Let f(x,y,z,p,q)≡2z+p2+qy+2y2=0 ............(1)
Charpit's auxiliary equation are
fx+pfzdp=fy+qfzdq=−pfp−qfqdz=−fpdx=−fqdy
....................(2)
From (1), fx=0, fy=q+4y, fz=2, fp=2p, fq=y .......(3)
Using (3),(2) reduce to
2pdp=3q+4ydq=−2p2−qydz=−2pdx=−ydy
Taking 1st and 4th fraction we get,
⟹2pdp=−2pdx
So that , P=−x ..............(4)
Now taking 2nd and 5th fraction ,we get
3q+4ydq=−ydy
⟹dydq=−y3q+4y
⟹dydq+y3q=−4
Which is a linear equation in first order.
∴I.F=e∫y3dy=e3logy=elogy3=y3
Now , q y3=∫(−4)y3dy=−y4
⟹q=−y .
Now putting the value of p and q in
dz=pdx+qdy we get dz=−xdx−ydy
⟹2z=−x2−y2+c
Where "c" is a integration constant.
Comments