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Question

Find the particular solution of the differential equation dxdy+x cot y=2y+y2 cot y, y ≠ 0 given that x = 0 when y=π2.

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Solution

We have,dxdy+x cot y=2y+y2cot y .....1Clearly, it is a linear differential equation of the form dxdy+Px=Qwhere P=cot y and Q=2y+y2cot y I.F.=eP dy =ecot y dy = elogsin y=sin yMultiplying both sides of 1 by I.F.=sin y, we getsin ydxdy+x cot y=sin yy2cot y+2ysin ydxdy+x cos y=y2cos y+2y sin yIntegrating both sides with respect to y, we getx sin y=y2Icos yIIdy+2y sin y dy+Cx sin y=y2cos ydy-ddyy2cos y dydy+2y sin y dy+Cx sin y=y2sin y-2y sin y dy+2ysin y dy+Cx sin y=y2sin y+CNow, y=π2 at x=0 0×sin π2=π42sin π2+CC=-π42Putting the value of C, we getx sin y=y2sin y-π42Hence, x sin y=y2sin y-π42 is the required solution.

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