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 is a linear differential equation of the form $\frac{dx}{dt} + P(t)x = Q(t)$, where $P(t) = \frac{1}{1 + t^2}$ and $Q(t) = \frac{e^{	an^{-1} t}}{1 + t^2}$.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 $f: R \rightarrow R$ be a continuous function which satisfies $f(x) = \int_{0}^{x} f(t) \, dt$. Then, the value of $f(\log_{e} 5)$ is:

The general solution of the differential equation $\sec y + (x - e^{\sin y}) \frac{dy}{dx} = 0$ is...

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

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=(y+1)((y+1)e^{x^{2}/2}-x)$,with $y(2)=0$. Then $y'(1)$ is equal to . . . .

If $\cos x$ and $\sin x$ are solutions of the differential equation $a_{0} \frac{d^{2} y}{d x^{2}}+a_{1} \frac{d y}{d x}+a_{2} y=0$ where $a_{0}, a_{1}$ and $a_{2}$ are real constants, then which of the following is/are always true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module