Question

solve the following P.D.E x(z²-y²)p+y(x²-z²)q = z(y²-x²)​

Answer 1

Step-by-step explanation:

Here p=∂z∂xp=∂z∂x and q=∂z∂yq=∂z∂y

Lagrange's equations are dxx2−yz=dyy2−zx=dzz2−xydxx2−yz=dyy2−zx=dzz2−xy

Let the general solution be ϕ(C1,C2)=0ϕ(C1,C2)=0

By Choosing multipliers x,y,zx,y,z and 1,1,11,1,1 we get C1C1 as below.

xdx+ydy+zdzx3+y3+z3−3xyz=dx+dy+dzx2+y2+z2−xy−yz−zx

xdx+ydy+zdzx3+y3+z3−3xyz=dx+dy+dzx2+y2+z2−xy−yz−zx

xdx+ydy+zdz(x+y+z)(x2+y2+z2−xy−yz−zx)=dx+dy+dzx2+y2+z2−xy−yz−zx

xdx+ydy+zdz(x+y+z)(x2+y2+z2−xy−yz−zx)=dx+dy+dzx2+y2+z2−xy−yz−zx

xdx+ydy+zdzx+y+z=dx+dy+dz

xdx+ydy+zdzx+y+z=dx+dy+dz

xdx+ydy+zdz=(x+y+z)d(x+y+z)

xdx+ydy+zdz=(x+y+z)d(x+y+z)

x22+y22+z22=(x+y+z)22+C

x22+y22+z22=(x+y+z)22+C

x2+y2+z2−(x+y+z)2=C1

x2+y2+z2−(x+y+z)2=C1

Now I am not able to find C2C2. The answer given in the textbook is (x−y)(xy+yz+zx)+y−z=0(x−y)(xy+yz+zx)+y−z=0