Question

A continuous function $f : R \rightarrow R$ satisfy the differential equation $f ( x )=\left(1+ x ^2\right)\left(1+\int\limits_0^{ x } \frac{ f ^2( t )}{1+ t ^2} dt \right)$ then the value of $f(-2)$ is

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

Answer 1

Answer: (D)

$\frac{ f ( x )}{1+ x ^2}=1+\int\limits_0^{ x } \frac{ f ^2( t ) dt }{1+ t ^2}$
[Note: $f (0)=1$ ]
differentiate both sides w.r.t. $x$
$\frac{\left(1+x^2\right) f^{\prime}(x)-f(x)(2 x)}{\left(1+x^2\right)^2}=\frac{f^2(x)}{\left(1+x^2\right)} $
$\frac{d y}{d x}-\left(\frac{2 x}{1+x^2}\right) y=y^2$
$\frac{1}{y^2} \frac{d y}{d x}-\left(\frac{2 x}{1+x^2}\right) \frac{1}{y}=1$
Let $\frac{-1}{y}=t$
$\frac{ dt }{ dx }+\left(\frac{2 x }{1+ x ^2}\right) t =1$
solving the above L.D.E,
$f(x)=\frac{-3\left(1+x^2\right)}{x^3+3 x-3}$
$f(-2)=\frac{-3(5)}{-8-6-3}=\frac{15}{17}$