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

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SUMMARY

The discussion focuses on verifying that the expression \( x = \frac{e^t - 1}{e^t - 2} \) is an implicit solution to the first-order differential equation \( \frac{dX}{dt} = (X-1)(1-2X) \) and finding explicit solutions. The user outlines the steps to derive the expression from the logarithmic form \( \ln\left(\frac{2X-1}{X-1}\right) = t \) by exponentiating both sides and performing algebraic manipulations. The verification process involves differentiating \( X \) and substituting back into the original differential equation to confirm the solution.

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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
 
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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:

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