Solve the following DE and hence find $f(x)$ Solve: $$
f(x)=\left(1+x^{2}\right)\left[1+\int_{0}^{x} \frac{f^{2}(t)}{1+t^{2}} d t\right]
$$
I have tried to solve this question by diffrentiating both sides with the help of $Lebnitz$ integral rule but I am stuck on the following ODE:
$$\frac{dy}{dx}=\frac{{y}^{2}+{(xy)}^{2}+2xy}{1+{x}^{2}}$$
Please help me after this step or provide an alternate approach.
My approach:
Diffrentiating both sides : $$
\frac{f^{2}(x)}{1+x^{2}}\cdot{1+{x}^{2}}+\frac{2 x f(x)}{1+x^{2}}=f'(x)
$$
Rearranging: $$f'(x)=\frac{x^{2} f^{2}(x)+f^{2}(x)+2xf(x)}{1+{x}^{2}}$$ I am stuck here
 A: Set $g(x)=\frac{f(x)}{1+x^2}$ then the integral equation simplifies to
$$
g(x)=1+\int_0^x(1+t^2)g(t)^2\,dt
$$
This now is equivalent to the initial value problem
$$
g(0)=1,~~g'(x)=(1+x^2)g(x)^2
$$
which is separable and leads to
$$
g(x)^{-1}=-x-\frac{x^3}2+1.
$$
A: This is a first-order Bernoulli differential equation of the form
$$y′+p(x)y=q(x)y^n$$
For Bernoulli DE's the general solution is obtained by substituting $\lambda =y^{1-n}$. Then $\lambda' = (1-n)y^{-n} \cdot y'$ so $$y'=\frac{1}{(1-n)}y^n\lambda'$$
and
$$y=\lambda y^n$$
Substituting this into our DE,
$$\frac{1}{(1-n)}y^n\lambda' + p(x)\lambda y^n = q(x)y^n$$
$$\lambda' + (1-n )p(x)\lambda = (1-n)q(x)$$
this of course being solved by an integrating factor
$$\mu(x) = \exp \bigg( (1-n) \int p(x)dx \bigg)$$
(from here, it's just solving a general first-order differential equation and then substituting the found $\lambda(x)$ and then finding $y(x)$)
Here,
$$y' = \frac{y^2(1+x^2)}{1+x^2} + \frac{2xy}{1+x^2}$$
so then
$$y' - \frac{2x}{1+x^2}y = y^2$$
Thus $p(x)=- \frac{2x}{1+x^2}$, $q(x)=1$ and $n=2.$
Our substitution will be $\lambda=\frac{1}{y}$.
Then performing the transformation in terms of $\lambda$,
$$\lambda' + \frac{2x }{1+x^2}\lambda = -1$$
This yields
$$\lambda = - \frac{x}{1+x^2}- \frac{x^3}{3(1+x^2)}+\frac{c}{1+x^2}$$
From here on it's very simple to find $y$, knowing that $\lambda=y^{-1}$.
