JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 12)
Let y = y(x) be the solution of the differential equation,
$${{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, such that y(0) = 1. Then :
$${{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, such that y(0) = 1. Then :
$$y\left( {{\pi \over 4}} \right) - y\left( { - {\pi \over 4}} \right) = \sqrt 2 $$
$$y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2 $$
$$y\left( {{\pi \over 4}} \right) + y\left( { - {\pi \over 4}} \right) = {{{\pi ^2}} \over 2} + 2$$
$$y'\left( {{\pi \over 4}} \right) + y'\left( { - {\pi \over 4}} \right) = - \sqrt 2 $$
Explanation
$${{dy} \over {dx}} + y(\tan x) = 2x + {x^2}\tan x$$
I.F. = $${e^{\int {\tan x\,dx} }}$$ = $${e^{\ln \sec x}}$$ = sec x
y. sec x = $$\int {(2x + {x^2}\tan x)\sec \,x\,dx} $$
$$ \Rightarrow y\sec x = {x^2}\sec x + \lambda $$
$$ \Rightarrow $$ y(0) = 0 + $$\lambda $$ = 1 $$ \Rightarrow $$ $$\lambda $$ = 1
$$ \Rightarrow $$ y = x2 + cos x
$$ \Rightarrow $$ y' = 2x – sinx
$$ \Rightarrow y'\left( {{\pi \over 4}} \right) = {\pi \over 2} - {1 \over {\sqrt 2 }}$$
$$ \Rightarrow y'\left( { - {\pi \over 4}} \right) = - {\pi \over 2} + {1 \over {\sqrt 2 }}$$
$$ \therefore $$ $$y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2 $$
I.F. = $${e^{\int {\tan x\,dx} }}$$ = $${e^{\ln \sec x}}$$ = sec x
y. sec x = $$\int {(2x + {x^2}\tan x)\sec \,x\,dx} $$
$$ \Rightarrow y\sec x = {x^2}\sec x + \lambda $$
$$ \Rightarrow $$ y(0) = 0 + $$\lambda $$ = 1 $$ \Rightarrow $$ $$\lambda $$ = 1
$$ \Rightarrow $$ y = x2 + cos x
$$ \Rightarrow $$ y' = 2x – sinx
$$ \Rightarrow y'\left( {{\pi \over 4}} \right) = {\pi \over 2} - {1 \over {\sqrt 2 }}$$
$$ \Rightarrow y'\left( { - {\pi \over 4}} \right) = - {\pi \over 2} + {1 \over {\sqrt 2 }}$$
$$ \therefore $$ $$y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2 $$
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