The differential equation satisfied by the equation y = ae^bx, where a and b are arbitrary constants, is Option 1: d^2y/dx^2 = y(dy/dx)^2 Option 2: yd^2y/dx^2 = (dy/dx)^2 Option 3: yd^2y/dx^2 = dy/dx Option 4: d^2y/dx^2 = ydy/dx

Answer: yd^2y/dx^2 = (dy/dx)^2. Solution: From y = ae^bx, dy/dx = by and d^2y/dx^2 = bdy/dx = b^2y. Then yd^2y/dx^2 = b^2y^2 and (dy/dx)^2 = b^2y^2, so yd^2y/dx^2 = (dy/dx)^2.

Calculus

Differential Equations

Medium

applied

The differential equation satisfied by the equation $y = ae^{bx}$ , where $a$ and $b$ are arbitrary constants, is

$\frac{d^2y}{dx^2} = y\left(\frac{dy}{dx}\right)^2$

$y\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$

$y\frac{d^2y}{dx^2} = \frac{dy}{dx}$

$\frac{d^2y}{dx^2} = y\frac{dy}{dx}$

✅ Correct Option: 2

From $y = ae^{bx}$ , $\frac{dy}{dx} = by$ and $\frac{d^2y}{dx^2} = b\frac{dy}{dx} = b^2y$ .

Then $y\frac{d^2y}{dx^2} = b^2y^2$ and $\left(\frac{dy}{dx}\right)^2 = b^2y^2$ , so $y\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$ .

2026: 15 May Shift 1

The solution of the differential equation $\dfrac{dy}{dx} = \dfrac{x^2+xy+y^2}{x^2}$ for $x>0, y>0$ at $x=1, y=2$ is

2026: 26 May Shift 2

Match the LIST-I with LIST-II

	LIST-I Differential equation		LIST-II Order and Degree
A.	$\left(\dfrac{dy}{dx}\right)^2 + 3y\left(\dfrac{d^2y}{dx^2}\right) = 0$	I.	$1,1$
B.	$\sqrt{1-\left(\dfrac{dy}{dx}\right)^2} = a\left(\dfrac{d^2y}{dx^2}\right)^{1/3}$	II.	$1,2$
C.	$y = \dfrac{dy}{dx} + \dfrac{5}{(dy/dx)}$	III.	$2,1$
D.	$\sqrt{1-y^2}\,dx + \sqrt{1-x^2}\,dy = 0$	IV.	$2,2$

Choose the correct answer from the options given below:

2022: 16 July Shift 2

Integrating factor of the differential equation $\frac{dy}{dx} - \frac{1}{x} y = 1$ is:

2025: 26 May Shift 2

Curd is at 80° F, five minutes later it came down at 60°F. After another 5 minutes, its temperature became 50° F. Given that the rate of change of temperature is proportional to (T - S), where S is temperature of the surroundings and T is temperature of the curd at any time t. Then the temperature of the surroundings is :

2025: 15 May Shift 2

Consider the curve which is represented by the differential equation $\frac{dy}{dx} = 1 + x + y + xy$ . If it passes through the point $(0,0)$ , then which of the following is/are true?

(A) it is a straight line.

(B) it is a parabola.

(D) Its equation is $xy(x + 1)\left(y - \frac{1}{\sqrt{e}} + 1\right) = 0$

Choose the correct answer from the options given below:

2026: 18 May Shift 1

Solution of differential equation $\frac{dy}{dx} = e^{x+y} - xe^y$ is

The differential equation satisfied by the equation y=aebxy = ae^{bx}y=aebx, where aaa and bbb are arbitrary constants, is

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CUET Mathematics Past Year Questions (Topic-Wise)

The differential equation satisfied by the equation $y = ae^{bx}$ , where $a$ and $b$ are arbitrary constants, is