Question

by Charpits AE sobylve z=p^2x+q^2y

Answer 1

We rewrite this equation in the form

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

which allows us to highlight the function $F(x,y,z,p,q)=z-p^2x-q^2y$ , which we need for Charpit's Method.

Charpit's equation is:

$$\frac{dx}{F_p}=\frac{dy}{F_q}=\frac{dz}{pF_p+qF_q}=\frac{dp}{-\left(F_x+pF_z\right)}= \frac{dq}{-\left(F_y+qF_z\right)}$$

In our case,

$$F(x,y,z,p,q)=z-p^2x-q^2y\longrightarrow\\[0.3cm] F_p=-2px\quad F_q=-2qy\\[0.3cm] pF_p+qF_q=-2p^2x-2q^2x=-2\left(p^2x+q^2y\right)\equiv-2z\\[0.3cm] F_x+pF_z=-p^2+p\quad F_y+qF_z=-q^2+q$$

Then, our equation takes the form

$$\frac{dx}{-2px}=\frac{dy}{-2qy}=\frac{dz}{-2z}=\frac{dp}{p^2-p}= \frac{dq}{q^2-q}$$

1) We are interested in the first and fourth element:

$$\frac{dx}{-2px}=\frac{dp}{p^2-p}\longrightarrow\frac{dx}{-2px}=\frac{dp}{p(p-1)}\longrightarrow\frac{dx}{-2x}=\frac{dp}{(p-1)}\\[0.3cm] -\frac{1}{2}\cdot\int\frac{dx}{x}=\int\frac{dp}{p-1}\longrightarrow\\[0.3cm] -\frac{1}{2}\ln|x|=\ln|p-1|-\ln|A|\longrightarrow \ln\left|\frac{A}{\sqrt{x}}\right|=\ln|(p-1)|\longrightarrow\\[0.3cm] \frac{A}{\sqrt{x}}=p-1\longrightarrow\boxed{p=\frac{A+\sqrt{x}}{\sqrt{x}}}$$

2) We are interested in the second and fifth element:

$$\frac{dy}{-2qy}=\frac{dq}{q^2-q}\longrightarrow\frac{dy}{-2qy}=\frac{dq}{q(q-1)}\longrightarrow\frac{dy}{-2y}=\frac{dq}{(q-1)}\\[0.3cm] -\frac{1}{2}\cdot\int\frac{dy}{y}=\int\frac{dq}{q-1}\longrightarrow\\[0.3cm] -\frac{1}{2}\ln|y|=\ln|q-1|-\ln|B|\longrightarrow \ln\left|\frac{B}{\sqrt{y}}\right|=\ln|(q-1)|\longrightarrow\\[0.3cm] \frac{B}{\sqrt{y}}=q-1\longrightarrow\boxed{q=\frac{B+\sqrt{y}}{\sqrt{y}}}$$

It remains to substitute the obtained results in the initial equation:

$$z=p^2x+q^2y\longrightarrow z=\left(\frac{A+\sqrt{x}}{\sqrt{x}}\right)^2x+\left(\frac{B+\sqrt{y}}{\sqrt{y}}\right)^2y\\[0.3cm] z=\frac{A^2+2A\sqrt{x}+x}{x}\cdot x+\frac{B^2+2B\sqrt{y}+y}{y}\cdot y\\[0.3cm] \boxed{z=x+y+2\left(A\sqrt{x}+B\sqrt{y}\right)+A^2+B^2}$$

ANSWER

$$z=x+y+2\left(A\sqrt{x}+B\sqrt{y}\right)+A^2+B^2$$