It is question from the 12th class cbse board 2013. And it from d. Equation
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Assuming this is supposed to be:

dx/dy+xcot(y)=2y+y²cot(y),

we can multiply through by sin(y):

sin(y)dx/dy+xcos(y)=2ysin(y)+y²cos(y).

That is:

d/dy(xsin(y))=d/dy(y²sin(y)).

Integrating both sides:

xsin(y)=y²sin(y)+C, where C is the constant of integration.

Therefore:

x=y²+Ccosec(y).

Plugging in x=0 and y=π/2:

0=π²/4+C, so C=-π²/4 and x=y²-¼π²cosec(y).

[The given equation dx/dy+cot(y)=2y+y²cot(y) would give us:

x=y²-ln|sin(y)|+∫y²cot(y)dy, which contains an indefinite integral that would be very difficult or impossible to calculate in general terms.]

by Top Rated User (1.2m points)

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