The integrating factor of the differential eq | vedclass

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

Question

(B) The given differential equation is $\frac{dy}{dx} = y 	an x - y^2 \sec x$.This is a Bernoulli's equation of the form $\frac{dy}{dx} + P(x)y = Q(x

Vedclass · Accepted Answer

$\sec x$

Vedclass · Answer

(B) The given differential equation is $\frac{dy}{dx} = y 	an x - y^2 \sec x$.This is a Bernoulli's equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$,where $n = 2$.Dividing both sides by $y^2$,we get:$y^{-2} \frac{dy}{dx} - y^{-1} 	an x = - \sec x$ ..... $(i)$Let $v = y^{-1} = \frac{1}{y}$. Then $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$,which implies $y^{-2} \frac{dy}{dx} = -\frac{dv}{dx}$.Substituting these into equation $(i)$,we get:$-\frac{dv}{dx} - v 	an x = - \sec x$Multiplying by $-1$,we obtain the linear differential equation:$\frac{dv}{dx} + v 	an x = \sec x$Comparing this with the standard linear form $\frac{dv}{dx} + P(x)v = Q(x)$,we have $P(x) = 	an x$.The integrating factor $(I.F.)$ is given by:$I.F. = e^{\int P(x) dx} = e^{\int 	an x dx} = e^{\ln |\sec x|} = \sec x$.Thus,the correct option is $B$.

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

Similar Questions

The general solution of $\cos ^2 x \frac{d y}{d x}+y=\tan x$ is

Find the solution of the differential equation $\frac{dy}{dx} = \frac{1}{xy(x^2 \sin y^2 + 1)}$,where $C$ is the integral constant.

Find the general solution of the differential equation: $x \frac{dy}{dx} + y - x + xy \cot x = 0$ $(x \neq 0)$.

The solution of $y' - y = 1, y(0) = -1$ is given by $y(x) = $

At any point on a curve, the slope of the tangent is equal to the sum of the abscissa and the product of the ordinate and abscissa of that point. If the curve passes through $(0, 1)$, then the equation of the curve is

Vedclass Test Series

Exam Paper Generator

Online Exam Module