Not just another DE

I found that $y = \sqrt{\frac{3}{x^2 + 3}}$ using a terrible brute force technique.

@Digvijay Singh , maybe you have a nice solution.

Let's first take derivative with respect to x, we get

$\displaystyle \dfrac{1}{x^{2}}\dfrac{d}{dx}x^{2}\dfrac{d}{dx}y+y^{5}=0$

Next, we substitute in $\displaystyle y=ux^{-1/2}$ (I got this substitution from considering "balance of variable" i.e. $\displaystyle \dfrac{1}{x^{2}}\dfrac{d}{dx}x^{2}\dfrac{d}{dx}y \sim y^{5}$ , so, $\displaystyle y \sim x^{-1/2}$ and hence, the guess.)

Now, $\displaystyle y'=u'x^{-1/2}-1/2ux^{-3/2}$ and $y"=u"x^{-1/2}-u'x^{-3/2}+3/4ux^{-5/2}$ and so,

$\displaystyle \dfrac{1}{x^{2}}\dfrac{d}{dx}x^{2}\dfrac{d}{dx}y+y^{5}=x^{2}u"+xu'-1/4u+u^{5}=0$

$\displaystyle x^{2}u"+xu'=\dfrac{d^{2}}{dlnx^{2}}u$ So, we have turned the equation into the form we are familiar with.

$\displaystyle (du/dlnx)^{2}=A+\dfrac{u^{2}}{4}-\dfrac{u^{6}}{3}$ but $du/dlnx=xu'=0=u$ at x=0, so, A=0

$\displaystyle du/dlnx=u\sqrt{\dfrac{1}{4}-\dfrac{u^{4}}{3}}$

$\displaystyle \int { \frac { du }{ u\sqrt { \frac { 1 }{ 4 } -\frac { { u }^{ 4 } }{ 3 } } } } =\frac { \sqrt { 3 } }{ 2 } \int { \frac { d{ u }^{ 2 } }{ { u }^{ 2 }\sqrt { \frac { 3 }{ 4 } -{ u }^{ 4 } } } }$

$\displaystyle u^{2}=\dfrac{\sqrt{3}}{2}sin\theta$

$\displaystyle \int { \frac { d\theta }{ \sin { \theta } } } =\frac { -1 }{ 2 } \ln { \frac { 1+\cos { \theta } }{ 1-\cos { \theta } } } =lnA+lnx$

$\displaystyle cos \theta=\dfrac{A-x^{2}}{A+x^{2}}=\dfrac{\sqrt{3/4-u^{4}}}{\sqrt{3/4}}$ so,

$\displaystyle u^{4}=\dfrac{3Ax^{2}}{(A+x^{2})^{2}}=y^{4}x^{2}$ now, y=1 at x=0 means A=3 so $\displaystyle y=\sqrt{\dfrac{3}{3+x^2}}$ now the integral asked is plain easy, so, I will just leave it at this stage.