If y = y(x) is the solution of the differential equation [dy dx = ( x - y ) ^2x], [x ( - 2…

Question

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

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

yetanx = tet -- et + c

yetanx = (tanx -- 1)etanx + c

y = (tanx -- 1) + c \cdot  e--tanx

Given y(0) = 0

$$ \Rightarrow $$ 0 = --1 + c $$ \Rightarrow $$ c = 1

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

Solutions · 1