I have the following ODE with initial/boundary value conditions:

$$\left. \begin{aligned} \left(x^2-10 x-y^2\right)y\, y'(x)+(x-5) y^2 y'(x)^2-(x-5) y^2=0 ;\qquad (\text{ODE})\\ y(0)^2=25;\qquad y'(0)^2=\frac{3-\sqrt{5}}{2} \qquad\qquad\qquad (\text{IBCs}) \end{aligned} \right\} $$

How to solve such a nonlinear ODE?

Additionally, how can I verify whether a special function is a potential solution or not, e.g., the one in implicit form as below:

$$\left(\sqrt{5}-1\right)\left(x-5\right)^2+2 y^2=25\left(\sqrt{5}+1\right)$$


By substituting $y^2$ and $y\,y'(x)$ obtained from the special solution into the original ODE and its IBCs, it seems this is the solution to the original nonlinear ODE problem.

So the only question remaining is how to solve it to obtain the known solution.


Below is another solution to the same ODE with the same IBCs:

$$\left(\sqrt{5}+1\right)\left(x-5\right)^2-2 y^2=25\left(\sqrt{5}-1\right)$$

The problem has double solutions:

enter image description here

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    $\begingroup$ Maybe defining $z := y^2$ and noting $x-5 = \frac{1}{2} \frac{d}{dx}(x^2-10x)$ helps a bit. $\endgroup$ – Dmoreno Sep 2 '14 at 8:26
  • $\begingroup$ Thank you; I now know how to verify the special solution. $\endgroup$ – LCFactorization Sep 2 '14 at 8:36
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    $\begingroup$ You're welcome. I was trying to solve the ODE but got stumped because of the minus sign of the third term. Quite complicated problem. $\endgroup$ – Dmoreno Sep 2 '14 at 8:38
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    $\begingroup$ This ODE is derived from another question: math.stackexchange.com/questions/904658/…, I am trying to see whether it is possible to solve ellipse curve by using only simple geometric rules. The IBCs here are different from the original problem; I suppose the ellipse curve is through $(0,\pm 5)$ $\endgroup$ – LCFactorization Sep 2 '14 at 8:49
  • $\begingroup$ So, please change the original ODE if you find any error in my deduction. I am using the properties of a bisection line of triangle. $\endgroup$ – LCFactorization Sep 2 '14 at 8:54

Let $U = x^2 + y^2$ and $V = (x-10)^2 + y^2$. Multiply the equation by $20$ and look at individual terms, we have:

$$\begin{align} 20(x-5) (yy')^2 &= \left[(x^2+y^2) - ((x-10)^2+y^2)\right] (yy')^2\\ &= (U-V)(yy')^2\\ \\ 20(x^2 - 10x - y^2) yy' &= 2yy'\left[(x-10)\left(x^2+y^2\right) - x\left((x-10)^2+y^2\right)\right]\\ &= 2yy'\left[(x-10)U - xV\right]\\ \\ -20(x-5) y^2 &= ((x-10)^2 - x^2)y^2\\ &= (x-10)^2 U - x^2 V \end{align}$$ So the ODE is equivalent to

$$U ( (yy')^2 + 2(x-10) + (x-10)^2 ) - V ( (yy')^2 + 2x + x^2 ) = 0\\ \iff U (x - 10 + yy')^2 - V(x + yy')^2 = 0$$

Since $U' = 2(x + yy')$ and $V' = 2(x - 10 + yy')$, this leads to

$$U (V')^2 - V(U')^2 = 0 \quad\iff\quad \frac{V'}{\sqrt{V}} \pm \frac{U'}{\sqrt{U}} = 0 \quad\iff\quad (\sqrt{V} \pm \sqrt{U})' = 0 $$

and hence the solution of the ODE is given by either $$ \sqrt{(x-10)^2 + y^2} + \sqrt{x^2+y^2} = \text{const}\\ \text{OR}\\ \sqrt{(x-10)^2 + y^2} - \sqrt{x^2+y^2} = \text{const}\\ $$ i.e. an ellipse or a hyperbola.


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