Verify an expression is an implicit solution to a first order DE.

Click For Summary
The discussion centers on verifying that a given expression is an implicit solution to a first-order differential equation and finding an explicit solution. The equation provided is ln((2X-1)/(X-1))=t, which is transformed into X=(e^t-1)/(e^t-2) through exponentiation and algebraic manipulation. Participants express confusion about the steps involved in deriving this explicit solution from the implicit form. The verification process includes differentiating X and substituting back into the original differential equation to confirm it satisfies the equation. The back of the book confirms that the derived expression is indeed a solution, prompting further exploration of its validity.
oddjobmj
Messages
305
Reaction score
0

Homework Statement


Verify that the indicated expression is an implicit solution of the given first-order differential equation. Find at least one explicit solution y=∅(x) in each case.

Homework Equations



\frac{dX}{dt}=(X-1)(1-2X); ln(\frac{2X-1}{X-1})=t

The Attempt at a Solution


I know I need to solve the 't=' portion for X and here it is:

ln(\frac{2X-1}{X-1})=t =>

\frac{e^t-1}{e^t-2}=X

In -explicit- cases I would differentiate X to find X' and replace the original X' from the given differential equation to see if it works out to 0.

I don't know how to go about proving it in the implicit case and everything I'm finding when I google is how to do implicit differentiation from a calc 2-3 perspective.

Thank you
 
Physics news on Phys.org
I don't see how you go from:

ln ( (2X-1)/(X-1))=t to (et-1)/(et -2) =X


I would think of raising e to both sides.
 
I don't see how you go from:

ln ( (2X-1)/(X-1))=t to (et-1)/(et -2) =X I would think of raising e to both sides.

Exactly. When you do that the ln goes away and you end up with e^t on the other side. We do some algebra and end up with the above result. I have verified this with wolframalpha.

Edit: Here are the steps for the curious-

Exponentiate both sides to get: (2X-1)/(X-1)=e^t
Multiply both sides by the denominator (x-1) to get: 2x-1=xe^t-e^t
Get the components with x's in them together: xe^t-2x=e^t-1
Factor out the x: x(e^t-2)=e^t-1
Divide both sides by (e^t-2) to get x by itself: x=(e^t-1)/(e^t-2)

Edit 2:
The back of the book says that x=(e^t-1)/(e^t-2) is a solution. I have to figure out why that's the case.
 
Last edited:
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
2K
Replies
8
Views
1K
Replies
2
Views
1K