i tried tried it using charpit method
$$f=z-pq$$
$$\frac{\partial f}{\partial x}=0,\frac{\partial f}{\partial y}=0,\frac{\partial f}{\partial p}=-q,\frac{\partial f}{\partial q}=-p,\frac{\partial f}{\partial z}=1$$
$$\frac{dx}{f_p}=\frac{dy}{f_q}=\frac{dz}{pf_p + qf_q}=\frac{dp}{−(f_x + pf_z)}=
\frac{dq}{−(f_y + qf_z)}$$
$$\frac{dx}{q}=\frac{dy}{p}=\frac{dz}{2pq }=\frac{dp}{p}=
\frac{dq}{q}$$
$\log p=\log qC$
,$p=qC$
now i can put the this in $z=pq,z=p^2C$
$p$ in terms of $(x,y,z,C)$
$$p=\sqrt{\frac{z}{C}},q=\sqrt{Cz}$$
writing $dz = p(x, y, z,C)dx + q(x, y, z, C)dy$
$$dz=\sqrt{\frac{z}{C}}dx+\sqrt{Cz}dy$$
$$\frac{dz}{\sqrt{z}}=adx+\frac{dy}{a}$$
$${2\sqrt{z}}=ax+\frac{y}{a}+b$$
$$4z=(ax+\frac{y}{a}+b)^2$$
Is my solution right ?
Best Answer
Charpit’s auxiliary equations are
$\dfrac{dp}{\dfrac{\partial f}{\partial x}+p{\dfrac{\partial f}{\partial z}}}=\dfrac{dq}{\dfrac{\partial f}{\partial y}+q{\dfrac{\partial f}{\partial z}}}=\dfrac{dz}{-p\dfrac{\partial f}{\partial p}-q{\dfrac{\partial f}{\partial q}}}=\dfrac{dx}{-p\dfrac{\partial f}{\partial p}}=\dfrac{dy}{-q\dfrac{\partial f}{\partial q}}=\dfrac{dF}{0}$
After getting all the required values, we have
$\dfrac{dp}{p}=\dfrac{dq}{q}=\dfrac{dz}{2pq}=\dfrac{dx}{q}=\dfrac{dy}{p}=\dfrac{dF}{0}$
Taking second and fourth factors, we get
$\dfrac{dq}{q}=\dfrac{dx}{q}\implies dq=dx$
Integrating, we get
$q=x+a$
After putting this value in the given equation, we get
$p=\dfrac{z}{x+a}$
Now $dz=pdx+qdy$ gives \begin{align} &dz =\dfrac{z}{x+a}dx+(x+a)dy\\ \implies &\dfrac{(x+a)dz-zdx}{(x+a)^2} =dy \end{align} On integration, we get
$\dfrac{z}{x+a}=y+b$
i.e $z=(x+a)(y+b)$
This is the required complete solution.