Let f be a differentiable function f : R righ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation: $\frac{f(x)}{1+x^2} = 1 + \int_{0}^{x} \frac{f(t)}{1+t^2} dt$. Let $g(x) = \int_{0}^{x} \frac{f(t)}{1+t^2} dt$. Then $g'(x) =

Vedclass · Accepted Answer

$2e$

Vedclass · Answer

(C) Given the equation: $\frac{f(x)}{1+x^2} = 1 + \int_{0}^{x} \frac{f(t)}{1+t^2} dt$. Let $g(x) = \int_{0}^{x} \frac{f(t)}{1+t^2} dt$. Then $g'(x) = \frac{f(x)}{1+x^2}$. The equation becomes $\frac{f(x)}{1+x^2} = 1 + g(x)$. Substituting $f(x) = (1+x^2)g'(x)$ into the equation: $g'(x) = 1 + g(x)$. This is a linear differential equation: $\frac{dg}{dx} - g = 1$. The integrating factor is $e^{\int -1 dx} = e^{-x}$. Multiplying by $e^{-x}$: $\frac{d}{dx}(g(x)e^{-x}) = e^{-x}$. Integrating both sides: $g(x)e^{-x} = -e^{-x} + C$. So,$g(x) = C e^x - 1$. Since $g(0) = \int_{0}^{0} \dots = 0$,we have $0 = C - 1$,so $C = 1$. Thus,$g(x) = e^x - 1$. Then $g'(x) = e^x$. Since $f(x) = (1+x^2)g'(x)$,we have $f(x) = (1+x^2)e^x$. Therefore,$f(1) = (1+1^2)e^1 = 2e$.

Let $f$ be a differentiable function $f : R \rightarrow R$ satisfying the equation $f(x) = (1+x^2) \left[ 1 + \int_{0}^{x} \frac{f(t)}{1+t^2} dt \right]$ for all $x \in R$. Then $f(1)$ is:

Similar Questions

An integrating factor of the equation $(1+y+x^2 y) dx+(x+x^3) dy=0$ is

The general solution of the differential equation $e^{x} dy + (ye^{x} + 2x) dx = 0$ is

If $f(x)$ is a differentiable function in the interval $(0, \infty)$ such that $f(1) = 1$ and $\mathop {\lim }\limits_{t \to x} \frac{{{t^2}f(x) - {x^2}f(t)}}{{t - x}} = 1$ for each $x > 0$,then $f(\frac{3}{2})$ is equal to

If $x \phi(x) = \int_{5}^{x} (3t^{2} - 2 \phi'(t)) dt$,$x > -2$,and $\phi(0) = 4$,then $\phi(2)$ is .... .

If $\int_{a}^{x} t y(t) dt = x^2 + y(x)$,then $y$ as a function of $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module