JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 4)
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 :
$${{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$,
such that y (0) = 0, then $$y\left( { - {\pi \over 4}} \right)$$ is equal to :
$${1 \over 2} - e$$
$$e - 2$$
$$2 + {1 \over e}$$
$${1 \over e} - 2$$
Explanation
$${{dy} \over {dx}} + y{\sec ^2}x = se{c^2}x\,tanx$$
$$ \to $$ 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
$$y.{e^{\tan x}} = \int {\mathop {{e^t}}\limits_{||} } \mathop t\limits_| dt$$
yetanx = tet – et + c
yetanx = (tanx – 1)etanx + c
y = (tanx – 1) + c · e–tanx
Given y(0) = 0
$$ \Rightarrow $$ 0 = –1 + c $$ \Rightarrow $$ c = 1
$$y\left( { - {\pi \over 4}} \right) = - 1 - 1 + e = - 2 + e$$
$$ \to $$ 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
$$y.{e^{\tan x}} = \int {\mathop {{e^t}}\limits_{||} } \mathop t\limits_| dt$$
yetanx = tet – et + c
yetanx = (tanx – 1)etanx + c
y = (tanx – 1) + c · e–tanx
Given y(0) = 0
$$ \Rightarrow $$ 0 = –1 + c $$ \Rightarrow $$ c = 1
$$y\left( { - {\pi \over 4}} \right) = - 1 - 1 + e = - 2 + e$$
Comments (0)
