1. ## Sep. by Var.

I have another seperation by variables problem, and the hardest part for me is separating the variables which is probably why we have this lesson.

This one seems complicated..I have to solve this dif eq by using sep. by variables:

y dy = t(1 + t^2)^(-1/2) * (1 + y^2)^(1/2) dt

The square root in the denominator for t and the square root in the numerator for y seems hard to separate them.

2. Originally Posted by Ideasman
I have another seperation by variables problem, and the hardest part for me is separating the variables which is probably why we have this lesson.

This one seems complicated..I have to solve this dif eq by using sep. by variables:

y dy = t(1 + t^2)^(-1/2) * (1 + y^2)^(1/2) dt

The square root in the denominator for t and the square root in the numerator for y seems hard to separate them.
$y~dy = \frac{t}{\sqrt{1 + t^2}} \cdot \sqrt{1 + y^2}~dt$

$\frac{y}{\sqrt{1 + y^2}}~dy = \frac{t}{\sqrt{1 + t^2}}~dt$

-Dan

3. Originally Posted by topsquark
$y~dy = \frac{t}{\sqrt{1 + t^2}} \cdot \sqrt{1 + y^2}~dt$

$\frac{y}{\sqrt{1 + y^2}}~dy = \frac{t}{\sqrt{1 + t^2}}~dt$

-Dan
Er duh! Thanks topsquark.

One final question:

Taking the integral of both sides yields:

sqrt(y^2 + 1) = sqrt(t^2 + 1) + C. Now is THIS the solution or can we infer that y and t must be equal and say more?

4. Originally Posted by Ideasman
Er duh! Thanks topsquark.

One final question:

Taking the integral of both sides yields:

sqrt(y^2 + 1) = sqrt(t^2 + 1) + C. Now is THIS the solution or can we infer that y and t must be equal and say more?
y = t only if C = 0

5. Originally Posted by Jhevon
y = t only if C = 0
Well, technically there is a "+ C" on both sides, but you can "group them up" and that's why my prof. says to just keep it on one side. So what if C was -5 on one side and 5 on the other. For simplicity, is my sol'n just sqrt(y^2 + 1) = sqrt(t^2 + 1) + C ?

6. Originally Posted by Ideasman
Well, technically there is a "+ C" on both sides, but you can "group them up" and that's why my prof. says to just keep it on one side. So what if C was -5 on one side and 5 on the other. For simplicity, is my sol'n just sqrt(y^2 + 1) = sqrt(t^2 + 1) + C ?
yes, but there is no guarantee that the C's on both sides are the same. i say play it safe. we cannot know for sure unless we were given some initial conditions