Aspire Faculty ID #13331 · Topic: JEE Main 2023 (1 February Evening Shift) · Just now
JEE Main 2023 (1 February Evening Shift)

Let $\alpha x = \exp(x^{\beta} y^{\gamma})$ be the solution of the differential equation $2x^{2}y\,dy - (1 - xy^{2})\,dx = 0,\ x>0,\ y(2)=\sqrt{\log_{e}2}.$ Then $\alpha + \beta - \gamma$ equals:

Previous 10 Questions — JEE Main 2023 (1 February Evening Shift)

Nearest first

Next 10 Questions — JEE Main 2023 (1 February Evening Shift)

Ascending by ID
Ask Your Question or Put Your Review.

loading...