SOLUTION TO PROBLEM 10

This solution is by Andrew Gacek

Let f(x) be the solution to

$$\frac{dy}{dx}=x^2+y^2,\qquad\qquad y(0)=1$$

and let g(x) be the solution to

$$\frac{dy}{dx}=y^2,\qquad\qquad y(0)=1$$

We can solve for g(x) using separation of variables.

$$\begin{array}{c} \displaystyle \int\frac{dy}{y^2}=\int dx\\ ... ...c{1}{y}=x+C\\ \ \\ \displaystyle y=\frac{1}{-C-x}.\end{array}$$

Since $y(0)=\frac{1}{-C}=1$, we have C=-1 and $g(x)=\frac{1}{1-x}$. We can apply the Race Track Principle to f(x) and g(x) since f(0)=g(0) and $f'(x)\geq g'(x)$ for all x>0. This yields $f(x)\geq g(x)$ for all x>0. Thus

$$f(x)\geq\frac{1}{1-x}$$

and as $\lim\limits_{x\to1^-}\frac{1}{1-x}=\infty$ we may conclude that $\lim\limits_{x\to 1^-}f(x)=\infty$.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Nov 2 14:32:08 CST 2001