Question

A continuous function $f:R \rightarrow R$ satisfies the differential equation $f\left(x\right)=\left(1 + x^{2}\right)\left[1 + \int\limits _{0}^{x} \frac{\left( f \left( t \right)\right)^{2}}{1 + t^{2}} d t\right],$ then $f\left(- 3\right)$ is

• A. $\frac{13}{10}$
• B. $\frac{8}{5}$
• C. $\frac{10}{13}$
• D. $\frac{5}{8}$

Answer 1

Answer: (C)

$\frac{f \left( x \right)}{1 + x^{2}}=1+\int\limits _{0}^{x}\frac{\left( f \left( t \right) \right)^{2}}{1 + t^{2}}dt$
Differentiating w.r.t. $x$
$\frac{\left(1 + x^{2}\right) f^{'} \left(\right. x \left.\right) - 2 x f \left(\right. x \left.\right)}{\left(1 + x^{2}\right)^{2}}=\frac{f^{2} \left(\right. x \left.\right)}{\left(1 + x^{2}\right)}\Rightarrow \frac{1}{y^{2}}\frac{d y}{d x}-\left(\frac{2 x}{1 + x^{2}}\right)\frac{1}{y}=1$
Put $-\frac{1}{y}=t$
$\therefore \frac{d t}{d x}+\left(\frac{2 x}{1 + x^{2}}\right)t=1,$
IF $=e^{\int \frac{2 x}{1 + x^{2}} d x}=1+x^{2}$
$\therefore $ Solution is $-\frac{\left(1 + x^{2}\right)}{y}=x+\frac{x^{3}}{3}+c$
Now $f\left(0\right)=1$
$\therefore c=-1\Rightarrow y=\frac{- 3 \left(1 + x^{2}\right)}{x^{3} + 3 x - 3}=f\left(x\right)$
$f\left(- 3\right)=\frac{- 3 \left(1 + 3^{2}\right)}{3^{3} + 3 \times 3 - 3}=\frac{10}{13}$