Trouble with non-linear ODE

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Homework Statement


Find the general solution of [tex]y'=\sqrt{2x-y}[/tex]


Homework Equations


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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 [tex]\int\sqrt{ax+b}[/tex], 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.
 

Answers and Replies

  • #2
Dick
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What's wrong with the obvious substitution u=(2x-y)? Or maybe even a little easier, u^2=(2x-y)?
 
  • #3
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What's wrong with the obvious substitution u=(2x-y)? Or maybe even a little easier, u^2=(2x-y)?

I've tried both of those. The problem is the y.

[tex]u=2x-y[/tex] when differentiated w.r.t. x gives [tex]du=(2-\frac{dy}{dx})dx[/tex]. Because of that [tex]\frac{dy}{dx}[/tex] term I see no way of rearranging this so that I can substitute back and get [tex]\int\sqrt{u} du[/tex]
 
  • #4
Dick
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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
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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.

Okay, I'm trying that now.

I get this:

[tex]2-u'=\sqrt{u}[/tex]
[tex]u'=2-\sqrt{u}[/tex]
[tex]\int u' dx=\int (2-\sqrt{u})dx[/tex]
[tex]u=\int 2 dx-\int\sqrt{u}dx[/tex]
[tex]u=2x-\int\sqrt{u}dx[/tex]

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
Dick
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Okay, I'm trying that now.

I get this:

[tex]2-u'=\sqrt{u}[/tex]
[tex]u'=2-\sqrt{u}[/tex]
[tex]\int u' dx=\int (2-\sqrt{u})dx[/tex]
[tex]u=\int 2 dx-\int\sqrt{u}dx[/tex]
[tex]u=2x-\int\sqrt{u}dx[/tex]

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.

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
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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.

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.

[tex]\frac{1}{2-\sqrt{u}}du=dx[/tex]
[tex]\int\frac{1}{2-\sqrt{u}}du=\int dx[/tex]
[tex]x=-\int\frac{1}{\sqrt{u}-2}du[/tex]
Let [tex]v=\sqrt{u}-2[/tex]
[tex]dv=\frac{1}{2\sqrt{u}}du[/tex]
[tex]\frac{1}{\sqrt{u}-2}/\frac{1}{2\sqrt{u}}=\frac{2\sqrt{u}}{\sqrt{u}-2}=\frac{4}{\sqrt{u}-2}+2[/tex]
[tex]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[/tex]
[tex]=-4ln|v|-2v+c[/tex]
[tex]=-4ln|\sqrt{u}-2|-2(\sqrt{u}-2)+c[/tex]
[tex]=-4ln|\sqrt{2x-y}-2|-2\sqrt{2x-y}+c[/tex]
Therefore
[tex]-4ln|\sqrt{2x-y}-2|-2\sqrt{2x-y}-x+c=0[/tex]
 
  • #8
Dick
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I think that looks about right. At least it agrees with what I got.
 
  • #9
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I think that looks about right. At least it agrees with what I got.

Awesome. Thanks for the help!
 

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