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Answer 1

Answer:

(1+y2)dx=(tan-1 y-x)dy

dx/dy= tan-1y/1+y2_x/1+y2

dx/dy+ x /1+y2= tan -1y/1+y2

Hence

1F=ef1/1+y2.dy

=e tan-1y

Hence the above differential equation changes to

e tan -1y .dx/dy+xetan-1/1+y2= e tan -1 y tan -1y / 1+ y2

e tan -1y.dx+ xetan-1y/1+ye dy=e tan -1 tan -1 y/1+y2 dye

d(e tan -1y. x) = d(e tan -1y)

Integrating both sides gives us

e tan -1. x= e tan -1+ c

Answer 2

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