JEE MAIN - Mathematics (2020 - 5th September Evening Slot - No. 2)
Let y = y(x) be the solution of the differential
equation
cosx$${{dy} \over {dx}}$$ + 2ysinx = sin2x, x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$.
If y$$\left( {{\pi \over 3}} \right)$$ = 0, then y$$\left( {{\pi \over 4}} \right)$$ is equal to :
cosx$${{dy} \over {dx}}$$ + 2ysinx = sin2x, x $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$.
If y$$\left( {{\pi \over 3}} \right)$$ = 0, then y$$\left( {{\pi \over 4}} \right)$$ is equal to :
$${1 \over {\sqrt 2 }} - 1$$
$${\sqrt 2 - 2}$$
$${2 - \sqrt 2 }$$
$${2 + \sqrt 2 }$$
Explanation
cosx$${{dy} \over {dx}}$$ + 2ysinx = sin2x
$$ \Rightarrow $$ $${{dy} \over {dx}} + 2y\tan x = 2\sin x$$
I.F = $${e^{\int {2\tan xdx} }}$$ = sec2 x
y.sec2 x = $${\int {2\sin x{{\sec }^2}xdx} }$$
$$ \Rightarrow $$ ysec2x = $${\int {2\tan x\sec xdx} }$$
$$ \Rightarrow $$ ysec2x = 2secx + c
Given at x = $${\pi \over 3}$$, y = 0
$$ \Rightarrow $$ 0 = $$2\sec {\pi \over 3} + c$$
$$ \Rightarrow $$ c = -4
ysec2x = 2secx - 4
Here put x = $${\pi \over 4}$$
y.2 = $$2\sqrt 2 $$ - 4
$$ \Rightarrow $$ y = $$\sqrt 2 - 2$$
$$ \Rightarrow $$ $${{dy} \over {dx}} + 2y\tan x = 2\sin x$$
I.F = $${e^{\int {2\tan xdx} }}$$ = sec2 x
y.sec2 x = $${\int {2\sin x{{\sec }^2}xdx} }$$
$$ \Rightarrow $$ ysec2x = $${\int {2\tan x\sec xdx} }$$
$$ \Rightarrow $$ ysec2x = 2secx + c
Given at x = $${\pi \over 3}$$, y = 0
$$ \Rightarrow $$ 0 = $$2\sec {\pi \over 3} + c$$
$$ \Rightarrow $$ c = -4
ysec2x = 2secx - 4
Here put x = $${\pi \over 4}$$
y.2 = $$2\sqrt 2 $$ - 4
$$ \Rightarrow $$ y = $$\sqrt 2 - 2$$
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