# What am I doing wrong? Integration and limits

I need some help identifying what I'm doing wrong here..

What is the limit of $$y(x)$$ when $$x→∞$$ if $$y$$ is given by:

$$y(x) = 10 + \int_0^x \frac{22(y(t))^2}{1 + t^2}\,dt$$ What i've done:

1) Integrating on both sides(and using the Fundamental Theorem of Calculus):

$$\frac{dy}{dx} = 0 + \frac{22(y(x))^2}{1 + x^2}$$

2) $$\frac{-1}{22y} = \arctan x$$ And after moving around and stuff I end up with the answer: $$\quad\dfrac{-1}{11 \pi}.$$

What's wrong?

• oops. edited now, thanks! :)
– Migr
Nov 9 '13 at 14:19
• As David said, it looks like you want 14 and not 22. But another point of concern is that it should be $\frac{-1}{22y} + \frac{1}{22\cdot 10} = \text{atan}(x)$. (Remember that you are integrating from 0 to t and y(0) = 10) Nov 9 '13 at 14:20

To add my comment above, the answer should be $\frac{-1}{22y} + \frac{1}{22\cdot 10} = \text{atan}(x)$. This reduces to $y(x) = \frac{-1}{22}\cdot (\text{atan}(x) - \frac{1}{220})^{-1}$. On a rigorous level, we know that this holds since $y(x)\geq 10\; \forall\; x$. Now $\text{lim}\; f(x) = \frac{-1}{22}\cdot (\frac{\pi}{2} - \frac{1}{220})^{-1}$.
$$-\frac{1}{y(x)}= \tan^{-1}(x) +c.$$
Now, use the initial condition $y(0)=10$ to find $c$ then you will be able to find the limit.