Questions
The solution of is given by
a
1+logx+y−log1+logx+y=x+c
b
1+logx+y−log1−logx+y=x+c
c
1+logx+y2−log1+logx+y=x+c
d
none of these
detailed solution
Correct option is A
dydx=(x+y−1)+x+yln(x+y) Let x+y=z⇒1+dydx=dzdx⇒dzdx−1=z−1+zlogz⇒dzdx=zlogz+z=z1+logzlogz⇒∫logzdzz(1+logz)=∫dx Let logz=t⇒∫t1+tdt=∫dx⇒t−ln(1+t)=x+c0⇒log(x+y)−log(1+log(x+y))=x+C0Regarding the expression by adding 1 both side, we get 1+logx+y−log1+logx+y=x+c , where c = 1+ C0Talk to our academic expert!
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