Test Index

Differential Equations

Section: Mathematics

© examsnet.com

Question : 3 of 25

Marks: +1, -0

Let

y(x)

be the solution of the differential equation

x^2 \;\frac{d y}{d x}+x y=x^2+y^2, \;\; x>\;\frac{1}{e},

y(1)=0

. Then the value of

2 \;\frac{(y(e))^2}{y (e^2) }

\_\_\_\_

[JEE Adv 2025 P2]

Your Answer:

Go to Question:

Prev Question

Next Question