If [dy dx + 3 ^2xy = 1 ^2x,,,x ( - 3, 3 )] and [y( 4 ) = 4 3,] then [y( - 4 )] equals -

Question

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 -

Accepted Answer

$${{dy} \over {dx}} + 3{\sec ^2}x.y = {\sec ^2}x$$

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

Solutions · 1