Given, $$y'(x)= |y(x)|(1-y(x)) \frac{x^3}{1+x^4}$$ with initial conditions $y(0) = 2$
I know that the y(x) can be found using Bernoulli's method such that
$$\frac{y'}{y^2} -\frac{x^3}{y(1+x^4)} = -\frac{x^3}{1+x^4} $$
Assuming $\frac{1}{y} = v$
Thus,
$$I.F ={\mathrm e}^{\int_0^x\frac{x^3}{1+x^4}} dx$$ or,$$ I.F. = (1+x^4)^\frac{1}{4} $$
Thus, $$ \frac{(1+x)^{\frac{1}{4}}}{y} =\int_0^x {\frac{x^3(1+x^4)^{\frac{1}{4}}}{1+x^4}} dx $$
and this is where I got stuck. How shall I proceed from here?
Somehow the answer is $$y(x) = \frac {(2^4)\sqrt{1+x^4}}{-1 + 2(1+x^4)^{\frac{1}4}}$$
I also observe that this can be solved using the separable variable method but I still have trouble getting to the answer. I will be grateful if anyone can help me out with this.
The derivative of $(1+x^4)^{1/4}$ is $x^3(1+x^4)^{-3/4}$, which is exactly the integrand you are in doubt of.
The equation is separable, so a better substitution would use and preserve that pattern by setting $v=y^{-1}-1$. Then $$ v'=-y^{-2}y'=-sv\frac{x^3}{1+x^4}, ~~~ s=sign(y(0)). $$ This is obviously again separable $$ \ln(v)=\frac{-s}4\ln(1+x^4)+c \\ v=C(1+x^4)^{-s/4} $$ etc.