Consider the differential equation, y^{2} d x+ (x- \frac{1}{y} ) d y=0 . If value of y is 1 when x = 1, then the value of x for which y = 2, is: [12 Apr 2019 Shift 1]

Correct Answer is : 32−1√ e

JEE Main 12 Apr 2019 Paper 1 Solved Paper

Section: Mathematics

Question : 68 of 90

Marks: +1, -0

Consider the differential equation, y2dx+ (x−

)dy=0 . If value of y is 1 when x = 1, then the value of x for which y = 2, is:

[12 Apr 2019 Shift 1]

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