Question

A continuous function $f:R\to R$ satisfies the differential equation $f\left(x\right)=(1+x^2)(1+\underset{0}{\overset{x}{\int }}\frac{f^2\left(t\right)}{1+t^2}dt).$ If area of triangle formed by tangent drawn to the curve $y=f\left(x\right)$ at $x=1$ with the co-ordinates axis is $△,$ then the value of $\left[\frac{△}{3}\right]$ is
[Note: [K] denotes the greatest integer less than or equal to $K$.]

Answer 1

$\frac{f\left(x\right)}{(1+x^2)}=1+\underset{0}{\overset{x}{\int }}\frac{f^2\left(t\right)}{1+t^2}dt$
Differentiate w.r.t. $x$
$\frac{(1+x^2)\cdot \left(f^{\prime }\right(x\left)\right)-2xf\left(x\right)}{(1+x^2{)}^2}=\frac{f^2\left(x\right)}{1+x^2}$
$\frac{dy}{dx}-\left(\frac{2x}{1+x^2}\right)y=y^2$
Let $-\frac{1}{y}=t$
$\frac{dt}{dx}+\left(\frac{2x}{1+x^2}\right)t=1$
Solution of the above equation is :
$-\frac{1}{y}(1+x^2)=\frac{x^3}{3}+x+C$
Since $f\left(0\right)=1$, we get $C=-1$
Hence, $f\left(x\right)=\frac{-3(1+x^2)}{x^3+3x-3}$
Equation of tangent at $(1,-6)$ to $y=f\left(x\right)$ is $y=30x-36$
$△=\frac{108}{5}$ sq.units