# How can I actually solve this kind of partial differential equations?

$$x \frac{\partial z}{\partial x}+t \frac{\partial z}{\partial t}+y \frac{\partial z}{\partial y}=xyt$$

I can see that one soultion for this equation is $$z=(1/3)xyt+ C$$ however how can one solve this kind of problems within a proper way and reach more general solutions?

• You can use the "method of characteristics" (see, e.g. Evans's book on PDEs, chapter 4). This amounts exactly to the computations in Hans Lundmark's answer. May 23 '16 at 15:48

You have found one particular solution, and since the PDE is linear any other solution will differ from that one by a solution of the homogeneous equation: $$x \frac{\partial z}{\partial x}+t \frac{\partial z}{\partial t}+y \frac{\partial z}{\partial y}=0 .$$ This equation says that the directional derivative of $z$ in the radial direction is zero; in other words, $z$ is constant on each ray from the origin.

So the general solution of your PDE is $$z(x,y,t) = \frac{xyt}{3} + g(x,y,t)$$ where $g$ is a function which depends only on the direction of the vector $(x,y,t)$, not on its magnitude. (Expressed in spherical coordinates, $g$ would depend only on the angles $\theta$ and $\phi$, not on the radial variable $r$.)

There are some issues to consider if you want a smooth function at the origin ($g$ must be constant in that case), but this is to be expected since the left-hand side of your PDE is zero there.

$$x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}+t \frac{\partial z}{\partial t}=xyt$$ Solving thanks to the method of characteristics :

The characteristic equations are : $$\frac{dx}{x}=\frac{dy}{y}=\frac{dt}{t}=\frac{dz}{xyt}$$

From $\frac{dx}{x}=\frac{dt}{t}$ the first characteristic : $\frac{x}{t}=c_1$

From $\frac{dy}{y}=\frac{dt}{t}$ the second characteristic : $\frac{y}{t}=c_2$

From the combination of the characteristic equations, $\frac{dx}{x}=\frac{dy}{y}=\frac{dt}{t}= \frac{ytdx+xtdy+xydt}{ytx+xty+xyt}=\frac{ytdx+xtdy+xydt}{3xyt}=\frac{d(xyt)}{3xyt} =\frac{dz}{xyt}$

$dz=\frac{d(xyt)}{3} \quad\to\quad$ the third characteristic : $\quad z-\frac{xyt}{3}=c_3$

Thus, the general solution on implicit form is :
$$\Phi\left(\frac{x}{t} \:,\: \frac{y}{t} \:,\: (z-\frac{xyt}{3})\right)=0$$ where $\Phi$ is any differentiable function of three variables.

Solving it for the third variable leads to the explicit form : $$z-\frac{xyt}{3}=F\left(\frac{x}{t} \:,\: \frac{y}{t}\right)$$ where $F$ is any differentiable function of two variables. $$z(x,y,t)=\frac{xyt}{3}+F\left(\frac{x}{t} \:,\: \frac{y}{t}\right)$$