Differential Equations Verifying Implicit Solution

In summary, we are given two equations and asked to verify that they are an implicit solution to a first order differential equation. After breaking down the second equation into two separate logarithmic equations, we are able to differentiate it implicitly and solve for dx/dt, successfully verifying that it is an implicit solution to the given differential equation.
  • #1
Lancelot59
646
1
I'm given:

1. [tex]\frac{dX}{dt}=(X-1)(1-2X)[/tex]
2. [tex]ln(\frac{2X-1}{X-1})=t[/tex]

and asked to verify that it is an implicit solution to the first order DE given.

I successfully derived the second equation there to get:

[tex]\frac{dX}{dt}=\frac{-1}{(2X-1)(X-1)}[/tex]

So now what? I tried several things and got nowhere.
 
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  • #2
Lancelot59 said:
I'm given:

1. [tex]\frac{dX}{dt}=(X-1)(1-2X)[/tex]
2. [tex]ln(\frac{2X-1}{X-1})=t[/tex]

and asked to verify that it is an implicit solution to the first order DE given.

I successfully derived the second equation there to get:

[tex]\frac{dX}{dt}=\frac{-1}{(2X-1)(X-1)}[/tex]

So now what? I tried several things and got nowhere.

Show us the work you did when you differentiated (not derived) the second equation.

Note that
[tex]ln(\frac{2X-1}{X-1})= ln(2X - 1) - ln(X - 1)[/tex]

so that should make differentiation a little easier.
 
  • #3
Mark44 said:
Show us the work you did when you differentiated (not derived) the second equation.

Note that
[tex]ln(\frac{2X-1}{X-1})= ln(2X - 1) - ln(X - 1)[/tex]

so that should make differentiation a little easier.

I did break it up initially:
[tex]\frac{dx}{dt}=\frac{d}{dt}( ln(2x-1) - ln(x-1))[/tex]
[tex]\frac{dx}{dt}=\frac{1}{2x-1}(2) - \frac{1}{x-1}(1)=\frac{2}{2x-1} - \frac{1}{x-1}[/tex]
Then subtracting the fractions:
[tex]\frac{dx}{dt}= \frac{2(x-1) - (2x-1)}{(2x-1)(x-1)}=\frac{-1}{(2x-1)(x-1)}[/tex]
 
  • #4
Lancelot59 said:
I did break it up initially:
[tex]\frac{dx}{dt}=\frac{d}{dt}( ln(2x-1) - ln(x-1))[/tex]
[tex]\frac{dx}{dt}=\frac{1}{2x-1}(2) - \frac{1}{x-1}(1)=\frac{2}{2x-1} - \frac{1}{x-1}[/tex]
Then subtracting the fractions:
[tex]\frac{dx}{dt}= \frac{2(x-1) - (2x-1)}{(2x-1)(x-1)}=\frac{-1}{(2x-1)(x-1)}[/tex]

Starting from your second equation, you have
[tex]t= ln(2x-1) - ln(x-1)[/tex]
[tex]\Rightarrow \frac{dt}{dt}=\frac{d}{dt}( ln(2x-1) - ln(x-1))[/tex]

You have to differentiate the right side implicitly, and then solve algebraically for dx/dt. Can you take it from here?
 
  • #5
I solved it! Thanks for the tip.
 

Related to Differential Equations Verifying Implicit Solution

1. What are differential equations?

Differential equations are mathematical equations that describe how a quantity changes over time, based on its rate of change and other variables. They are commonly used in physics, engineering, and other scientific fields to model and predict real-world phenomena.

2. What is an implicit solution to a differential equation?

An implicit solution to a differential equation is a general solution that is expressed in terms of both the independent variable and the dependent variable. It is often written in the form of an equation, rather than a specific function, and can be found by manipulating the original differential equation.

3. How do you verify an implicit solution to a differential equation?

To verify an implicit solution, you can substitute it into the original differential equation and see if it satisfies the equation for all values of the independent and dependent variables. If it does, then the solution is valid.

4. What is the difference between an explicit and implicit solution?

An explicit solution to a differential equation is a specific function that can be written in terms of the independent variable only. It is usually easier to work with and can provide more detailed information about the behavior of the system. In contrast, an implicit solution is more general and can be harder to interpret.

5. Can differential equations have more than one implicit solution?

Yes, differential equations can have multiple implicit solutions. This is because the original equation can often be manipulated in different ways to find different solutions. However, only one of these solutions may accurately model the real-world phenomenon being studied.

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