Non linear diff. eq. first order

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TL;DR

How can I solve
(x^2 - R^2) y'^2 - 2xyy' + y^2 - R^2 = 0 ?

I tried to solve for y' the algebric second grade equation but I don't know how to solve the diff. equation either.
Any help?
Thanks.

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[anuttarasammyak]

(x^2 - R^2) y'^2 - 2xyy' + y^2 - R^2 = 0
I observe
x=R\ \sin t , y=R\ \cos t
satisfy the equation.

 

[BvU]

Is this homework ?

 

[wrobel]

It does not look like a homework.
I see the equation $$(xy'−y)^2=R^2(y'^2+1).$$ Differentiate it in x and you will get a linear first order ode
and $y''=0$

 

[lightarrow]

(x^2 - R^2) y'^2 - 2xyy' + y^2 - R^2 = 0
I observe
x=R\ \sin t , y=R\ \cos t
satisfy the equation.

Thank you.
That is a solution, but not the only one. I know which is the [general] solution (^*) because I started from that to derive the differ. eq. , but I don't know how to do solve this; of course I could in principle reverse all the computations from the [general] solution to here, but those computations, altough not difficult, would seem so unusual, peculiar, that no one could think of them without already knowing the result.
Of course I'll give the [general] solution, in the end, but my concern is how to tackle the equation.
(^*) Edit: I know "a family of curves" which are solutions, but actually I don't know if they are all the solutions!

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[wrobel]

I know "a family of curves" which are solutions, but actually I don't know if they are all the solutions!

yes, and the answer has already been provided

 

[lightarrow]

It does not look like a homework.
I see the equation $$(xy'−y)^2=R^2(y'^2+1).$$ Differentiate it in x and you will get a linear first order ode
and $y''=0$

Thanks.
Differentiating
$$(xy'−y)^2=R^2(y'^2+1)$$
we have
$$y''[x(xy'-y)-R^2y']=0$$
=>
$$y_1''=0$$
$$(x^2-R^2)y_2'=xy_2$$
Excluding |x|=R and since y_2=0 is not solution of the initial equation:
$$(x^2-R^2)y'^2-2xyy'+y^2-R^2=0 \ (1)$$
we can write:
$$\frac{dy_2}{y_2}=\frac{xdx}{x^2-R^2}$$
and integrate, which gives:
$$y_2=A\sqrt{x^2-R^2}$$
which, however, doesn't satisfy (1).

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[wrobel]

you have missed something
think again
did you know that $\int \frac{dx}{x}=\ln|x|+c$ ?
and thus $y_2=A\sqrt{|x^2-R^2|}$
and do not forget $y_1$

 

[lightarrow]

you have missed something
think again
did you know that $\int \frac{dx}{x}=\ln|x|+c$ ?

Yes, but I uncorrectly considered irrelevant here. However even considering it, I can't make it coming back, see down.

and thus $y_2=A\sqrt{|x^2-R^2|}$

Ok

and do not forget $y_1$

I intended to consider it later;
$$y_1(x)=ax+b$$
and there have to be a relation between a and b to satisfy (1):
$$b^2=R^2(1+a^2)$$
which brings to... something interesting (that is, the family of curves from which I started to find the diff. eq.) However now let's come back to y_2.
$$y_2=A\sqrt{|x^2-R^2|}$$
$$y_2'=\frac{Ax}{\sqrt{x^2-R^2}}\ ; \ when\ x^2>R^2$$
$$y_2'=-\frac{Ax}{\sqrt{R^2-x^2}}\ ;\ when\ x^2<R^2$$
Substituting in (1) we have, in the first case:
$$R^2(A^2+1)=0$$
which is meaningless; in the second case:
$$4A^2x^2=R^2(A^2-1)$$
which is meaningless either.
Unless I made some other mistake.
Anyway it's disturbing to me:
$$y_2=A\sqrt{|x^2-R^2|}$$
have to be a solution! Why I can't make it coming back?

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[anuttarasammyak]

Hi. After scaling of dividing x and y by R, the equation of second order infinitesimals is
(1-x^2)dy^2+2xy\ dx\ dy+(1-y^2)dx^2=0
dx^2+dy^2=(xdy-ydx)^2
|\mathbf{dl}|=|\mathbf{r}\times \mathbf{dl}|
r\ sin\theta = \pm 1
where $r=\sqrt{x^2+y^2},\theta$ is angle between $\mathbf{r}$ and tangential vector.
The figure is symmetric for x and y. $r=1,\theta=\pm \pi/2$ gives unit circle $x^2+y^2=1$ satisfy the above equation. You may be able to find another solutions if any from the above equation.

 

[wrobel]

e:

have to be a solution! Why I can't make it coming back?

I do not know,
I have come back with $|A|=1$
$y_2=\pm\sqrt{R^2-x^2}$

 

[lightarrow]

I do not know,
I have come back with $|A|=1$
$y_2=\pm\sqrt{R^2-x^2}$

You're right! I made a sign mistake...
Thanks!
So this solution is such that:
$$y_2^2+x^2=R^2$$
which is the circle centered in the origin, with radius R of which anuttarasammyak has written.
Now let's consider
$$y_1$$
because it's more interesting: it contains the solution
$$y_2$$
we have (at last :-) ) found!
In my previous posts I wrote that
$$y_1(x) =ax+b$$
and, substituting in
$$(x^2-R^2)y'^2-2xyy'+y^2-R^2=0$$
we find this relation between a and b:
$$b^2=R^2(1+a^2)$$
$$b=\pm R\sqrt{1+a^2}$$
so:
$$y_1=ax \pm R\sqrt{1+a^2}$$
dividing by
$$\sqrt{1+a^2}:$$
$$\frac{a}{\sqrt{1+a^2}}x-\frac{y_1}{\sqrt{1+a^2}} \pm R=0$$
which we could also write as the set of the following 2 equations:
$$cx+sy_1 \pm R=0$$
$$c^2+s^2=1$$
where:
$$c=\frac{a}{\sqrt{1+a^2}}$$
$$s=-\frac{1}{\sqrt{1+a^2}}$$
That is the family of straight lines which are tangent to the circle:
$$x^2+y^2=R^2$$
and so they contains that circle in particular.
Thanks to all who answered.

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[anuttarasammyak]

Congratulations! I observe both the solutions are written as
\cos t \ x + \sin t\ y=R
where $t [0,2\pi]$ , for any fixed constant t and for variable t.