The integrating factor of the differential equation (x log_e x)dy/dx + y = 2log_e x is Option 1: log_e x Option 2: x Option 3: 1/x Option 4: 1/(log_e x)

Answer: log_e x. Solution: To find the integrating factor, the differential equation must be in standard form: dy/dx + P(x) * y = Q(x) The integrating factor is I.F. = e^( P(x)dx) Given equation: (x _e x)dy/dx + y = 2_e x Dividing by (x _e x): dy/dx + y/(x _e x) = (2_e x)/(x _e x) dy/dx + 1/(x _e x) * y = 2/x Therefore, P(x) = 1/(x _e x) The integrating factor is: I.F. = e^( P(x)dx) I.F. = e^( 1/(x _e x)dx) To solve 1/(x _e x)dx, let u = _e x Then du/dx = 1/x, so du = 1/xdx 1/(x _e x)dx = 1/udu = _e u = _e(_e x) I.F. = e^(_e(_e x)) I.F. = _e x Therefore, the integrating factor is _e x.

Calculus

Differential Equations

Medium

core

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

$log_e x$

$x$

$\frac{1}{x}$

$\frac{1}{log_e x}$

✅ Correct Option: 1

To find the integrating factor, the differential equation must be in standard form:

$\frac{dy}{dx} + P(x) \cdot y = Q(x)$

The integrating factor is $I.F. = e^{\int P(x)dx}$

Given equation:

$(x \log_e x)\frac{dy}{dx} + y = 2\log_e x$

Dividing by $(x \log_e x)$ :

$\frac{dy}{dx} + \frac{y}{x \log_e x} = \frac{2\log_e x}{x \log_e x}$

$\frac{dy}{dx} + \frac{1}{x \log_e x} \cdot y = \frac{2}{x}$

Therefore, $P(x) = \frac{1}{x \log_e x}$

The integrating factor is:

$I.F. = e^{\int P(x)dx}$

$I.F. = e^{\int \frac{1}{x \log_e x}dx}$

To solve $\int \frac{1}{x \log_e x}dx$ , let $u = \log_e x$

Then $\frac{du}{dx} = \frac{1}{x}$ , so $du = \frac{1}{x}dx$

$\int \frac{1}{x \log_e x}dx = \int \frac{1}{u}du$

$= \log_e u$

$= \log_e(\log_e x)$

$I.F. = e^{\log_e(\log_e x)}$

$I.F. = \log_e x$

Therefore, the integrating factor is $\log_e x$ .

2025: 2 June Shift 1

Match List-I with List-II

List-I	List-II
(A) The degree of differential equation $\frac{d^3y}{dx^3} = e^{\frac{dx}{dy}}$	(I) 2
(B) The order of differential equation $\left(\frac{dy}{dx}\right)^2 + \frac{d^3y}{dx^3} = 0$	(II) 4
(C) The sum of order and degree of differential equation $\frac{d}{dx}\left(\frac{d^2y}{dx^2}\right) + \left(\frac{dy}{dx}\right)^5 = x$	(III) not defined
(D) The number of arbitrary constants in the general solution of a differential equation of order 2	(IV) 3

Choose the correct answer from the options given below:

2026: 30 May Shift 1

Integrating factor of the differential equation $x\log x\frac{dy}{dx}+y=2\log x$ is equal to:

(Consider $\log x=\log_e x$ )

2026: 15 May Shift 1

The solution of the differential equation $\dfrac{dy}{dx} = 2^{x+y}$ is

2025: 3 June Shift 1

Solution of the differential equation $y\log_e y dx - x dy = 0$ is (Where c is an arbitrary constant)

2025: 15 May Shift 1

Let $y(x) = a(x + 1) \log(x + 1) + bx + 5$ be the solution of the differential equation e $^\frac{dy}{dx} = x + 1_{;}y(0) = 5$ , then the value of $(a + b)$ is:

2024: 16 May Shift 1

For the differential equation $\left(x \log _{e} x\right) d y=\left(\log _{e} x-y\right) d x$

(A) Degree of the given differential equation is $1$ .

(B) It is a homogeneous differential equation.

(C) Solution is $2y \log _{\mathrm{e}} \mathrm{x}+A=\left(\log _{\mathrm{e}} \mathrm{x}\right)^{2}$ , where $A$ is an arbitrary constant

(D) Solution is $2 y \log _{e} x+A=\log _{e}\left(\log _{e} x\right)$ , where $A$ is an arbitrary constant

Choose the correct answer from the options given below :

The integrating factor of the differential equation (xlogex)dydx+y=2logex(x log_e x)\frac{dy}{dx} + y = 2log_e x(xloge​x)dxdy​+y=2loge​x is

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

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