Let y = y(x) be the solution of the differential equation, \(\frac{{dy}}{{dx}} + y\;tan\;x = 2x + {x^2}\;tan\;x,\;\)\(x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\), such that y(0) = 1. Then:

\(y\left( {\frac{\pi }{4}} \right) + y\left( { - \frac{\pi }{4}} \right) = \frac{{{\pi ^2}}}{2} + 2\)

\(y'\left( {\frac{\pi }{4}} \right) + y'\left( { - \frac{\pi }{4}} \right) = - \sqrt 2 \)

\(y\left( {\frac{\pi }{4}} \right) - y\left( { - \frac{\pi }{4}} \right) = \sqrt 2 \)

\(y'\left( {\frac{\pi }{4}} \right) - y'\left( { - \frac{\pi }{4}} \right) = \pi - \sqrt 2 \)

Let y = y(x) be the solution of the differential equation, \(\frac{{d

1.	Let y = y(x) be the solution of the differential equation, \(\frac{{dy}}{{dx}} + y\;tan\;x = 2x + {x^2}\;tan\;x,\;\)\(x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\), such that y(0) = 1. Then:
A.	\(y\left( {\frac{\pi }{4}} \right) + y\left( { - \frac{\pi }{4}} \right) = \frac{{{\pi ^2}}}{2} + 2\)
B.	\(y'\left( {\frac{\pi }{4}} \right) + y'\left( { - \frac{\pi }{4}} \right) = - \sqrt 2 \)
C.	\(y\left( {\frac{\pi }{4}} \right) - y\left( { - \frac{\pi }{4}} \right) = \sqrt 2 \)
D.	\(y'\left( {\frac{\pi }{4}} \right) - y'\left( { - \frac{\pi }{4}} \right) = \pi - \sqrt 2 \)
Answer» E.

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Let y = y(x) be the solution of the differential equation, \(\frac{{dy}}{{dx}} + y\;tan\;x = 2x + {x^2}\;tan\;x,\;\)\(x \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\), such that y(0) = 1. Then:

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