JEE MAIN - Mathematics (2019 - 10th January Morning Slot - No. 19)
If $${{dy} \over {dx}} + {3 \over {{{\cos }^2}x}}y = {1 \over {{{\cos }^2}x}},\,\,x \in \left( {{{ - \pi } \over 3},{\pi \over 3}} \right)$$ and $$y\left( {{\pi \over 4}} \right) = {4 \over 3},$$ then $$y\left( { - {\pi \over 4}} \right)$$ equals -
$${1 \over 3} + {e^6}$$
$${1 \over 3}$$
$${1 \over 3}$$ + e3
$$-$$ $${4 \over 3}$$
Explanation
$${{dy} \over {dx}} + 3{\sec ^2}x.y = {\sec ^2}x$$
I.F. = $${e^{3\int {{{\sec }^2}xdx} }} = {e^{3\tan x}}$$
or $$y.e{}^{3\tan x} = \int {{{\sec }^2}x.{e^{3\tan x}}} $$
or $$y.{e^{3\tan x}} = {1 \over 3}{e^{3\tan x}} + C$$
Given
$$y\left( {{\pi \over 4}} \right) = {4 \over 3}$$
$$ \therefore $$ $${4 \over 3}.{e^3} = {1 \over 3}{e^3} + C$$
$$ \therefore $$ C = e3
I.F. = $${e^{3\int {{{\sec }^2}xdx} }} = {e^{3\tan x}}$$
or $$y.e{}^{3\tan x} = \int {{{\sec }^2}x.{e^{3\tan x}}} $$
or $$y.{e^{3\tan x}} = {1 \over 3}{e^{3\tan x}} + C$$
Given
$$y\left( {{\pi \over 4}} \right) = {4 \over 3}$$
$$ \therefore $$ $${4 \over 3}.{e^3} = {1 \over 3}{e^3} + C$$
$$ \therefore $$ C = e3
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