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 :

${{dy} \over {dx}} + 2{\sin ^2}x = 1 + y\cos 2x$<br><br>$\Rightarrow {{dy} \over {dx}} + ( - \cos 2x)y = \cos 2x$<br><br>$I.F. = {e^{\int { - \cos 2xdx} }} = {e^{ - {{\sin 2x} \over 2}}}$<br><br>Solution of D.E.<br><br>$$y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = \int {(\cos 2x)\left( {{e^{ - {{\sin 2x} \over 2}}}} \right)} dx + c$$<br><br>$$ \Rightarrow y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = - {e^{ - {{\sin 2x} \over 2}}} + c$$<br><br>Given<br><br>$y\left( {{\pi \over 4}} \right) = 0$<br><br>$$ \Rightarrow 0 = - {e^{{{ - 1} \over 2}}} + c \Rightarrow c = {e^{{{ - 1} \over 2}}}$$<br><br>$$ \Rightarrow y\left( {{e^{ - {{\sin 2x} \over 2}}}} \right) = - {e^{ - {{\sin 2x} \over 2}}} + {e^{{{ - 1} \over 2}}}$$<br><br>at x = 0<br><br>$y = - 1 + {e^{{{ - 1} \over 2}}}$<br><br>$$ \Rightarrow y(0) = - 1 + {e^{{{ - 1} \over 2}}} \Rightarrow {(y(0) + 1)^2} = {e^{ - 1}}$$