0

On p. 11, the last paragraph says: Confirm by direct integration that $x(t)=1/(1+t)$ is indeed a root of the equation $$ x(t)=1-\int_0^t [x(u)]^2 \mathrm{d}u. $$ I am a little confused on how to carry this out.

Update: thought I'd carry it out explicitly for those that come later:

$$ \begin{align*} x(t)&=1-\int_0^t [x(u)]^2\mathrm{d}u\\ \frac{1}{1+t}&=1-\int_0^t \frac{1}{1+u}^2\mathrm{d}u\\ \frac{1}{1+t}&=\frac{1}{1+t}\quad\text{(integration by substitution steps left out for brevity)} \end{align*} $$

Joe
  • 489

2 Answers2

2

What is

$$ \int_0^t \frac{1}{(1+u)^2} \mathrm{d}u $$

??????????????

Will Jagy
  • 139,541
  • 1
    Thank you, this helped me. When I carried out the integration I saw that LHS and RHS are equal. – Joe Aug 15 '15 at 17:05
1

$x(u) = \frac{1}{1+u}$ and $x(t) = \frac{1}{1+t}$, you just check if these satisfy the equation.

  • And of course you find that they do. –  Aug 15 '15 at 16:54
  • Thanks for your reply, but how do I check that they they satisfy the equation? – Joe Aug 15 '15 at 16:55
  • By replacing $x(u)$'s expression and integrating, then comparing to $x(t)$. –  Aug 15 '15 at 16:57
  • You literally just take these two expressions and put them in the equation, then verify that the LHS = RHS. The LHS is ready, but you have to do some calculation in the RHS. –  Aug 15 '15 at 16:58
  • Ah, the simple things I forget when I don't do calculus for too long... – Joe Aug 15 '15 at 17:34