An integrating factor of the differential equ | vedclass

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

Question

(C) The given differential equation is $(1 - x^2)\frac{dy}{dx} - xy = 1$.Dividing both sides by $(1 - x^2)$,we get $\frac{dy}{dx} - \frac{x}{1 - x^2}y

Vedclass · Accepted Answer

$\sqrt{1 - x^2}$

Vedclass · Answer

(C) The given differential equation is $(1 - x^2)\frac{dy}{dx} - xy = 1$.Dividing both sides by $(1 - x^2)$,we get $\frac{dy}{dx} - \frac{x}{1 - x^2}y = \frac{1}{1 - x^2}$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{x}{1 - x^2}$ and $Q = \frac{1}{1 - x^2}$.The integrating factor $(I.F.)$ is given by $e^{\int P dx}$.$I.F. = e^{\int -\frac{x}{1 - x^2} dx}$.Let $u = 1 - x^2$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$.$I.F. = e^{\int \frac{1}{2u} du} = e^{\frac{1}{2}\log|u|} = e^{\log|u|^{1/2}} = \sqrt{1 - x^2}$.

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

Similar Questions

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

The general solution of the differential equation $(1+y^2) dx = (\tan^{-1} y - x) dy$ is

If the solution of the differential equation $\frac{dy}{dx} + e^x(x^2 - 2)y = (x^2 - 2x)(x^2 - 2)e^{2x}$ satisfies $y(0) = 0$,then the value of $y(2)$ is

The general solution of the differential equation $dx = (2x + 3y - 4) dy$ 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