Trouble with non-linear ODE

In summary, the student is having difficulty solving an equation for y'=\sqrt{2x-y} and is looking for help. He has tried a couple of substitutions but they haven't worked. He has found a solution to the equation for u=(2x-y), but he's not sure how to apply it. He's started to get the idea but still has a way to go.
  • #1
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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.
 
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  • #2
What's wrong with the obvious substitution u=(2x-y)? Or maybe even a little easier, u^2=(2x-y)?
 
  • #3
Dick said:
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
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
Dick said:
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
Esker said:
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
Dick said:
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
I think that looks about right. At least it agrees with what I got.
 
  • #9
Dick said:
I think that looks about right. At least it agrees with what I got.

Awesome. Thanks for the help!
 

1. What is a non-linear ODE?

A non-linear ODE (ordinary differential equation) is a type of mathematical equation that involves both dependent variables and their derivatives in a non-linear fashion. This means that the relationship between the variables and their derivatives is not a simple linear one, making the equation more complex and difficult to solve.

2. What causes trouble with non-linear ODEs?

Non-linear ODEs can be difficult to solve because they do not have a general solution like linear equations do. This means that each individual non-linear ODE must be solved in its own unique way, making it more time-consuming and challenging for scientists to work with.

3. How are non-linear ODEs used in science?

Non-linear ODEs are commonly used in science to model complex systems and phenomena. They can help scientists understand and predict the behavior of systems such as weather patterns, chemical reactions, and population dynamics. Non-linear ODEs are also used in fields such as physics, biology, and engineering.

4. What are some techniques for solving non-linear ODEs?

There are several techniques that can be used to solve non-linear ODEs, including numerical methods, perturbation methods, and power series methods. Each technique has its own advantages and limitations, and the specific method chosen will depend on the nature of the equation and the desired level of accuracy.

5. What are some challenges in dealing with non-linear ODEs?

Dealing with non-linear ODEs can be challenging due to their complexity and the lack of a general solution. In many cases, the equations must be approximated or solved numerically, which can be time-consuming and may not always provide accurate results. Additionally, non-linear ODEs can exhibit chaotic behavior, making it difficult to predict their long-term behavior.

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