How to Solve the Differential Equation \(\frac{dx}{dt} = \frac{x+t}{x-t}\)?

[retracell]

Homework Statement

Find the general solutions to \frac {dx}{dt} = \frac {x+t}{x-t}
Edit: Sorry I messed up the original equation.

Homework Equations

No idea.

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.

 

[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.

 

[retracell]

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

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

 

[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?

 

[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