Question #f0f5a

Question

1838 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-14T11:36:48

#dy/dx=(1/ysec^2(x/y))/(1+x/y^2sec^2(x/y))#

#sec^2(x/y)*(y-dy/dxx)/y^2=dy/dx#

Answer 2 · 2016-11-14T12:10:00

#(1+y^2)/(y+(1+y^2)tan^(-1)y)#

You can try this differentiation, instead.

Solving for x,

#x = y tan^(-1)y#

#(dx)/(dy)=tan^(-1)y+y/(1+y^2)#.

#y'=1/((dx)/(dy))=(1+y^2)/(y+(1+y^2)tan^(-1)y)#