The solution of the differential equation cosx cosy dx + sinx siny dy = cos2y dx, is
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cosx cos y dy+sinx siny dx=0cosx cos y dydx+sinxsiny =0cosysinydydx+sinxcosx=0cotydy+tanxdx=0Integrating both sides we hav∫cotydy+∫tanxdx=∫0⇒log(siny)−log(cosx)=logC⇒logsinycosx=logC⇒sinycosx=C⇒siny=C cosx
cosx cos y dy+sinx siny dx=0cosx cos y dydx+sinxsiny =0cosysinydydx+sinxcosx=0cotydy+tanxdx=0Integrating both sides we hav∫cotydy+∫tanxdx=∫0⇒log(siny)−log(cosx)=logC⇒logsinycosx=logC⇒sinycosx=C⇒siny=C cosx
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Nikki bhai sorry mujhe ye answer nahi aata hai
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