F=px⁵-4q³x²+6x²z-3=0 where p=∂z/∂x, q=∂z/∂y for z=f(x,y).
Charpit’s method:
∂F/∂p=x⁵, ∂F/∂q=-12q²x², ∂F/∂x=5px⁴-8q³x+12xz, ∂F/∂y=0, ∂F/∂z=6x².
Now we introduce the integrables:
①dx/(∂F/∂p)=dx/x⁵=
②dy/(∂F/∂q)=-dy/(12q²x²)=
③dz/(p∂F/∂p+q∂F/∂q)=dz/(px⁵-12q³x²)=
④-dp/(∂F/∂x+p∂F/∂z)=-dp/(5p⁴-8q³x+12xz+6px²)=
⑤-dq/(∂F/∂y+q∂F/∂z)=-dq/(6qx²).
From ② and ⑤ we can reduce to two variables:
-dy/(12q²x²)=-dq/(6qx²), dy/(2q)=dq, so dy=2qdq which can be integrated: y=q²+a where a is constant.
From ① and ② we can find dy/dx:
dx/x⁵=-dy/(12q²x²), so dx/x³=-dy/(12q²)=-dy/(12(y-a)), which can be integrated:
12∫(dx/x³)=-∫(dy/(y-a))⇒-6/x²=-ln(y-a)+constant. [dy/dx=-12(y-a)/x³.]
So, 6/x²=ln(y-a)+k, where k is an integration constant, which can be absorbed into the log argument. So if k=ln(A), 6/x²=ln(Ay-B) where B=Aa. From this e^(6/x²)=Ay-B, and y=(e^(6/x²)+B)/A. Again the constants can be replaced with other constants:
y=be^(6/x²)+c (that is, b=1/A and c=B/A=a).
[Also, since q²=y-a, 6/x²=ln(q²)+k, 6/x²=2ln(q)+k, or 3/x²=ln(q)+k₁ where k₁=k/2, or q=e^(3/x²-k₁) or q=k₂e^(3/x²) where constant k₂=1/e^k₁. The subscripts are just used to show that the constants are related. We can replace k₂ with constant d:
q=de^(3/x²). The same result is obtained from ① and ⑤.]
The standard differentiation process is:
dz/dx=∂z/∂x+(∂z/∂y)dy/dx=p+qdy/dx. We can substitute for dy/dx=-12q²/x³ to get:
dz/dx=p-12q³/x³. (The same result can be obtained from the Charpit’s equations ➀ and ➂.) We have q in terms of x, but we need to find p in terms of x in order to consider integration.
However, z=f(x,y) can be reduced to z=g(x) because we know y in terms of x already. p=∂z/∂x=dz/dx when there are no y terms, and q=∂z/∂y=0. So F can be reduced to px⁵+6x²z-3=0⇒x⁵dz/dx+6x²z=3. This is a 1st order DE, and we can write it: dz/dx+6z/x³=3/x⁵ and using the integrating factor e^∫((6/x³)dx)=e^(-3/x²) we can try to solve the DE.
d(ze^(-3/x²))/dx=3e^(-3/x²)/x⁵; ze^(-3/x²)=3∫((e^(-3/x²)/x⁵)dx).
To integrate the expression on the right, let s=-3/x² then ds=6/x³dx.
We have 3∫((eˢ/x⁵)(x³/6)ds)=½∫(eˢds/x²)=-⅙∫(seˢds), when we substitute for x²=-3/s. Integrating by parts we get -⅙(seˢ-eˢ)=⅙(e^(-3/x²)+3e^(-3/x²)/x²)+r, where r is constant of integration.
So z=⅙(1+3/x²)+re^(3/x²), y=be^(6/x²)+c, where b, c, r are constants of integration. Since c=a we could write y=be^(6/x²)+a. Note that e^(6/x²)=(y-a)/b, and e^(3/x²)=√e^(6/x²)=√((y-a)/b), so z can be written in terms of x and y, z=f(x,y)=⅙(1+3/x²)+r√((y-a)/b).