If curve represented by differential equation x dy/dx + y = e^x passes through (1, 1), then y(-1) is : Option 1: e Option 2: e - 1 Option 3: e^2 - (e + 1) Option 4: (e^2 - (e + 1))/e

Answer: (e^2 - (e + 1))/e. Solution: x dy/dx + y = e^x d/dx(xy) = e^x. Integrating: xy = e^x + C. At (1, 1): 1 = e + C C = 1 - e. So xy = e^x + 1 - e. At x = -1: -y = e^(-1) + 1 - e y = e - 1 - 1/e = (e^2 - e - 1)/e = (e^2 - (e+1))/e.

Calculus

Differential Equations

Medium

If curve represented by differential equation $x \frac{dy}{dx} + y = e^x$ passes through (1, 1), then $y(-1)$ is :

$e$

$e - 1$

$e^2 - (e + 1)$

$\frac{e^2 - (e + 1)}{e}$

✅ Correct Option: 4

$x \frac{dy}{dx} + y = e^x \Rightarrow \frac{d}{dx}(xy) = e^x$ .

Integrating: $xy = e^x + C$ .

At $(1, 1)$ : $1 = e + C \Rightarrow C = 1 - e$ .

So $xy = e^x + 1 - e$ .

At $x = -1$ : $-y = e^{-1} + 1 - e \Rightarrow y = e - 1 - \frac{1}{e} = \frac{e^2 - e - 1}{e} = \frac{e^2 - (e+1)}{e}$ .

2025: 13 May Shift 1

The particular solution of the differential equation $e^x\sqrt{1-y^2}dx + \frac{y}{x}dy = 0$ , given that $y = 1$ , when $x = 0$ is:

2022: 23 Aug Shift 1

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

2025: 27 May Shift 1

Match List-I with List-II

List-I	List-II
(A) The degree of the differential equation $\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = x\sin\left(\frac{dy}{dx}\right)$	(I) 4
(B) The degree of differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^{1/4} + x^{1/5} = 0$	(II) 1
(C) The degree of differential equation $\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^3 + 6y^5 = 0$	(III) Not defined
(D) The degree of differential equation $1 + \left(\frac{dy}{dx}\right)^4 = 7\left(\frac{d^2y}{dx^2}\right)^3$	(IV) 3

Choose the correct answer from the options given below:

2025: 21 May Shift 1

The differential equation representing the curve $y = e^{2x}(a + bx)$ , where a, b are arbitrary constants is

2025: 30 May Shift 1

The differential equation representing the family of curves $y = Ax + \frac{B}{x}$ , $x \neq 0$ where A and B are arbitrary constants, is given by

2025: 27 May Shift 1

Which one of the following equations is a homogeneous differential equation?

If curve represented by differential equation xdydx+y=exx \frac{dy}{dx} + y = e^xxdxdy​+y=ex passes through (1, 1), then y(−1)y(-1)y(−1) is :

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

If curve represented by differential equation $x \frac{dy}{dx} + y = e^x$ passes through (1, 1), then $y(-1)$ is :