How to solve an ODE to find general and particular solutions

Question

Salutations, I have a problem with this ODE: $$5y^2(y')^2+2xy(y')+x^2+4y^2-4=0$$

I tried to find a general solution through substitutions, but I haven´t seen an ODE like this before, I got complex expressions (procedure in next picture).

My Procedure 1

So, I read about singular solutions and I tried to apply P-discriminant ($p=\frac{\partial y}{\partial x}$) to this ODE, but I'm not sure if the procedure is correct because I'm starting to approach these concepts and topics about ODEs and I want to understand for mathematical curiosity. I tried to find the singular solution in this procedure:

My procedure 2

So, I noticed there is a relationship between general and singular solution through C-discriminant and P-discriminant, but I'm confused how to apply them to solve this ODE.

My question is:

-How would be the mathematical steps to apply to find the general and singular solution of the ODE?

I would like any guidance or starting steps or explanations to find the solutions because it's an interesting problem.

Thanks for your attention.

Answer 1

Maple finds the following solutions, where $c$ is an arbitrary constant: $$ \eqalign{y \left( x \right) &= \pm \sqrt {1-x^2/5}\cr y \left( x \right) &=\pm \sqrt {-4+5\,c-{x}^{2}\pm 4\,\sqrt {1-c }\,x}\cr }$$ Of course there will be restrictions on $x$ and $c$ if you want real-valued solutions.

Answer 2

$$5(yy')^2+2x(yy')+x^2+4y^2-4=0$$ Let $Y=y^2 \quad\to\quad Y'=2yy'$ $$5(Y')^2+4xY'+4x^2+16Y-16=0$$ $5(Y'+\frac25 x)^2-\frac45 x^2+4x^2+16Y-16=0$

Let $u=Y+\frac15 x^2 \quad\to\quad u'=Y'+\frac25 x$

$5(u')^2-\frac45 x^2+4x^2+16(u-\frac15 x^2 )-16=0$ $$5(u')^2+16u-16=0$$ This is a separable ODE, easy to solve :

Obvious particular solution : $\quad u=1\quad\to\quad Y=1-\frac15 x^2\quad\to\quad y=\pm\sqrt{1-\frac15 x^2}$

$\frac{du}{\sqrt{1-u}}=\pm\frac{4}{\sqrt{5}}dx\quad\to\quad u=1-(C+\frac{2}{\sqrt{5}}x)^2 \quad\to\quad y=\pm\sqrt{1-(C+\frac{2}{\sqrt{5}}x)^2-\frac15 x^2}$ $$y(x)=\pm\sqrt{1-x^2-\frac{4C}{\sqrt{5}}x-C^2}$$ This is equivalent to MAPLE's solution :

$y(x) =\pm \sqrt {-4+5\,c-{x}^{2}\pm 4\,\sqrt {1-c}\,x}\quad$ with relationship $\quad C=\pm\sqrt{5(1-c)}$