Question

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

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