Solve the following DE and hence find $f(x)$

Question

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

Answer 1

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 $$ g(x)^{-1}=-x-\frac{x^3}2+1. $$

Answer 2

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 $\lambda=y^{-1}$.