Differential Equations Verifying Implicit Solution

[Lancelot59]

I'm given:

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

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:

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

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

 

[Mark44]

I'm given:

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

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:

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

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

so that should make differentiation a little easier.

 

[Lancelot59]

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

Note that
$$ln(\frac{2X-1}{X-1})= ln(2X - 1) - ln(X - 1)$$

so that should make differentiation a little easier.

I did break it up initially:
$$\frac{dx}{dt}=\frac{d}{dt}( ln(2x-1) - ln(x-1))$$
$$\frac{dx}{dt}=\frac{1}{2x-1}(2) - \frac{1}{x-1}(1)=\frac{2}{2x-1} - \frac{1}{x-1}$$
Then subtracting the fractions:
$$\frac{dx}{dt}= \frac{2(x-1) - (2x-1)}{(2x-1)(x-1)}=\frac{-1}{(2x-1)(x-1)}$$

 

[Mark44]

I did break it up initially:
$$\frac{dx}{dt}=\frac{d}{dt}( ln(2x-1) - ln(x-1))$$
$$\frac{dx}{dt}=\frac{1}{2x-1}(2) - \frac{1}{x-1}(1)=\frac{2}{2x-1} - \frac{1}{x-1}$$
Then subtracting the fractions:
$$\frac{dx}{dt}= \frac{2(x-1) - (2x-1)}{(2x-1)(x-1)}=\frac{-1}{(2x-1)(x-1)}$$

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

You have to differentiate the right side implicitly, and then solve algebraically for dx/dt. Can you take it from here?

 

[Lancelot59]

I solved it! Thanks for the tip.