Questions
The solution of differential equation is
a
−1xy+C
b
−1xy+logy=C
c
1xy+logy=C
d
logy=Cx
detailed solution
Correct option is B
Given, ydx+x+x2ydy=0⇒ dydx=−yx+x2y⇒ dxdy=−x+x2yy⇒ dxdy+xy=−x2⇒−1x2⋅dxdy−1xy=1 Put 1x=z⇒ −1x2⋅dxdy=dzdy⇒ dzdy−zy=1which is a linear differential equation of the formdzdy+P(y)z=Q(y)where ,P=−1y and Q=1IF=e∫−1ydy=e−logy=1yHence, the solution of given differential equation is zy=∫1ydy+C⇒ zy=logy+C⇒ 1xy=logy+C ∵z=1x⇒ −1xy+logy=CTalk to our academic expert!
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