Non linear differential equation , instead of y it is $ y^2$ $$ \frac{dy} {dx}  +x \sin ^2 y = x ^3 \cos ^2 y$$
I attempted like this, 
divide by $ \cos ^2y$ 
and putting $\tan x = t$  it converted to 
$$ \frac{dt} {dx}  + x t ^ 2 = x ^3 $$ 
now it is not LDE of first order so i am not able to use INTEGRATING FACTOR method.
 how to further solve?
 A: $$\frac{dy} {dx}  +x \sin ^2 y = x ^3 \cos ^2 y$$
divide by $\cos^2 y$
$$\frac 1 {\cos^2 y}\frac{dy} {dx}  +x \tan ^2 y = x ^3 $$
$$(\tan y)'+ x \tan ^2 y = x ^3 $$
Substitute $z=\tan y$
$$z'=- x z^2 + x ^3 $$
This is Riccati 's equation
https://en.wikipedia.org/wiki/Riccati_equation
A: Set $t$ resp. $z=\frac{u'}{xu}$ in the resulting Riccati equation, then
$$
x^3=\frac{u''}{xu}-\frac{u'}{x^2u}-\frac{u'^2}{xu^2}+x\frac{u'^2}{x^2u^2}\implies xu''-u'-x^5u=0
$$
which gives power series coefficients the relation
$$
x^{n-1}:~~n(n-2)a_n=a_{n-6}
$$
which gives non-zero coefficients for $n=6k$ and $n=6k+2$, $u(x)=f(x^6)+x^2g(x^6)$, $f$,$g$ power series.
The coefficients of $f(x^6)=\sum a_{6k}x^{6k}$ are 
$$
a_{6k}=\frac{1}{6k(6k-2)}a_{6(k-1)}=\frac{\Gamma(2/3)}{6^{2k}k!\Gamma(k+2/3)}a_0.
$$
Compare this with the Bessel function
$$
 J_{\alpha }(w)=\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{m!\,\Gamma (m+\alpha +1)}}{\left({\frac {w}{2}}\right)}^{2m+\alpha }
$$
to identify $\alpha=-\frac13$, $w=\frac{ix^3}3$ so that
$$
f(x^6)=-i\frac{\Gamma(2/3)}{\sqrt[3]6} \,x\,J_{-\frac13 }\left(\frac{ix^3}3\right)
$$
which can be more properly expressed using modified Bessel or Hankel functions,...

Test this via $u(x)=xv(x^3/3)$, $u'(x)=x^3v'(x^3/3)+v(x^3/3)$, $u''(x)=x^5u''(x^3/3)+4x^2v'(x^3/3)$ so that indeed
$$
s^2v''(s)+sv'(s)-(\tfrac19+s^2)v(s)=0
$$
is the modified Bessel equation for $\alpha=\pm\frac13$.
A: You obtained this Riccati equation :
$$ \frac{dt} {dx}  + x t ^ 2 = x ^3 \tag 1$$
In order to solve it, when no particular solution can be guessed, the usual change of variable is :
$$t(x)=\frac{z'(x)}{x\:z(x)} \tag 2$$
See Eqs.$(4)$ and $(5)$ in http://mathworld.wolfram.com/RiccatiDifferentialEquation.html
This transforms the first order non-linear ODE into a second order linear ODE. Generally a linear ODE is easier to solve than a non-linear ODE even if the degree is higher.
$t'=\frac{z''}{x\:z}-z'\frac{z+x\:z'}{x^2z^2}$
$\frac{z''}{x\:z}-z'\frac{z+x\:z'}{x^2z^2}+x\left(\frac{z'}{xz}\right)^2=x^3$
$$z''-\frac{1}{x}z'-x^4z=0$$
This is a generalized Bessel ODE. To transform it into the usual Bessel ODE, change of variables $z(x)=xf(\tau)$ and $\tau=\frac13x^3$. 
See Eq.$(6)$ and $(7)$ in http://mathworld.wolfram.com/BesselDifferentialEquation.html
Finally, after boring calculus the result is 
$$z=c_1xI_{1/3}\left(\frac13x^3\right)+c_2xI_{-1/3}\left(\frac13x^3\right)$$
Where $I$ denotes the modified Bessel function of first kind.
Then we have to express explicitly $z'(x)$ and $t(x)$ from Eq.$(2)$.
