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

[Ciaran]

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?

 

[I like Serena]

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 $(3)$?

 

[Ciaran]

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!

 

[I like Serena]

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...

 

[Ciaran]

I got (rr')^2-r^2=0. Is this correct?

 

[I like Serena]

I got (rr')^2-r^2=0. Is this correct?

It is correct yes... but you can simplify (and solve) further...

 

[Ciaran]

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?

 

[I like Serena]

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?

 

[Ciaran]

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

 

[I like Serena]

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 $y'=2$?

 

[Ciaran]

I'd say u= +- 1 and y= 2x (+C)

 

[I like Serena]

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...

 

[Ciaran]

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 like Serena]

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?

 

[Ciaran]

So that means that y^2= (x+C)^2-x^2 or y^2=(-x+D)^2-x^2?

 

[I like Serena]

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?

 

[Ciaran]

They can indeed- thank you very much for your help!