Question

Using Charpit’s method, solve the equation:

zp² -y²p +y²q =0

Answer 1

Using the Charpit's method, we shall solve PDE

$zp²-y²p+y²q$

Consider

$f(x,y,z,p,q)=0$

Given the PDE $zp²-y²p+y²q$

We have that

$f(x,y,z,p,q)$ $=zp²-y²p+y²q=0$

We have the formula

$\displaystyle\frac{dp}{\frac{\partial f}{\partial x}+p\frac{\partial f}{\partial z}}=\frac{dq}{\frac{\partial f}{\partial y}+q\frac{\partial f}{\partial z}}=\frac{dz}{-p\frac{\partial f}{\partial p}-q\frac{\partial f}{\partial q}}=\frac{dx}{-\frac{\partial f}{\partial p}}=\frac{dy}{-\frac{\partial f}{\partial q}}$

It follows that

$\frac{\partial f}{\partial x}=0,\frac{\partial f}{\partial y}=-2yp+2yq,\frac{\partial f}{\partial z}=p²,\frac{\partial f}{\partial p}=2pz-2py,\frac{\partial f}{\partial q}=y²$

Substituting, we have

$\displaystyle \implies\frac{dp}{p³}=\frac{dq}{2py+2qy+p²q}=\frac{dz}{-p(2pz-y²)-q(y²)}=\frac{dy}{-y²}$

Consider

$\frac{dp}{p³}=\frac{dy}{-y²}$

We have that

$\displaystyle p^{-3}dp=-y^{-2}dy$

Integrating, we have

$\implies\int p^{-3}dp=\int -y^{-2}dy\\\implies\int p^{-3}dp=-\int y^{-2}dy$

$\implies \frac{p^{-2}}{-2}=y^{-1}+c$

Where $c$ is the integrating constant.

$\implies \frac{1}{p²}=\frac{-2}{y}+c\\\implies y=c-2p²$

by making $p$ the subject of the relation

$\implies p=\sqrt{\frac{c-y}{2}}$

Now, Substituting the value of $p$ into the given PDE

We have

$\implies z(\frac{a-y}{2})-y²\sqrt{\frac{a-y}{2}}+y²q=0$

$\implies y²q=y²\sqrt{\frac{a-y}{2}}-z(\frac{a-y}{2})$

$\implies q=\sqrt{\frac{a-y}{2}}-\frac{z}{y²}(\frac{a-y}{2})$

we have been able to get values for $p\hspace{0.1cm}and\hspace{0.1cm}q$

NOW,

we know that $dz=p\hspace{0.05cm}dx+q\hspace{0.05cm}dy$

By substituting $p\hspace{0.1cm}and\hspace{0.1cm}q\hspace{0.1cm}in\hspace{0.1cm}dz$

We have

$dz=\sqrt{\frac{a-y}{2}}dx+\left\{\sqrt{\frac{a-y}{2}}-\frac{z}{y²}(\frac{a-y}{2})\right\}dy$

It follows that,

$\displaystyle dz=\int\sqrt{\frac{a-y}{2}}dx+\int \left\{\sqrt{\frac{a-y}{2}}-\frac{z}{y²}(\frac{a-y}{2})\right\}dy$

By integration, we have our final result to be

$\displaystyle z=\frac{x\sqrt{c-y}}{\sqrt{2}}+\left[\frac{\sqrt{2(c-y)}}{3}(y-c)+z\frac{c+y\hspace{0.05cm}log(y)}{2y}\right]$

Which is the solution of the PDE

$zp²-y²p+y²q$