Let f: R rightarrow R be a continuous functio | vedclass

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

Question

(B) Given the equation: $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0$. Differentiating both sides with respect to $x$ using Leibniz's rule: $f'(x) + x f(x

Vedclass · Accepted Answer

$\lim_{x \rightarrow -\infty} f(x) = -2$

Vedclass · Answer

(B) Given the equation: $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0$. Differentiating both sides with respect to $x$ using Leibniz's rule: $f'(x) + x f(x) + 2x = 0$. Rearranging the terms: $f'(x) = -x(f(x) + 2)$. This is a first-order linear differential equation. Separating variables: $\frac{f'(x)}{f(x) + 2} = -x$. Integrating both sides: $\ln|f(x) + 2| = -\frac{x^2}{2} + C$. Thus,$f(x) + 2 = A e^{-x^2/2}$,or $f(x) = A e^{-x^2/2} - 2$. Using the initial condition from the original equation at $x = 0$: $f(0) + \int_{0}^{0} t f(t) dt + 0^2 = 0 \Rightarrow f(0) = 0$. Substituting $x = 0$ into $f(x) = A e^{-x^2/2} - 2$: $0 = A(1) - 2 \Rightarrow A = 2$. So,$f(x) = 2 e^{-x^2/2} - 2$. Now,evaluating the limits: $\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} (2 e^{-x^2/2} - 2) = 0 - 2 = -2$. $\lim_{x \rightarrow -\infty} f(x) = \lim_{x \rightarrow -\infty} (2 e^{-x^2/2} - 2) = 0 - 2 = -2$. Since $f(x) = 2(e^{-x^2/2} - 1)$,$f(x) = f(-x)$,so it is an even function. Comparing with the options,$\lim_{x \rightarrow -\infty} f(x) = -2$ is correct.

Let $f: R \rightarrow R$ be a continuous function satisfying $f(x) + \int_{0}^{x} t f(t) dt + x^2 = 0$ for all $x \in R$. Then:

Similar Questions

The solution of $\frac{dy}{dx} + \frac{1}{3}y = 1$ is

The integrating factor of the differential equation $\frac{dy}{dx}(x \log x) + y = 2 \log x$ is given by

Let $y=y(x)$ be the solution of the differential equation $(x^2+4)^2 dy + (2x^3y+8xy-2) dx = 0$. If $y(0)=0$,then $y(2)$ is equal to

If $y=y(x)$ is the solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ satisfying $y(1) = 1$,then the value of $y\left(\frac{1}{2}\right)$ is

The equation of the curve passing through the point $(0, \pi)$ and satisfying the differential equation $y dx = (x + y^3 \cos y) dy$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module