Particular solution of the differential equation (dy/dx) = x + y, given that when x = 0, y = 0 is: Option 1: e^x + e^-y = 2 Option 2: e^-x + e^y = 2 Option 3: e^x + e^y = 2 Option 4: e^-x + e^-y = 2

The answer is 'e^x + e^-y = 2'. View full solution:

CUET Mathematics - Particular solution of the differential equation (dy/dx) = x + y, given that when x = 0, y = 0 is: | PYQs + Solutions

3 June Shift 2

For the differential equation $x\frac{dy}{dx} + 2y = x^2\log_e x$

(A) Integrating factor is $2x$

(B) Integrating factor is $x^2$

(C) General Solution is $y = \frac{x^2}{16}(4\log_e|x| - 1) + Cx^{-2}$ Where C is an arbitrary constant.

(D) General Solution is $y = \frac{x^4}{16}(4\log_e|x| - 1) + C$ Where C is an arbitrary constant.

Choose the correct answer from the options given below:

15 May Shift 1

Consider the differential equation $\frac{dy}{dx} + y \tan x = \sec x$ , then which of the following statements are correct?

(A) It is homogeneous

(B) It has $\sec x$ as its integrating factor

(C) It's general solution is $y \sec x = \tan x + c$ , where c is arbitary constant.

(D) It's degree is not defined

Choose the correct answer from the options given below:

10 Aug Shift 1

General solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} = 0$ is

A. $\tan^{-1} x + \tan^{-1} y = C$

B. $\sin^{-1} x - \cos^{-1} y = C$

C. $x\sqrt{1 - y^2} - y\sqrt{1 - x^2} = C$

D. $\sin^{-1} x + \sin^{-1} y = C$

E. $\cos^{-1} x + \cos^{-1} y = C$

(where C is arbitrary constant)

Choose the correct answer from the options given below