Differential Equations_JEE-Main-2021_010921_S2 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2021

01-09-2021 S2

Question

If $y=y(x)$ is the solution curve of the differential equation $x^2 d y+\left(y-\frac{1}{x}\right) d x=0 ; x>0$ and $y(1)=1$, then $\left(\frac{1}{2}\right)$ is equal to :

Select the correct option:

$\frac{3}{2}-\frac{1}{\sqrt{\mathrm{e}}}$

$3+\frac{1}{\sqrt{\mathrm{e}}}$

$3+e$

$3-e$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Hard

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