JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 3)
Let y = y(x) be the solution of the differential equation $$\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx$$, with $$y\left( {{\pi \over 4}} \right) = 0$$. Then, the value of $${(y(0) + 1)^2}$$ is equal to :
e1/2
e$$-$$1/2
e$$-$$1
e
Explanation
$${{dy} \over {dx}} + 2{\sin ^2}x = 1 + y\cos 2x$$
$$ \Rightarrow {{dy} \over {dx}} + ( - \cos 2x)y = \cos 2x$$
$$I.F. = {e^{\int { - \cos 2xdx} }} = {e^{ - {{\sin 2x} \over 2}}}$$
Solution of D.E.
$$y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = \int {(\cos 2x)\left( {{e^{ - {{\sin 2x} \over 2}}}} \right)} dx + c$$
$$ \Rightarrow y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = - {e^{ - {{\sin 2x} \over 2}}} + c$$
Given
$$y\left( {{\pi \over 4}} \right) = 0$$
$$ \Rightarrow 0 = - {e^{{{ - 1} \over 2}}} + c \Rightarrow c = {e^{{{ - 1} \over 2}}}$$
$$ \Rightarrow y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = - {e^{ - {{\sin 2x} \over 2}}} + {e^{{{ - 1} \over 2}}}$$
at x = 0
$$y = - 1 + {e^{{{ - 1} \over 2}}}$$
$$ \Rightarrow y(0) = - 1 + {e^{{{ - 1} \over 2}}} \Rightarrow {(y(0) + 1)^2} = {e^{ - 1}}$$
$$ \Rightarrow {{dy} \over {dx}} + ( - \cos 2x)y = \cos 2x$$
$$I.F. = {e^{\int { - \cos 2xdx} }} = {e^{ - {{\sin 2x} \over 2}}}$$
Solution of D.E.
$$y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = \int {(\cos 2x)\left( {{e^{ - {{\sin 2x} \over 2}}}} \right)} dx + c$$
$$ \Rightarrow y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = - {e^{ - {{\sin 2x} \over 2}}} + c$$
Given
$$y\left( {{\pi \over 4}} \right) = 0$$
$$ \Rightarrow 0 = - {e^{{{ - 1} \over 2}}} + c \Rightarrow c = {e^{{{ - 1} \over 2}}}$$
$$ \Rightarrow y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = - {e^{ - {{\sin 2x} \over 2}}} + {e^{{{ - 1} \over 2}}}$$
at x = 0
$$y = - 1 + {e^{{{ - 1} \over 2}}}$$
$$ \Rightarrow y(0) = - 1 + {e^{{{ - 1} \over 2}}} \Rightarrow {(y(0) + 1)^2} = {e^{ - 1}}$$
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