Recent content by Esker

Awesome. Thanks for the help!

Okay, I think I have it now. It's so simple I feel quite dumb for having totally missed it. Thanks a bunch for the help. I'd appreciate it if you could look over my working from this point to the finish and tell me if I've made any other glaringly stupid mistakes. \frac{1}{2-\sqrt{u}}du=dx...

Okay, I'm trying that now. I get this: 2-u'=\sqrt{u} u'=2-\sqrt{u} \int u' dx=\int (2-\sqrt{u})dx u=\int 2 dx-\int\sqrt{u}dx u=2x-\int\sqrt{u}dx And then I'm basically back to where I was when I first introduced the substitution. There's clearly some kind of basic principle here...

I've tried both of those. The problem is the y. u=2x-y when differentiated w.r.t. x gives du=(2-\frac{dy}{dx})dx. Because of that \frac{dy}{dx} term I see no way of rearranging this so that I can substitute back and get \int\sqrt{u} du

Homework Statement Find the general solution of y'=\sqrt{2x-y} Homework Equations -- The Attempt at a Solution I've tried approaching this equation through several methods, but I can't separate the variables, I can't make it fit the pattern for linear, I can't make it fit the...