Trouble with non-linear ODE

1. Mar 13, 2010

Esker

1. The problem statement, all variables and given/known data
Find the general solution of $$y'=\sqrt{2x-y}$$

2. Relevant equations
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3. 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 Bernoulli equation, and as best I can tell it isn't homogeneous. I've tried a couple of substitutions but I've either failed to pick the right ones or that's the wrong method.

I found in a table of integrals an integral for $$\int\sqrt{ax+b}$$, but I'm rather doubtful about actually applying it with y in the mix. Besides, I'm required to show working.

Mostly I'd just like to know where to start on this one, since everything I've tried has led directly to a dead end.

2. Mar 13, 2010

Dick

What's wrong with the obvious substitution u=(2x-y)? Or maybe even a little easier, u^2=(2x-y)?

3. Mar 13, 2010

Esker

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 $$\int\sqrt{u} du$$

4. Mar 13, 2010

Dick

u'=2-y'. y'=2-u'. Substitute that into the left side of your equation. The right side is sqrt(u). There, no more y's. And it's separable.

5. Mar 13, 2010

Esker

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 that I'm failing to understand. That or I'm making it a lot harder for myself than I should be.

6. Mar 13, 2010

Dick

You aren't quite getting the 'separation of variables' thing. You've got du/dx=2-sqrt(u). Separate that into du/(2-sqrt(u))=dx. Now integrate both sides, one side du and the other dx.

7. Mar 13, 2010

Esker

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.

$$\frac{1}{2-\sqrt{u}}du=dx$$
$$\int\frac{1}{2-\sqrt{u}}du=\int dx$$
$$x=-\int\frac{1}{\sqrt{u}-2}du$$
Let $$v=\sqrt{u}-2$$
$$dv=\frac{1}{2\sqrt{u}}du$$
$$\frac{1}{\sqrt{u}-2}/\frac{1}{2\sqrt{u}}=\frac{2\sqrt{u}}{\sqrt{u}-2}=\frac{4}{\sqrt{u}-2}+2$$
$$x=-\int\frac{1}{\sqrt{u}-2}du=-\int(\frac{4}{\sqrt{u}-2}+2)\frac{1}{2\sqrt{u}}du=-\int(\frac{4}{v}+2)dv=-4\int\frac{1}{v}dv-\int 2dv$$
$$=-4ln|v|-2v+c$$
$$=-4ln|\sqrt{u}-2|-2(\sqrt{u}-2)+c$$
$$=-4ln|\sqrt{2x-y}-2|-2\sqrt{2x-y}+c$$
Therefore
$$-4ln|\sqrt{2x-y}-2|-2\sqrt{2x-y}-x+c=0$$

8. Mar 13, 2010

Dick

I think that looks about right. At least it agrees with what I got.

9. Mar 14, 2010

Esker

Awesome. Thanks for the help!