212 Dr. A. A. Robb: Graphical Solution of Differential 

 We shall then have 



r+Mf(p)dp = C. 



s + 



Let the transparent squared paper now be placed on the 

 top of the plain sheet so that one of the lines on the squared 

 paper parallel to OX passes through the point P and let 

 this line intersect the curve DE in the point R . 



£ i 



i 











^A R n 





























' x E / 





Y 

















' s / 











' 1 N / 





















^ ' ' ^ / 'i 



















' ' \ / '< n y 











• ' ' N / '' 





















' 1 s / ' 





















• ' ' ' X /- ' 











,' ' / *" 











' /' Mn 











• i /' 











p ^'' d'\ _v 











y 



AnV x / 



K 











/£ 



- sS 1 











/ 





\ -a &z^?' / 















A £ _3^>V' D/ 









D 





No 



V \ A A, x"A 2 A n M M n 



X. 



















Ni^v /Jn 





\ 







nXc\ ,'' 





\ 







/ >^n 





\ 



. 





./if • '' 









\ 





/ s' 







Let the perpendicular through P on OX intersect OX in 

 N and the curve (5) in T , and let the perpendicular through 

 R on OX intersect OX in M . 



Let a length M A equal to PT (or r ) be measured off 

 along the axis of a, and let it be measured from M in the 

 negative direction if r be positive, or in the positive direc- 

 tion if r be negative ; and let a pin be stuck through A 

 into the drawing-board. 



Let the point of the lower paper thus marked be denoted 

 byA . 



Further, let a pin-prick be made through N into the low T er 

 paper and let the point of it thus marked be denoted by N . 



Next let the squared paper be turned through a very 

 small angle about the pin through A until another of the 

 lines on it parallel to OX passes through the point P. 



