JEE Main 2019 — Math solution | Fundamenthol

JEE Main2019Math

Q.

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 :

A
$\frac{5}{2} + \frac{1}{e}$
B
$\frac{3}{2} - e$
C
$\frac{3}{2} - \frac{1}{e}$
D
$\frac{1}{2} + \frac{1}{e}$

Solution

← Back to full paper

Practice more Math

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →