Solve ODE: y'(x)=-y(x)/√a^2-y(x)^2

  • Thread starter justaboy
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In summary: In that case, the substitution might work better like this:z^2=a^2-y^2After making the substitution and solving for y, you should have found that y=a sin(u).In summary, my teacher suggests that the substitution z^2=a^2-y^2 could be helpful, but I can't seem to take the next step. I'm having trouble even classifying this ODE.
  • #1
justaboy
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Homework Statement



Solve the ODE:

[tex] y'(x) = - \frac{y(x)}{\sqrt{a^2-y(x)^2}} [/tex]

The Attempt at a Solution



To be honest I'm having trouble even classifying this ODE. My teacher hinted that the substitution [tex]z^2=a^2-y^2[/tex] could be helpful, but once I make the substitution, I can't seem to take the next step. I know I haven't shown much work, but if you could nudge me in the right direction I would be extremely grateful. Thanks.
 
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  • #2
Isn't this just http://en.wikipedia.org/wiki/Method_of_separation_of_variables" ?
 
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  • #3
snipez90 said:
Isn't this just http://en.wikipedia.org/wiki/Method_of_separation_of_variables" ?

If you separate it, you get
[tex]\frac{\sqrt{a^2-y^2}}{y}dy=-dx[/tex]
but I still have the same problem... how do I integrate the left side?
 
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  • #4
I would start with a trig substitution, say y = a sin(u).
 
  • #5
Okay... after many trig subs, I found

[tex]a\log(a^3 y)+\sqrt{a^2-y^2}-a\log(2a\sqrt{a^2-y^2}+a) = -x[/tex]

How do I go about finding a y(x)?
 
  • #6
That looks fine, but you forgot to add the constant of integration.
I don't believe that you can solve this equation for y as an explicit function of x. That's OK - just leave it in the form you have.
 
  • #7
The question asks to solve for a curve y=y(x) explicitly.
 
  • #8
I don't see any way of solving for y, so I don't have any advice for getting y = f(x), but you could write x = g(y). That would be pretty simple.
 
  • #9
Does anyone see a way of solving for y=f(x)?

My teacher is quite positive that it can be done. Apparently this is an old geometry problem that Fermat, Descartes, Bernoulli and Leibniz worked on in the late 17th century... the curve y(x) should be such that the length of the segment of the tangent
to the curve between the curve and the x-axis is equal to a constant "a" at all points on the curve.
 
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  • #10
Why don't you just make the substitution like your teacher suggested?

Hint: Differentiate both sides of [itex]z^2=a^2-y^2[/itex] with respect to [itex]x[/itex].
 
  • #11
On the right hand side you need to differentiate z with respect to x, not y. Forget about dz/dy. You want an expression that involves dz/dx. On the left hand side, you already know what dy/dx is. Use that knowledge.
 
  • #12
Thanks very much. I end up with this:

[tex]\frac{dz}{dx}=\frac{a^2-z^2}{z^2}[/tex]

Which has turned out to be difficult to solve.
 
  • #13
Tough, but doable. You should get x as a function of z, and from that, x as a function of y. Inverting this to get y as a function of x: Hmmmm.
 
  • #14
Maybe he didn't mean for the z in the substitution to be squared.
 

Related to Solve ODE: y'(x)=-y(x)/√a^2-y(x)^2

What is an ODE?

An ODE, or Ordinary Differential Equation, is a mathematical equation that describes the relationship between a function and its derivatives. It is commonly used in scientific fields, particularly in physics and engineering, to model physical systems.

What does the notation y'(x) represent?

The notation y'(x) represents the derivative of the function y(x) with respect to x. In other words, it tells us how the function changes as the independent variable x changes.

What does the equation y'(x)=-y(x)/√a^2-y(x)^2 mean?

This equation is a first-order ODE that describes the relationship between the function y(x) and its derivative. It is a separable equation, meaning that it can be solved by separating the variables and integrating both sides.

How do you solve this ODE?

To solve this ODE, we can use the method of separation of variables. This involves isolating the y and y' terms on opposite sides of the equation and then integrating both sides. Once we have the general solution, we can use initial conditions to find the particular solution.

What are the applications of solving ODEs?

Solving ODEs has many real-world applications, particularly in scientific fields. It is used to model physical systems such as motion, population growth, and chemical reactions. It is also used in engineering to design and analyze systems such as circuits and control systems.

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