Question

A continuous function $f:R\to R$ satisfies the differential equation $f\left(x\right)=\left(1+x^2\right)\left[1+\underset{0}{\overset{x}{\int }}\frac{{\left(f\left(t\right)\right)}^2}{1+t^2}dt\right],$ 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)

$\frac{f\left(x\right)}{1+x^2}=1+\underset{0}{\overset{x}{\int }}\frac{{\left(f\left(t\right)\right)}^2}{1+t^2}dt$
Differentiating w.r.t. $x$
$\frac{\left(1+x^2\right)f^{{}^{\prime }}(x)-2xf(x)}{{\left(1+x^2\right)}^2}=\frac{f^2(x)}{\left(1+x^2\right)}\Rightarrow \frac{1}{y^2}\frac{dy}{dx}-\left(\frac{2x}{1+x^2}\right)\frac{1}{y}=1$
Put $-\frac{1}{y}=t$
$\therefore \frac{dt}{dx}+\left(\frac{2x}{1+x^2}\right)t=1,$
IF $=e^{\int \frac{2x}{1+x^2}dx}=1+x^2$
$\therefore$ Solution is $-\frac{\left(1+x^2\right)}{y}=x+\frac{x^3}{3}+c$
Now $f\left(0\right)=1$
$\therefore c=-1\Rightarrow y=\frac{-3\left(1+x^2\right)}{x^3+3x-3}=f\left(x\right)$
$f\left(-3\right)=\frac{-3\left(1+3^2\right)}{3^3+3\times 3-3}=\frac{10}{13}$