The particular solution of the differential equation dy/dx + 3y/x = 0, y(1) = 1 is Option 1: y = 1/x^3 Option 2: y = 1/x^2 Option 3: x = 1/y Option 4: x = 1/y^2

Answer: y = 1/x^3. Solution: Given differential equation: dy/dx + 3y/x = 0 with initial condition y(1) = 1. dy/dx = -3y/x dy/y = -3/xdx dy/y = -3/xdx, y, = -3, x, + C Using the property n(x) = (x^n):, y, =, x^(-3), + C, y, = (1/x^3) + C Taking exponential on both sides: y = e^C * 1/x^3 Let e^C = A: y = A/x^3 Using the initial condition y(1) = 1: 1 = A/1^3 1 = A/1 A = 1 Therefore, the particular solution is: y = 1/x^3

Calculus

Differential Equations

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The particular solution of the differential equation $\frac{dy}{dx} + \frac{3y}{x} = 0$ , $y(1) = 1$ is

$y = \frac{1}{x^3}$

$y = \frac{1}{x^2}$

$x = \frac{1}{y}$

$x = \frac{1}{y^2}$

✅ Correct Option: 1

Given differential equation: $\frac{dy}{dx} + \frac{3y}{x} = 0$ with initial condition $y(1) = 1$ .

$\frac{dy}{dx} = -\frac{3y}{x}$

$\frac{dy}{y} = -\frac{3}{x}dx$

$\int \frac{dy}{y} = \int -\frac{3}{x}dx$

$\ln|y| = -3\ln|x| + C$

Using the property $n\ln(x) = \ln(x^n)$ :

$\ln|y| = \ln|x^{-3}| + C$

$\ln|y| = \ln\left(\frac{1}{x^3}\right) + C$

Taking exponential on both sides:

$y = e^C \cdot \frac{1}{x^3}$

Let $e^C = A$ :

$y = \frac{A}{x^3}$

Using the initial condition $y(1) = 1$ :

$1 = \frac{A}{1^3}$

$1 = \frac{A}{1}$

$A = 1$

Therefore, the particular solution is:

$y = \frac{1}{x^3}$

2026: 29 May Shift 2

In reference to the differential equation : $y''+y=(e)^{y'}$ , which of the following statements are correct ?

A. The order of the differential equation is 1.

B. The degree of the differential equation is 1.

C. The order of the differential equation is 2.

D. The degree of the differential equation is not defined.

Choose the correct answer from the options given below:

2026: 14 May Shift 2

Match the LIST-I with LIST-II

	LIST-I (Differential Equation)		LIST-II (Order and degree)
A.	$\log\left(\dfrac{dy}{dx}\right) + \dfrac{d^2y}{dx^2} = 0$	I.	Order : 3 and degree : not defined
B.	$\dfrac{d^3y}{dx^3} + y^2 + e^{\frac{dy}{dx}} = 4$	II.	Order : 2 and degree : 2
C.	$\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)^2 + \dfrac{dy}{dx} = \sqrt{x}$	III.	Order : 2 and degree : not defined
D.	$\left(1+\left(\dfrac{dy}{dx}\right)^2\right)^{3/2} = \dfrac{d^2y}{dx^2}$	IV.	Order : 2 and degree : 1

Choose the correct answer from the options given below:

2023: 22 May Shift 3

Integerating factor of $(x \log_e x) \frac{dy}{dx} + y = 2 \log_e x$ is :

2026: 18 May Shift 1

The solution of the differential equation $x\frac{dy}{dx} = 2x^2 + 1$ ; $x \neq 0$ with $y = 1$ when $x = 1$ , is

2026: 31 May Shift 2

Match the LIST-I with LIST-II

	LIST-I Differential equation		LIST-II Integrating factor
A.	$\dfrac{dy}{dx} + 3y = e^{-2x}$	I.	$\dfrac{1}{x}$
B.	$x\dfrac{dy}{dx} + y = 3x^2$	II.	$e^{-x}$
C.	$x\dfrac{dy}{dx} - y = 3x^2$	III.	$x$
D.	$\dfrac{dy}{dx} - y = x$	IV.	$e^{3x}$

Choose the correct answer from the options given below:

2025: 3 June Shift 2

The sum of the order and degree of the differential equation representing the family of curves $y = mx + m^4$ , where m is arbitrary constant, is

The particular solution of the differential equation dydx+3yx=0\frac{dy}{dx} + \frac{3y}{x} = 0dxdy​+x3y​=0, y(1)=1y(1) = 1y(1)=1 is

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The particular solution of the differential equation $\frac{dy}{dx} + \frac{3y}{x} = 0$ , $y(1) = 1$ is