Given cos−1(yb)=log(xn)n
(yb)=cos{nlog(xn)}
Differentiate with respect to x we get
⇒y1=−bsin{nlog(xn)}×nx
⇒xy1=−bnsin{nlog(xn)}
Again differentiate with respect to x we get
⇒xy2+y1=−bncos{nlog(xn)}×nx
⇒x2y2+xy1=−bn2cos{nlog(xn)}
Since y=bcos{nlog(xn)}, we get
x2y2+xy1=−n2y