Solving a non-linear Partial differential equation $px^5-4q^3x^2+6x^2z-2=0$

Question

I have to find out the complete integral of :
$$px^5-4q^3x^2+6x^2z-2=0$$
My attempt: Let $$f(x,y,z,p,q)=px^5-4q^3x^2+6x^2z-2$$
So, $$f_p=x^5,f_q=-12q^2x^2,f_x=5px^4-8q^3x+12xz,f_y=0,f_z=6x^2$$
Applying Charpit's method:
$$\frac{dx}{x^5}=\frac{dy}{-12q^2x^2}=\frac{dz}{px^5-12q^3x^2}=\frac{-dp}{5px^4-8q^3x+12xz+6px^2}=\frac{-dq}{6x^2q} $$ $$\implies \frac{dy}{-12q^2x^2}=\frac{dq}{-6x^2q}$$ $$\implies q=\sqrt{(y+c)}$$
where $c$ is an arbitrary constant.

I don't see here a way to find out $p$. Please help me solve this problem. Even a hint would be much help.

I'll appreciate any help towards it. Thanks in advance.

Answer 1

Hint:

$x^5z_x-4x^2z_y^3+6x^2z-2=0$

$x^3z_x-4z_y^3+6z-\dfrac{2}{x^2}=0$

$x^3z_{xy}-12z_y^2z_{yy}+6z_y=0$

Let $u=z_y$ ,

Then $x^3u_x-12u^2u_y+6u=0$

$12u^2u_y-x^3u_x=6u$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{du}{dt}=6u$ , letting $u(0)=1$ , we have $u=e^{6t}$

$\dfrac{dy}{dt}=12u^2=12e^{12t}$ , we have $y=e^{12t}+y_0=u^2+y_0$

$\dfrac{dx}{dt}=-x^3$ , we have $\dfrac{1}{x^2}=2t+f(y_0)=\dfrac{\ln u}{3}+f(y-u^2)$ , i.e. $u=e^\frac{3}{x^2}F(y-u^2)$ , i.e. $u^2=y-g\left(ue^{-\frac{3}{x^2}}\right)$

Answer 2

To find the complete integral of $$px^5-4q^3x^2+6x^2z-2=0 $$ we can use Charpit's method.
$$ Let~~f(x,y,z,p,q)= px^5-4q^3x^2+6x^2z-2=0 ~ \hspace{35mm}\textbf{(1)}$$ The auxiliary equations are $$\frac{dx}{f_p}= \frac{dy}{f_q}=\frac{dz}{pf_p+qf_q}=\frac{-dp}{f_x+pf_z}=\frac{-dq}{f_y+qf_z}$$ We have, $$ f_x = 5px^4-8q^3x+12xz, ~~~f_y=0,~~~f_z=6x^2,~~~f_p=x^5,~~~f_q=-12q^2x^2$$ Computing these values in auxiliary equations we get, $$\frac{dx}{x^5}= \frac{dy}{-12q^2x^2}=\frac{dz}{px^5-12q^3x^2}=\frac{-dp}{5px^4-8q^3x+12xz+6x^2z}=\frac{-dq}{0+6qx^2}$$ $$\Longrightarrow \frac{dy}{-12q^2x^2}=\frac{-dq}{6qx^2}$$ $$\Longrightarrow dy= \frac{-12q^2x^2 }{-6qx^2} dq = 2q~ dq$$ Integrating both sides we get, $$\Longrightarrow y+a = q^2 $$ $$\Longrightarrow q = \sqrt{y+a}$$ where, $a$ is an arbitrary constant.
Now substituting the value of $q$ in $\textbf{(1)}$ and solving for the value of $p$ we get, $$px^5=4q^3x^2-6x^2z+2 $$ $$\Longrightarrow p=\frac{4q^3x^2}{x^5}-\frac{6x^2z}{x^5}+\frac{2}{x^5} = \frac{4(a+y)^\frac{3}{2}}{x^3}-\frac{6z}{x^3}+\frac{2}{x^5}$$ Thus we have obtained the values of $p$ and $q$ as follows, $$p= \frac{4(a+y)^\frac{3}{2}}{x^3}-\frac{6z}{x^3}+\frac{2}{x^5} \hspace{20mm} q = (a+y)^\frac{1}{2}$$ Substituting values of $p$ and $q$ in $dz=p~dx+q~dy$ we get, $$dz=\left(\frac{4(a+y)^\frac{3}{2}}{x^3}-\frac{6z}{x^3}+\frac{2}{x^5}\right) ~dx+ (a+y)^\frac{1}{2}~dy$$ Multiplying by $\displaystyle{e^\frac{-3}{x^2}}$ on both sides we get, $$\displaystyle{e^\frac{-3}{x^2}}dz=\displaystyle{e^\frac{-3}{x^2}}\left(\frac{4(a+y)^\frac{3}{2}}{x^3}-\frac{6z}{x^3}+\frac{2}{x^5}\right) ~dx+ \displaystyle{e^\frac{-3}{x^2}}(a+y)^\frac{1}{2}~dy$$ $$\Longrightarrow \displaystyle{e^\frac{-3}{x^2}}dz=\displaystyle{e^\frac{-3}{x^2}}\frac{4(a+y)^\frac{3}{2}}{x^3}~dx-\displaystyle{e^\frac{-3}{x^2}} \frac{6z}{x^3}~dx+\displaystyle{e^\frac{-3}{x^2}}\frac{2}{x^5}~dx+ \displaystyle{e^\frac{-3}{x^2}}(a+y)^\frac{1}{2}~dy$$ $$\Longrightarrow \displaystyle{e^\frac{-3}{x^2}}dz -\displaystyle{e^\frac{-3}{x^2}} \frac{6z}{x^3}~dx =\displaystyle{e^\frac{-3}{x^2}}\frac{4(a+y)^\frac{3}{2}}{x^3}~dx+\displaystyle{e^\frac{-3}{x^2}}\frac{2}{x^5}~dx+ \displaystyle{e^\frac{-3}{x^2}}(a+y)^\frac{1}{2}~dy$$ $$\Longrightarrow \left(\displaystyle{e^\frac{-3}{x^2}}dz -z \displaystyle{e^\frac{-3}{x^2}} \frac{6}{x^3}~dx\right) =\left(\displaystyle{e^\frac{-3}{x^2}}\frac{2}{x^5}~dx\right) +\left(\displaystyle{e^\frac{-3}{x^2}}\frac{4(a+y)^\frac{3}{2}}{x^3}~dx+ \displaystyle{e^\frac{-3}{x^2}}(a+y)^\frac{1}{2}~dy\right) \hspace{5mm} \textbf{(2)}$$ Put $\displaystyle{\frac{-3}{x^2}}=u$ $\Longrightarrow \displaystyle{\frac{6}{x^3}}dx=du$
Using this substitution in $\textbf{(2)}$ we obtain, $$\left(\displaystyle{e^u}dz -z \displaystyle{e^u} ~du\right) =\left(\displaystyle{\frac{-1}{9}ue^u}~du\right) +\left(\displaystyle{\frac{2e^u(a+y)^\frac{3}{2}}{3}}~du+ \displaystyle{e^u}(a+y)^\frac{1}{2}~dy\right)$$ $$\Longrightarrow d(ze^u) = \frac{-1}{9} d \left( (u-1) e^u \right) + \frac{2}{3} d\left( (a+y)^\frac{3}{2} e^u \right) $$ On integrating we obtain, $$\Longrightarrow ze^u = \frac{-1}{9} \left( (u-1) e^u \right) + \frac{2}{3} \left( (a+y)^\frac{3}{2} e^u \right) +b $$ $$\Longrightarrow z = \frac{-1}{9} \left( (u-1) \right) + \frac{2}{3} \left( (a+y)^\frac{3}{2} \right) +be^{-u} $$ But $u=\frac{-3}{x^2}$,
$$ \therefore ~~ z = \frac{2}{3} \left( (a+y)^\frac{3}{2} \right) +be^{\frac{3}{x^2}} + \frac{3}{x^2} + \frac{1}{9} $$

Answer 3

Since $q:=\frac{\partial z}{\partial y}$, it follows from $q=\sqrt{y+c}$ that $$ \frac{\partial z}{\partial y}=\sqrt{y+c} \implies z(x,y)=\frac{2}{3}(y+c)^{3/2}+f(x), \tag{1} $$ hence $$ p=\frac{\partial z}{\partial x}=f'(x). \tag{2} $$ Substituting $z, p,$ and $q$ in the PDE $px^5-4q^3x^2+6x^2z-2=0$, we obtain the following ODE: $$ f'(x)+\frac{6}{x^3}f(x)=\frac{2}{x^5}. \tag{3} $$ To solve it, we multiply both sides of $(3)$ by $e^{-3/x^2}$ and integrate the result: $$ e^{-3/x^2}\left(f'(x)+\frac{6}{x^3}f(x)\right)=\frac{d}{dx}\left(e^{-3/x^2}f(x)\right)=\frac{2}{x^5}e^{-3/x^2} $$ $$ \implies f(x)=e^{3/x^2}\int\frac{2}{x^5}e^{-3/x^2}dx=\frac{x^2+3}{9x^2}+be^{3/x^2}. \tag{4} $$ Substituting $(4)$ in $(1)$ we finally obtain $$ z(x,y)=\frac{2}{3}(y+c)^{3/2}+\frac{x^2+3}{9x^2}+be^{3/x^2}. \tag{5} $$

Answer 4

From given pde and by using the value q, we get $P=\frac{4(y+c)^{\frac{3}{2}}}{x^3}-\frac{6z}{x^3}+\frac{2}{x^5}$ and subtitute the values $p$ and $q$ in $Z=pdx+qdy$ and use multiplier $e^{-\frac{3}{x^2}}$ and on integration we get complete required complete integral