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

The answer is 'y = 1/x^2+1'. View full solution:

CUET Mathematics - Particular solution of the differential equation dy + 2xy^2 dx = 0, given that y=1, when x=0, is | PYQs + Solutions

CUET Mathematics 2026 14th May Shift 2

Calculus

Easy

common

$y = x^2+1$

$y = x^2-1$

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

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

✅ Correct Option: 3

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Match the LIST-I with LIST-II

LIST-I Differential equation	LIST-II Order and degree
A. $y'' = [1+(y')^2]^{3/2}$	I. 3, 4
B. $y''' + y' + y = 2x$	II. 2, 2
C. $2(y'')^3 - (y')^4 + y = 0$	III. 3, 1
D. $(y''')^4 + (y'')^3 + y' + y = 0$	IV. 2, 3

Choose the correct answer from the options given below: