# Homework Help: How does one solve this differential equation?

1. Oct 2, 2011

### retracell

1. The problem statement, all variables and given/known data
Find the general solutions to $$\frac {dx}{dt} = \frac {x+t}{x-t}$$
Edit: Sorry I messed up the original equation.

2. Relevant equations
No idea.

3. The attempt at a solution
I tried to use substitution u=x+t to change it into a separable equation but I only got to:
$$\frac {du}{dt} = \frac {2u-2t}{u-2t}$$
which I don't think is separable.

Last edited: Oct 2, 2011
2. Oct 2, 2011

### lineintegral1

Actually, the original differential equation is separable. Try separating and integrating accordingly. After all, your right hand side is a function only of x.

3. Oct 2, 2011

### retracell

Sorry I messed up the original equation. The one I need help on isn't as simple. :/

4. Oct 2, 2011

### retracell

Actually I just solved this question after much head scratching. I'm going to post it here for reference if needed because I know a lot of Google searches point to Physics Forums.

Letting $$u=\frac{x}{t}$$ then $$\frac{dx}{dt}=t\frac{du}{dt}+u$$
Substituting into original DE you'll get $$t\frac{du}{dt} = \frac{-u^2+2u+1}{u-1}$$
This then becomes separable and simply solve for $$\int\frac{u-1}{-u^2+2u+1}du=\int\frac{1}{t}dt$$ and make sure to back-substitute.

I'm pretty sure this is right but I'm not 100%. I still do not understand how do we choose the substitution equation for DE or is it similar to choosing substitution for integration?

5. Oct 2, 2011

### ehild

It is right.

Your original equation can be written in the form $x'=\frac{1+x/t}{1-x/t}$

When the DE is the form of y'=F(y/x) substitute y/x =u.

Then y=ux, y'=u'+u, so u'+u=F(u), u'=F(u)-u, a separable equation.

See this place: http://www.math.hmc.edu/calculus/tutorials/odes/odes.pdf

ehild

Last edited: Oct 2, 2011