Differential Equations Question of the Day

1. $y' = \sqrt{x^2+y^2}, \; y(0)=0$ (correct answer)
2. $y' = x^2 + \arctan(y), \; y(1)=\pi$
3. $(y-1)y' = x, \; y(0)=0$
4. $y' = y \ln(x+1), \; y(1)=2$

Explanation: We analyze each case by checking the continuity of $f(x,y)$ and $\frac{\partial f}{\partial y}$ at the initial point $(x_0, y_0)$. A) $f(x,y) = \sqrt{x^2+y^2}$. This is continuous everywhere. However, $\frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2+y^2}}$, which is undefined and thus not continuous at the initial point $(0,0)$. The theorem does not guarantee uniqueness. B) $f(x,y) = x^2 + \arctan(y)$ and $\frac{\partial f}{\partial y} = \frac{1}{1+y^2}$ are both continuous everywhere. A unique solution is guaranteed. C) Rewriting gives $y' = \frac{x}{y-1}$. Here $f(x,y)=\frac{x}{y-1}$ and $\frac{\partial f}{\partial y} = -\frac{x}{(y-1)^2}$. At the initial point $(0,0)$, both are continuous. A unique solution is guaranteed. D) $f(x,y) = y \ln(x+1)$ and $\frac{\partial f}{\partial y} = \ln(x+1)$ are both continuous in a neighborhood of $(1,2)$ (specifically for $x>-1$). A unique solution is guaranteed.