that $(yx)'=\pm1 \implies xy=\pm x+c \implies c=0$ from initial condition $y(0)=0$ so that $y=\pm1$ Mar 31, 2021 at 22:24
• 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. Mar 31, 2021 at 22:55

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.

• Your explanation is very clear to me. Thanks for sharing your time. I wish you a good day. Mar 31, 2021 at 23:29

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