QJEEM19Maf96c: JEE Main 2019 Mathematics - Class 12 JEEM19Maf96c Solution: If y = y(x) is the solution of the differential equation , ,…

Question

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If y = y(x) is the solution of the differential equation

{{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x

,

x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)

,
such that y (0) = 0, then

y\left( { - {\pi \over 4}} \right)

is equal to :

Accepted Answer

{{dy} \over {dx}} + y{\sec ^2}x = se{c^2}x\,tanx

\to

This is linear differential equation

IF = {e^{\int {{{\sec }^2}xdx} }} = {e^{\tan x}}

Now solution is

y.{e^{\tan x}} = \int {{e^{\tan x}}} {\sec ^2}x\tan xdx

\therefore

Let tanx = t

sec²xdx = dt

y.{e^{\tan x}} = \int {\mathop {{e^t}}\limits_{||} } \mathop t\limits_| dt

ye^tanx = te^t – e^t + c

ye^tanx = (tanx – 1)e^tanx + c

y = (tanx – 1) + c · e^–tanx

Given y(0) = 0

\Rightarrow

0 = –1 + c

\Rightarrow

c = 1

y\left( { - {\pi \over 4}} \right) = - 1 - 1 + e = - 2 + e

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