An integrating factor of the differential equ | vedclass

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

Question

(D) The given differential equation is $(1-x^2) \frac{dy}{dx} + xy = \frac{x^4}{(1+x^5)} (\sqrt{1-x^2})^3$. Dividing by $(1-x^2)$,we get: $\frac{dy}{d

Vedclass · Accepted Answer

$\frac{1}{\sqrt{1-x^2}}$

Vedclass · Answer

(D) The given differential equation is $(1-x^2) \frac{dy}{dx} + xy = \frac{x^4}{(1+x^5)} (\sqrt{1-x^2})^3$. Dividing by $(1-x^2)$,we get: $\frac{dy}{dx} + \frac{x}{1-x^2} y = \frac{x^4 (\sqrt{1-x^2})^3}{(1+x^5)(1-x^2)} = \frac{x^4 \sqrt{1-x^2}}{1+x^5}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{x}{1-x^2}$. The integrating factor $(IF)$ is given by $e^{\int P dx}$. $IF = e^{\int \frac{x}{1-x^2} dx}$. Let $u = 1-x^2$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$. $IF = e^{-\frac{1}{2} \int \frac{1}{u} du} = e^{-\frac{1}{2} \ln|u|} = e^{\ln|u|^{-1/2}} = u^{-1/2} = \frac{1}{\sqrt{1-x^2}}$.

An integrating factor of the differential equation $(1-x^2) \frac{dy}{dx} + xy = \frac{x^4}{(1+x^5)} (\sqrt{1-x^2})^3$ is

Similar Questions

Which of the following is a correct statement?

$ABC$ is a triangular plate of uniform thickness. Its sides are in the ratio as shown in the figure. Let $I_{AB}$,$I_{BC}$,and $I_{CA}$ be the moments of inertia of the plate about the sides $AB$,$BC$,and $CA$ respectively. Which of the following relations is true for this arrangement?

$A$ particle of mass $m$ and charge $q$ is placed at rest in a uniform electric field $E$ and then released. The kinetic energy attained by the particle after moving a distance $y$ is

Which of the following products is formed when $SiF_4$ reacts with water?

The ratio of angular speeds of the minute hand and the hour hand of a watch is

Vedclass Test Series

Exam Paper Generator

Online Exam Module