QJEEM21Ma0af0: JEE Main 2021 Mathematics - Class 12 JEEM21Ma0af0 Solution: Let y = y(x) be the solution of the differential equation ,…

Question

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Let y = y(x) be the solution of the differential equation

{e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0

, y(1) =

-

1. Then the value of (y(3))² is equal to :

Accepted Answer

{e^x}\sqrt {1 - {y^2}} dx + {y \over x}dy = 0

\Rightarrow {e^x}\sqrt {1 - {y^2}} dx + {{ - y} \over x}dy

\Rightarrow \int {{{ - y} \over {\sqrt {1 - {y^2}} }}} dy = \int_{II}^{{e^x}} {_1^xdx}

\Rightarrow \sqrt {1 - {y^2}} = {e^x}(x - 1) + c

Given : At x = 1, y =

-

1

\Rightarrow

0 = 0 + c

\Rightarrow

c = 0

\therefore

\sqrt {1 - {y^2}} = {e^x}(x - 1)

At x = 3

1 - {y^2} = {({e^3}2)^2} \Rightarrow {y^2} = 1 - 4{e^6}

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