JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 1)
If y = y(x) is the solution of the differential equation
$${{dy} \over {dx}}$$ + (tan x) y = sin x, $$0 \le x \le {\pi \over 3}$$, with y(0) = 0, then $$y\left( {{\pi \over 4}} \right)$$ equal to :
$${{dy} \over {dx}}$$ + (tan x) y = sin x, $$0 \le x \le {\pi \over 3}$$, with y(0) = 0, then $$y\left( {{\pi \over 4}} \right)$$ equal to :
$${1 \over 2}$$loge 2
$$\left( {{1 \over {2\sqrt 2 }}} \right)$$ loge 2
loge 2
$${1 \over 4}$$ loge 2
Explanation
Integrating Factor $$= {e^{\int {\tan x\,dx} }} = {e^{\ln (\sec x)}} = \sec x$$
$$y\sec x = \int {(\sin x)\sec x\,dx = \ln (\sec x) + C} $$
$$y(0) = 0 \Rightarrow C = 0$$
$$ \therefore $$ $$y = \cos x\ln |\sec x|$$
$$y\left( {{\pi \over 4}} \right) = {1 \over {\sqrt 2 }}\ln \left( {\sqrt 2 } \right) = {1 \over {2\sqrt 2 }}\ln 2$$
$$y\sec x = \int {(\sin x)\sec x\,dx = \ln (\sec x) + C} $$
$$y(0) = 0 \Rightarrow C = 0$$
$$ \therefore $$ $$y = \cos x\ln |\sec x|$$
$$y\left( {{\pi \over 4}} \right) = {1 \over {\sqrt 2 }}\ln \left( {\sqrt 2 } \right) = {1 \over {2\sqrt 2 }}\ln 2$$
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