x(y2+z)dx∂z−y(x2+z)dy∂z=(x2−y2)z
x+y=0,z=1
The auxiliary equations is:
x(y2+z)dx=−y(x2+z)dy=(x2−y2)zdz
A first characteristic equation comes from
x2(y2+z)−y2(x2+z)xdx+ydy=(x2−y2)zdz
(x2−y2)zxdx+ydy=(x2−y2)zdz
xdx+ydy=dz
2x2+y2=z+2C1
C1=x2+y2−2z
A second characteristic equation comes from
xy(y2+z)−xy(x2+z)ydx+xdy=(x2−y2)zdz
−xy(x2−y2)ydx+xdy=(x2−y2)zdz
−xyd(xy)=zdz
−ln∣xy∣+lnC2=ln∣z∣
xyC2=z
C2=xyz
General solution of the PDE on the form of implicit equation:
Φ(C1,C2)=0.
C1=x2+y2−2z=x2+y2+2xy−2xy−2z=(x+y)2−2xy−2z (1)
C2=xyz (2)
Substitute (2) into (1):
C1=(x+y)2−2zC2−2z
Use x+y=0,z=1 to exclude C1 and C2 :
C1=(0)2−21C2−2⋅1=−2C2−2
−2C2−2=x2+y2−2z
−2xyz−2=x2+y2−2z
z=2x2+y2+xyz+1
Answer: z=2x2+y2+xyz+1 .
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