# Thread: 1st order Separable Equation

1. ## 1st order Separable Equation

$\displaystyle dx/dt-x^3=x$
$\displaystyle dx/dt=(x^3+x)$
$\displaystyle dt=(x^3+x)^{-1}dx$

So I haven't been able to figure out how to do the integral.

I don't think I can do fraction decomposition. At least when I try I don't get anything that makes a whole lot of sense.

I've tried integration by parts and I don't see any thing that looks remotely like the answer. Which is $\displaystyle x= \pm(Ce^{2t}/(1-Ce^{2t}))^{1/2} ,C\geq0$

2. ## Re: 1st order Separable Equation

Originally Posted by bkbowser
$\displaystyle dx/dt-x^3=x$
$\displaystyle dx/dt=(x^3+x)$
$\displaystyle dt=(x^3+x)^{-1}dx$

So I haven't been able to figure out how to do the integral.

I don't think I can do fraction decomposition. At least when I try I don't get anything that makes a whole lot of sense.

I've tried integration by parts and I don't see any thing that looks remotely like the answer. Which is $\displaystyle x= \pm(Ce^{2t}/(1-Ce^{2t}))^{1/2} ,C\geq0$
$\displaystyle \frac{1}{x^3 - x} = \frac{1}{x(x + 1)(x - 1)} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1}$

I get the equation
$\displaystyle A(x + 1)(x - 1) + Bx(x - 1) + Cx(x + 1) = 1$

Thus (after some work)
$\displaystyle A + B + C = 0$

$\displaystyle -B + C = 0$

$\displaystyle -A = 1$

Can you solve it from there?

-Dan

3. ## Re: 1st order Separable Equation

AH, OK. I see it now. Thanks!

4. ## Re: 1st order Separable Equation

Originally Posted by topsquark
$\displaystyle \frac{1}{x^3 - x} = \frac{1}{x(x + 1)(x - 1)} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1}$

I get the equation
$\displaystyle A(x + 1)(x - 1) + Bx(x - 1) + Cx(x + 1) = 1$

Thus (after some work)
$\displaystyle A + B + C = 0$

$\displaystyle -B + C = 0$

$\displaystyle -A = 1$

Can you solve it from there?

-Dan
There is a mistake here. It is $\displaystyle (x^3+x)^{-1}$ it is not $\displaystyle (x^3-x)^{-1}$.

Fraction decomposition on $\displaystyle (x^3+x)^{-1}$ leads me to $\displaystyle Ax^2+Bx+A=1$ so A has to both equal 0 and 1 which is impossible.

Can I instead take the log of each side as so;

$\displaystyle ln dt = ln (x^3+x)^{-1}$

$\displaystyle ln dt = -1*ln (x^3+x)$

exponentiation;
$\displaystyle e^{ln dt}= -1(x^3+x)$

and then integrating;

$\displaystyle t= -1\int(x^3+x)dx$

5. ## Re: 1st order Separable Equation

Originally Posted by bkbowser
There is a mistake here. It is $\displaystyle (x^3+x)^{-1}$ it is not $\displaystyle (x^3-x)^{-1}$.

Fraction decomposition on $\displaystyle (x^3+x)^{-1}$ leads me to $\displaystyle Ax^2+Bx+A=1$ so A has to both equal 0 and 1 which is impossible.

Can I instead take the log of each side as so;

$\displaystyle ln dt = ln (x^3+x)^{-1}$

$\displaystyle ln dt = -1*ln (x^3+x)$

exponentiation;
$\displaystyle e^{ln dt}= -1(x^3+x)$

and then integrating;

$\displaystyle t= -1\int(x^3+x)dx$

Whoops! Sorry about that. In this case we get:
$\displaystyle \frac{1}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$

So
$\displaystyle 1 = A(x^2 + 1) + (Bx + C)x$

etc.

The logarithm argument isn't going to work. What, for example, is ln(dt)? The only way to get rid of the dt and dx is to integrate over them.

-Dan

6. ## Re: 1st order Separable Equation

$\displaystyle Ax^2+A+Bx^2+Cx=x^2(A+B)+Cx+A$

so A=1 and B=-1 while C=0

$\displaystyle \int x^{-1} dx + \int -1x(x^{2}+1)^-1$

$\displaystyle ln|x|+ -1*\int \frac{1}{2}u^{-1} du$

$\displaystyle t= ln|x|+ - \frac{1}{2} ln|x^2+1|+c$

The text asks me to "solve the equation" which I take to mean solve for the dependent variable "x"

$\displaystyle e^t= |x| + -\frac{1}{2} |x^2+1| +e^c$
$\displaystyle e^t+\frac{1}{2}-e^c=x-x^2$ Provided $\displaystyle x\geq0$

This still looks pretty wrong though.

7. ## Re: 1st order Separable Equation

Originally Posted by bkbowser
$\displaystyle Ax^2+A+Bx^2+Cx=x^2(A+B)+Cx+A$

so A=1 and B=-1 while C=0

$\displaystyle \int x^{-1} dx + \int -1x(x^{2}+1)^-1$

$\displaystyle ln|x|+ -1*\int \frac{1}{2}u^{-1} du$

$\displaystyle t= ln|x|+ - \frac{1}{2} ln|x^2+1|+c$
Try this: Let c = ln|A|. Then your equation becomes:
$\displaystyle t = ln|x| - \frac{1}{2} ln|x^2+1| + ln|A|$

"Collapsing" the logarithms:
$\displaystyle t = ln \left | \frac{Ax}{(x^2 + 1)^{1/2}} \right |$

Now how does it turn out?

-Dan

Addendum: This is also a Bernoulli equation. The solution is a bit simpler doing it that way.