I have 2 questions
- Suppose i have this pde
$x\dfrac{\partial z}{\partial x}+y\dfrac{\partial z}{\partial y}+t\dfrac{\partial z}{\partial t}=xyt$
And the auxiliary equation is
$\dfrac{\partial x}{x}=\dfrac{\partial y}{y}=\dfrac{\partial t}{t}=\dfrac{\partial z}{xyt}$
I know that the last solution will be has this form :
$\Phi(a,b,c)=0$
So, i have to find out what $a$, $b$, and $c$ are.
From $\dfrac{\partial x}{x}=\dfrac{\partial y}{y}$, i got $a=\dfrac{x}{y}$
From $\dfrac{\partial y}{y}=\dfrac{\partial t}{t}$, i got $b=\dfrac{y}{t}$
- And i want to write $\Phi(a,b,c)=0$ as explicit in $z$ so, i have to solve for $dz$
Suppose i want to solve for
$\dfrac{\partial t}{t}=\dfrac{\partial z}{xyt}$
And i'm stuck here. I was confused with the denominator $xyt$ that is not has $z$ term. How do i solve this? And we know, if i have a solution $F(a,b)=0$, it can be written in explicit term as $F(a)=b$.
But how about 3 variables that is in my case i have a solution $\Phi(a,b,c)=0$?
Please help me to solve this. Thanks.
$$x\dfrac{\partial z}{\partial x}+y\dfrac{\partial z}{\partial y}+t\dfrac{\partial z}{\partial t}=xyt$$
$$\dfrac{d x}{x}=\dfrac{d y}{y}=\dfrac{d t}{t}=\dfrac{d z}{xyt}$$
A first characteristic equation comes from $\dfrac{d x}{x}=\dfrac{d y}{y}$ $$\frac{x}{y}=c_1 \tag 1$$ A second characteristic equation comes from $\dfrac{d y}{y}=\dfrac{d t}{t}$ $$\frac{y}{t}=c_2 \tag 2$$
A third characteristic equation comes from
$$\dfrac{d x}{x}=\dfrac{d y}{y}=\dfrac{d t}{t}= \frac{(yt)dx+(xt)dy+(xy)dt}{(yt)x+(xt)y+(xy)t}=\dfrac{d (xyt)}{3(xyt)}=\dfrac{d z}{xyt}\qquad\text{See note below.}$$ $$z-\frac13 xyt=c_3 \tag 3$$
General solution of the PDE on the form of implicit equation $\Phi(c_1, c_2,c_3)=0$
Or equivalently on the form : $c_3=F(c_1, c_2)=z-\frac13 xyt=F\left(\frac{x}{y}\:,\:\frac{y}{t} \right)$
$$\boxed{z(x,y,t)=\frac13 xyt+F\left(\frac{x}{y}\:,\:\frac{y}{t} \right)}$$ This consistent with the result from Mathos $z(x,y,t)=\frac13 xyt+f\left(\frac{x}{t}\:,\:\frac{y}{t} \right)$ since $F$ and $f$ are both arbitrary functions : $F(X,Y)=f(XY,Y)$.
NOTE : The relationship between equal fractions is a well known property in elementary fraction calculus. If $\frac{A}{B}=\frac{C}{D}$ then $\frac{A}{B}=\frac{C}{D}=\frac{k_1A+k_2C}{k_1B+k_2D}$ with any $k_1, k_2$ not both nul.