The differential equation representing the family of curves y=xecx, where c is a constant, is
A
dydx=yx(1−lnyx)
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B
dydx=yxln(yx)+1
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C
dydx=yx(1+lnyx)
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D
dydx+1=yxlnyx
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Solution
The correct option is Cdydx=yx(1+lnyx) y=xecx ⇒yx=ecx
Taking ln both sides, we get lnyx=lnecx ⇒lnyx=cx ⇒1xlnyx=c
Differentiating both sides w.r.t. x : 1x(xy×1x2(xdydx−y))+lnyx(−1x2)=0 ⇒1xy(dydx)=1x2(lnyx+1) ⇒dydx=yx(lnyx+1)