Solving y-y' equation with variable change r^2-x^2

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In summary, the conversation involves solving a differential equation by substituting a new variable and then simplifying to find the solutions. The process is further explained by taking the derivative and making substitutions. The conversation also touches on the inclusion of an integration constant in the final solutions.
  • #1
Ciaran
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Hello!

I've to solve y(y')^2+2xy' -y=0 by letting y^2= r^2-x^2 but this change of variable has thrown me. Could someone please help?
 
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  • #2
Ciaran said:
Hello!

I've to solve y(y')^2+2xy' -y=0 by letting y^2= r^2-x^2 but this change of variable has thrown me. Could someone please help?

Hi Ciaran! ;)

Taking the derivative of:
$$y^2= r^2-x^2 \tag 1$$
we find:
$$2yy' = 2rr' - 2x \Rightarrow yy' = rr' - x \tag 2$$

To make it fit in the DE, we multiply the DE by $y$:
$$y^2(y')^2+2xyy' -y^2=0 \tag 3$$

What do you get if you substitute $(1)$ and $(2)$ in $(3)$?
 
Last edited:
  • #3
Thanks for your swift reply! Should it not be yy'= rr'-x instead of 2x? I got

(rr'-x)^2 +2(rr'-2x)x-(r^2-x^2)=0

after substituting!
 
  • #4
Ciaran said:
Thanks for your swift reply! Should it not be yy'= rr'-x instead of 2x? I got

(rr'-x)^2 +2(rr'-2x)x-(r^2-x^2)=0

after substituting!

Yep. You're right. It should be without the 2.
I've edited my post to correct that.

Unfortunately I see it already made it in your new equation, in the second substitution. (Doh)

Anway, next step is to simplify...
 
  • #5
I got (rr')^2-r^2=0. Is this correct?
 
  • #6
Ciaran said:
I got (rr')^2-r^2=0. Is this correct?

It is correct yes... but you can simplify (and solve) further...
 
  • #7
So just taking out a common factor of r^2 giving me r^2=0 or (r')^2=1 then substituting x and y back in?
 
  • #8
Ciaran said:
So just taking out a common factor of r^2 giving me r^2=0 or (r')^2=1 then substituting x and y back in?

Yes, but I wouldn't substitute x and y back in just yet.
Can you solve $(r')^2=1$ first?
 
  • #9
I'm not actually sure how to solve that- is it something really simple that I'm missing?

- - - Updated - - -

Is it just r or negative r? Because differentiating these gives either 1 or -1 and squaring gives 1
 
  • #10
Ciaran said:
I'm not actually sure how to solve that- is it something really simple that I'm missing?

I think it is quite doable.

How would you solve $u^2=1$ for $u$?
And how would you solve $y'=2$?
 
  • #11
I'd say u= +- 1 and y= 2x (+C)
 
  • #12
Ciaran said:
I'd say u= +- 1 and y= 2x (+C)

Exactly.
So the solution of $(r')^2=1$ is $r=x+C$ or $r=-x+C$.

Btw, we can assume $r\ge 0$, since it is intended as a polar coordinate - the distance to the origin.
However, we can't make any such assumption about $r'$.
$r'$ is how $r$ changes along a solution curve. It can be both positive and negative.

Anyway, now is the time to back substitute x and y again...
 
  • #13
Do you need to include the constant? I was once told that it was unnecessary. If you do not need to include it, that means y=0 as y= r^2-x^2 and x^2=r^2
 
  • #14
Ciaran said:
Do you need to include the constant? I was once told that it was unnecessary. If you do not need to include it, that means y=0 as y= r^2-x^2 and x^2=r^2

I'm afraid that we really need to include the integration constant.

A DE typically has solution curves that cover the plane.
A boundary condition effective selects one of them.
Now you have only one curve, y=0, and no option to apply a boundary condition any more.

What are the solutions if you do include an integration constant?
 
  • #15
So that means that y^2= (x+C)^2-x^2 or y^2=(-x+D)^2-x^2?
 
  • #16
Ciaran said:
So that means that y^2= (x+C)^2-x^2 or y^2=(-x+D)^2-x^2?

Yep.

Perhaps those can be simplified?
 
  • #17
They can indeed- thank you very much for your help!
 

Related to Solving y-y' equation with variable change r^2-x^2

1. What is the purpose of solving a y-y' equation with variable change r^2-x^2?

The purpose of solving a y-y' equation with variable change r^2-x^2 is to find the solution for a differential equation that involves the variables y and y' (y prime) in terms of another variable, r^2-x^2. This type of variable change can help simplify the equation and make it easier to solve.

2. How do you perform a variable change r^2-x^2 on a y-y' equation?

To perform a variable change r^2-x^2 on a y-y' equation, you can substitute r^2-x^2 for the original variable in the equation. This can also involve using the chain rule to differentiate the equation with respect to the new variable.

3. What are the benefits of using a variable change r^2-x^2 in solving a y-y' equation?

Using a variable change r^2-x^2 can help simplify the differential equation and make it easier to solve. It can also help to eliminate any complex or nested functions that may be present in the original equation.

4. Are there any limitations to using a variable change r^2-x^2 in solving a y-y' equation?

There may be limitations to using a variable change r^2-x^2 in solving a y-y' equation, depending on the specific equation and situation. In some cases, the substitution may not lead to a simpler equation or may not be applicable at all. It is important to carefully consider the equation before using this variable change.

5. Can a variable change r^2-x^2 be used for any type of differential equation?

No, a variable change r^2-x^2 may not be applicable to all types of differential equations. It is most commonly used for solving equations involving the variables y and y' (y prime), but may not work for equations with other variables or more complex functions. It is important to carefully consider the equation before using this variable change.

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