# differential equation

• Nov 12th 2006, 07:39 PM
viet
differential equation
a) Find the general solution to the differential equation:
(Write the answer in a form such that its numerator is 1 and its integration constant is C)
$\displaystyle \frac{du}{dt} = u^3(t^3 - t)$

b) Find the particular solution of the above differential equation that satisfies the condition $\displaystyle u = 4$ at $\displaystyle t = 0$
• Nov 12th 2006, 07:49 PM
Soltras
Quote:

Originally Posted by viet
a) Find the general solution to the differential equation:
(Write the answer in a form such that its numerator is 1 and its integration constant is C)
$\displaystyle \frac{du}{dt} = u^3(t^3 - t)$

b) Find the particular solution of the above differential equation that satisfies the condition $\displaystyle u = 4$ at $\displaystyle t = 0$

Solve by separation of variables. Move the $\displaystyle u^3$ and the $\displaystyle dt$ to the other sides:

Integrate:

$\displaystyle \int u^{-3}\,du = \int t^3 - t\,dt$

Get:

$\displaystyle -\frac{1}{2} u^{-2} = \frac{1}{4} t^4 - \frac{1}{2} t^2 + C$

Solve for u (conveniently, we are allowed to leave in the form where 1 is in the denominator):

$\displaystyle u = $$\displaystyle \frac{1}{\sqrt{t^2 - \frac{1}{2}t^4 - 2C}} That is the general solution. Now, we know that \displaystyle 4 = \frac{1}{\sqrt{0^2 - \frac{1}{2}0^4 - 2C}} = \frac{1}{\sqrt{-2C}} Solve for C and substitute your result back into the general solution to get the particular solution. • Nov 13th 2006, 03:26 AM ThePerfectHacker Quote: Originally Posted by Soltras That is the general solution. . What about \displaystyle u=0:eek: • Nov 13th 2006, 05:55 AM Soltras Quote: Originally Posted by ThePerfectHacker What about \displaystyle u=0:eek: Oh yeah, and the equilibrium solution \displaystyle u = 0, although looking ahead that wouldn't have satisfied the initial condition. Still, by dividing by \displaystyle u^3, I needed to say I was temporarily assuming it wasn't 0, which I completely forgot all about. Thanks, Hacker. • Nov 13th 2006, 07:31 AM ThePerfectHacker Quote: Originally Posted by Soltras Oh yeah, and the equilibrium solution \displaystyle u = 0, although looking ahead that wouldn't have satisfied the initial condition. Still, by dividing by \displaystyle u^3, I needed to say I was temporarily assuming it wasn't 0, which I completely forgot all about. Thanks, Hacker. I purposely did that. My point was that when people solve differencial equations they do not consider intervals where the function is somewhere zero. I made a post about this some time ago. What turns out are 3 possible cases (though I did not prove it completely what I am about to say I think is true). 1)The function is nowhere zero, thus division by the function is valid and produces this solution (your solution) on this open differenciable interval. 2)The function is somewhere zero, then if there is a point on the open differenciable interval where the function is not zero that leads to an non-differenciable point. A contradiction to the hypothesis that such a function exists throughout the interval. Thus, having a case where is function is zero at one point and non-zero at some other point is not possible. 3)The function is always zero. Unlike in case #2 where we can pick a non-zero point and lead ourselves to a contradiction here such a point does not exist. And thus the trivial zero function solution is indeed a solution. Does that make sense. You argree that those are the possible 3 cases. • Nov 13th 2006, 07:11 PM thedoge I'm still confused on this problem. ---> Find the general solution to the above differential equation. (Instruction: Write the answer in a form such that its numerator is 1 and its integration constant is C -- rename your constant if necessary. • Nov 13th 2006, 07:16 PM viet yah... i was able to get the first part of the question, but not the second • Nov 13th 2006, 07:27 PM thedoge What'd you end up for on the first part? I'm stuck on how to deal with these diff equations. • Nov 13th 2006, 07:36 PM viet \displaystyle u =$$\displaystyle \frac{1}{\sqrt{- \frac{t}{2}^4 + t^2 + C}}$
• Nov 13th 2006, 07:41 PM
viet
nvm i got it the second part :)
• Nov 13th 2006, 07:42 PM
killasnake
Yup I am also having problems getting the second part of these problems. But the first part is 1/((sqrt((-t^4/2)+t^2+C))) You wouldn't to happen to have the answer for 13 or 14 would you?
• Nov 13th 2006, 07:54 PM
thedoge
Other than the last one, those are the ones I don't have.

but for #14
-->the middle 4 are
0, 1000
10, 1700

*year: 2014