3
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ oops. edited now, thanks! :) $\endgroup$
    – Migr
    Nov 9 '13 at 14:19
  • $\begingroup$ 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) $\endgroup$ Nov 9 '13 at 14:20
1
$\begingroup$

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 $\text{lim}\; f(x) = \frac{-1}{22}\cdot (\frac{\pi}{2} - \frac{1}{220})^{-1}$.

$\endgroup$
0
$\begingroup$

Here is how you finish your problem. You are missing the constant of integration

$$ -\frac{1}{y(x)}= \tan^{-1}(x) +c. $$

Now, use the initial condition $y(0)=10$ to find $c$ then you will be able to find the limit.

$\endgroup$
1
  • $\begingroup$ One thing though. The initial condition has nothing to do with the integrand, right? If the lower boundary of the integrand was 1 - the initial condition y(0) = 10 would still be the same, right? $\endgroup$
    – Migr
    Nov 10 '13 at 12:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.