The integrating factor of the differential eq | vedclass

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

Question

(A) The given differential equation is $(1 + t^2) + (x - e^{	an^{-1} t}) \frac{dt}{dx} = 0$.Rearranging the terms, we get $(x - e^{	an^{-1} t}) \fra

Vedclass · Accepted Answer

$e^{	an^{-1} t}$

Vedclass · Answer

(A) The given differential equation is $(1 + t^2) + (x - e^{	an^{-1} t}) \frac{dt}{dx} = 0$.Rearranging the terms, we get $(x - e^{	an^{-1} t}) \frac{dt}{dx} = -(1 + t^2)$.Taking the reciprocal, $\frac{dx}{dt} = -\frac{x - e^{	an^{-1} t}}{1 + t^2} = -\frac{x}{1 + t^2} + \frac{e^{	an^{-1} t}}{1 + t^2}$.This can be written as $\frac{dx}{dt} + \frac{1}{1 + t^2} x = \frac{e^{	an^{-1} t}}{1 + t^2}$.This is a linear differential equation of the form $\frac{dx}{dt} + P(t)x = Q(t)$, where $P(t) = \frac{1}{1 + t^2}$.The integrating factor $(IF)$ is given by $e^{\int P(t) dt} = e^{\int \frac{1}{1 + t^2} dt} = e^{	an^{-1} t}$.

The integrating factor of the differential equation $(1 + t^2) + (x - e^{\tan^{-1} t}) \frac{dt}{dx} = 0$ is

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $\cos x(3 \sin x+\cos x+3) dy = (1+y \sin x(3 \sin x+\cos x+3)) dx$; $0 \leq x \leq \frac{\pi}{2}, y(0)=0$. Then,$y\left(\frac{\pi}{3}\right)$ is equal to ..... .

The curve satisfying the differential equation $y \, dx - (x + 3y^2) \, dy = 0$ and passing through the point $(1, 1)$ also passes through the point

The integrating factor of the differential equation $(x^2 + 1)\frac{dy}{dx} + 2xy = x^2 - 1$ is

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

If $y=y(x)$ is the solution of the differential equation $\frac{dy}{dx} + 2y \tan x = \sin x$ with the condition $y(\frac{\pi}{3}) = 0$,then the maximum value of the function $y(x)$ over $\mathbb{R}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module