Question

A continuous function $f:R\to R$ satisfy the differential equation $f\left(x\right)=(1+x^2)(1+{\int }_0^x\frac{f^2\left(t\right)}{1+t^2}dt)$ then the value of $f(-2)$ is

• A. $0$
• B. $\frac{17}{15}$
• C. $\frac{-17}{15}$
• D. $\frac{15}{17}$

Answer 1

The correct option is D $\frac{15}{17}$

By the fundamental theorem of calaries $f$ is differentiate

$f\left(x\right)=(1+{\lambda }^2)(1+\left({\int }_0^2\left(\frac{f^2\left(t\right)}{1+t^2}\right)dt\right))$

$\Rightarrow f^1\left(x\right)=2x(1+{\int }_0^2\frac{f^2\left(t\right)}{1+t^2}dt)+(1+2^2)\left(\frac{f^2\left(x\right)}{1+x^2}\right)$$

$\Rightarrow f^1\left(x\right)=\frac{2xf\left(x\right)}{1+x^2}+f^2\left(x\right)$

This is a non linear first order differential equation

$\Rightarrow f\left(x\right)=\frac{-3(x^2+1)}{x^3+3x+C},c$ is the whstat of

$f\left(0\right)=1\Rightarrow C=-3\Rightarrow f\left(x\right)=\frac{-3(x^2+1)}{x^3-3x-3}$

$\therefore f(-2)=\frac{15}{17}$