Implicit Solution To Differential Equation

[Bashyboy]

Homework Statement

Verify that the indicated expression is an implicit solution of the given first order differential equation. Find at least one explicit solution in each case. Give an interval I of definition of each solution.

The differential equation is: $\displaystyle \frac{dX}{dt} = (X -1)(1-2X)$

and the solution is $\displaystyle \ln \left( \frac{2X-1}{X-1} \right) = t$

Homework Equations

The Attempt at a Solution

Implicitly differentiating gives

$\displaystyle \ln(2X -1) - \ln(X-1) = t$

$\displaystyle \frac{\dot{X}}{2X-1} - \frac{\dot{X}}{X-1} = 1$

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

$\displaystyle \frac{\dot{X}(2X-1-2X +1)}{(2X-1)(X-1)}$

$\displaystyle \frac{\dot{X} \cdot 0}{(2X-1)(X-1)} = 1$

What happened? What did I do wrong? According to this link http://rmower.com/s_diff_eq/Examples/0101p2.pdf I am incorrect.

 

[DrClaude]

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

$\displaystyle \frac{\dot{X}(2X-1-2X +1)}{(2X-1)(X-1)}$

You error is in this step. You didn't distribute the factor of 2 correctly.

What happened? What did I do wrong? According to this link http://rmower.com/s_diff_eq/Examples/0101p2.pdf I am incorrect.

In that link, I don't understand how they can go from
$$
\frac{2 dX}{2X-1} - \frac{dX}{X-1} = dt
$$
to
$$
2dX - dX = (2X-1)(X-1) dt
$$

 

[vanceEE]

Implicitly differentiating gives

$\displaystyle \ln(2X -1) - \ln(X-1) = t$

$\displaystyle \frac{\dot{X}}{2X-1} - \frac{\dot{X}}{X-1} = 1$

Try leaving $$ ln(\frac{1-2X}{X-1}) $$ as a a single logarithm and implicitly differentiating by the quotient rule. You should end up with
$$ 1 = (\frac{(-2X+2)-(1-2X)}{(1-2X)(X-1)})\frac{dX}{dt} $$
where you can simplify to prove that $$ t = ln(\frac{1-2X}{X-1})$$ is a particular solution where C = 0

 

[vanceEE]

You error is in this step. You didn't distribute the factor of 2 correctly.
In that link, I don't understand how they can go from
$$
\frac{2 dX}{2X-1} - \frac{dX}{X-1} = dt
$$
to
$$
2dX - dX = (2X-1)(X-1) dt
$$

$$\frac{-2dX}{1-2X} - \frac{dX}{X-1} \equiv \frac{dX}{(1-2X)(X-1)}$$
by partial fractions.

Therefore, by multiplying by $$(1-2X)(X-1)$$ we're left with
$$-2dX(X-1)-dX(1-2X) = (1-2X)(X-1)dt $$
$$ -2XdX + 2dX - dX + 2XdX = (1-2X)(X-1)dt $$ and our 2XdX's cancel out.

 

[DrClaude]

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

Geez. I read $(2X-1)$ where it was written $(1-2X)$.

 

[vanceEE]

Geez. I read $(2X-1)$ where it was written $(1-2X)$.

The initial differential equation, independent of the assumption that X < 0 was $$ \frac{dX}{dt} = (1-2X)(X-1) $$