# An IVP with No Solutions but also a Unique Solution

If I try to solve the ODE $y'=y\sqrt {\vert x \vert}, y(0)=0$, I get $$\frac{y'}{y}=\sqrt{\vert x \vert}\\ \Rightarrow \log\vert y\vert=\int_0 ^x\sqrt{\vert x \vert}\\\Rightarrow y=\pm e^{\int_0 ^x\sqrt{\vert x \vert}},$$ but for neither of these solutions $y$ do we have $y(0)=0$. So there's no solution.

On the other hand, Picard's Theorem on Existence and Uniqueness applies: since $y\sqrt{\vert x \vert}$ is continuous near $(0,0)$ and since $\frac{\partial}{\partial y} y\sqrt{\vert x \vert}=\sqrt{\vert x \vert}$ is continuous near $(0,0)$, there exists a unique solution.

Why wasn't I able to find it using separable variables?

• When you divide by $y$, you're kicking out the potential solution of $y(x)=0$, which turns out to be the true solution in this case. – πr8 Jan 9 '16 at 20:18
• Mathematica gives a very Long solution – Dr. Sonnhard Graubner Jan 9 '16 at 20:37
• Despite what the accepted answer says, the trouble is not that you divide by $y$ since this step can be circumvented easily as follows. Consider $z(x)=\exp(2x\sqrt{|x|}/3)$, then $z(x)\ne0$ for every $x$ and $z'(x)=z(x)\sqrt{|x|}$ hence the differential equation is equivalent to $y'(x)z(x)=y(x)\sqrt{|x|}z(x)=y(x)z'(x)$, that is, $(y(x)/z(x))'=0$, that is, $y(x)/z(x)=c$ for some constant $c$. Thus, without any division by zero, one sees that the solutions to the differential equation are exactly $y(x)=cz(x)$, for some constant $c$ (which is easily seen to be $c=y(0)$ ... – Did Jan 10 '16 at 12:18
• ... since $z(0)=1$). To sum up, the trouble is not that solving the equation kicked out a solution but that you solved incorrectly the differential equation, whose solutions are $y(x)=cz(x)$, not $y(x)=\pm z(x)$. (User @πr8 more or less points at this in a comment.) – Did Jan 10 '16 at 12:18

The problem is that at 0 the expression $\frac{y'}{y}$ is meaningless, therefore you cannot integrate it. If you had $y(0) \neq 0$ you would have been able to roceed and find a solution (maybe for a small time interval). By the way, $y(x) = 0$ is a soluion, and, by Picard's theorem, it must be the unique solution.
• Can also note that if we do assume $y(0)=p\neq 0$, then we arrive at the solution $y=p e^{\int_0 ^x\sqrt{\vert x \vert}dx}$. If we then take $p\to0$, we retrieve the $y(x)=0$ solution, as we'd hope. – πr8 Jan 9 '16 at 20:23