1. ## 1st order DE

Solve the following differential equation:

$\displaystyle x\frac{dy}{dx} = ye^{\frac{x}{y}} - x$

Well, an appropriate substitution would be $\displaystyle y = ux$, so that $\displaystyle y' = u'x + u.$

After dividing both sides by x, and substituting the above, we get

$\displaystyle u'x =ue^{\frac{1}{u}} - u - 1$.

This is separable, but how do you integrate the RHS? Or could we just say, like what we do in linear DE's, that

$\displaystyle (ux)' = ue^{\frac{1}{u}} - 1$, and then integrate both sides?

Although unusual (and gives what looks like a simpler integral to solve), I don't think $\displaystyle ue^{\frac{1}{u}}$ can be integrated using elementary functions.

Can anyone see a way out?

2. Originally Posted by nocturnal
Solve the following differential equation:

$\displaystyle x\frac{dy}{dx} = ye^{\frac{x}{y}} - x$

Well, an appropriate substitution would be $\displaystyle y = ux$, so that $\displaystyle y' = u'x + u.$

After dividing both sides by x, and substituting the above, we get

$\displaystyle u'x =ue^{\frac{1}{u}} - u - 1$.

This is separable, but how do you integrate the RHS? Or could we just say, like what we do in linear DE's, that

$\displaystyle (ux)' = ue^{\frac{1}{u}} - 1$, and then integrate both sides? Mr F says: No. The integration would have to be wrt x, but the right hand side is a function of u. You have to seperate and integrate. The integral wrt u is certainly non-elementary. I'd re-check the question for a typo.

Although unusual (and gives what looks like a simpler integral to solve), I don't think $\displaystyle ue^{\frac{1}{u}}$ can be integrated using elementary functions.

Can anyone see a way out?
..