If x = sint and y = sinpt prove that :

(1 - xsquare)d2y/dx2 - xdy/dx + psquarey = 0

As x = sint, y = sinptThese have parameters tSo dydx=dydtdxdtHence dydt=pcospt, dxdt=costSo dydx=dydtdxdt=pcosptcost (i)And for d2ydx2, differentiate both sides of (i), wrt x and multiplying by dtdx on the RHSAs dydx is in the form of uv, so d2ydx2=vu'-uv'v2Hence d2ydx2=cost(-p2sinpt)-pcospt(-sint)cos2t×dtdx=-p2costsinpt + pcosptsintcos2t×1cost=-p2costsinpt + pcosptsint1-sin2t×1costHence (1-x2)d2ydx2-xdydx+p2y =(1-sin2t)d2ydx2-sintdydx+p2sinpt=(1-sin2t)×-p2costsinpt + pcosptsint1-sin2t×1cost-sint×pcosptcost+p2sinpt=-p2sinpt+sint×pcosptcost-sint×pcosptcost+p2sinpt=0 (hence proved)

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