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

1. Aug 31, 2012

### oddjobmj

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

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

3. 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

2. Aug 31, 2012

### Bacle2

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.

3. Aug 31, 2012

### oddjobmj

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: Aug 31, 2012