Answer to Question #86487 in Differential Equations for Kalipada

Question

P^2x+q^2y-z=0

Answer 1

Solution:

If

$$p^2x+q^2y-z=0$$

then

$$z=p^2x+q^2y.(1)$$

Find p and q the Charpit"s method:

$$\frac{dp}{\frac{df}{dx}+p\frac{df}{dz}}=\frac{dq}{\frac{df}{dy}+q\frac{df}{dz}}=\frac{dz}{-p\frac{df}{dp}-q\frac{df}{dq}}=\frac{dx}{-\frac{df}{dp}}=\frac{dy}{-\frac{df}{dq}},$$

where

$$f(x,y,z,p,q)=p^2x+q^2y-z,$$

$$\frac{df}{dx}=p^2,\frac{df}{dy}=q^2,\frac{df}{dz}=-1,\frac{df}{dp}=2xp,\frac{df}{dq}=2yq.$$

Then

$$\frac{dp}{p(p-1)}=\frac{dq}{q(q-1)}=\frac{dz}{-2(xp^2+yq^2)}=\frac{dx}{-2xp}=\frac{dy}{-2yq},$$

from here we have

$$\frac{dp}{p(p-1)}=\frac{dx}{-2xp},$$

$$\int\frac{dp}{p-1}=\int\frac{dx}{-2x},$$

$$ln(p-1)=ln(\frac{1}{\sqrt{x}})+ln(a),$$ $$p=\frac{a}{\sqrt{x}}+1;$$

$$\frac{dq}{q(q-1)}=\frac{dy}{-2yq},$$

$$\int\frac{dq}{q(q-1)}=\int\frac{dy}{-2yq},$$

$$ln(q-1)=ln(\frac{1}{\sqrt{y}})+ln(b),$$

$$q=\frac{b}{\sqrt{y}}+1.$$

Put p and q in (1):

$$z=(\frac{a}{\sqrt{x}}+1)^2x+(\frac{b}{\sqrt{y}}+1)^2y$$

or

$$z=x+y+2(a\sqrt{x}+b\sqrt{y})+a^2+b^2.$$

Answer:

$$z=x+y+2(a\sqrt{x}+b\sqrt{y})+a^2+b^2.$$