# A Simple PDE

1. May 18, 2005

### Oxymoron

Just need some verification.

Question 1

Find the general solutions of the following first order PDE

$$z_x - yz_y = z$$

Question 2

Find the general solution of the following first order PDE

$$x^2z_x+y^2z_y = xy$$

Last edited: May 18, 2005
2. May 18, 2005

### Oxymoron

Solution to Question 1

The first thing I did was find the characteristic equations of which I want to combine to form a surface of solutions.

$$\frac{dy}{dx} = -y$$
$$\frac{dz}{dx} = z$$

Solving these gives me

$$y = c_1e^{-x}$$
$$z = c_2e^x$$

In proper form...

$$c_1 = ye^x$$
$$c_2 = ze^{-x}$$

So the general solution is

$$c_2 = g(c_1)$$

$$z = e^xg(ye^x)$$

Checking this solution is correct, I computed

$$z_x = e^xg(ye^x) + e^xg'(ye^x)ye^x = e^xg(ye^x) + ye^{2x}g'(ye^x)$$

$$z_y = e^xg'(ye^x)e^x = e^{2x}g'(ye^x)$$

And substitute these into the initial PDE we get

$$z_x - yz_y = e^xg(ye^x) + ye^{2x}g'(ye^x) - y(e^{2x}g'(ye^x))$$
$$\quad = e^xg(ye^x) + ye^{2x}g'(ye^x) - ye^{2x}g'(ye^x)$$
$$\quad = e^xg(ye^x)$$
$$\quad = z$$

Note that I had earlier defined $z = e^xg(ye^x)$

3. May 18, 2005

### dextercioby

Yes,it looks okay.Can u do the same with the second...?

Daniel.

4. May 18, 2005

### Oxymoron

Characteristic Equations are

$$\frac{dy}{dx} = \frac{y^2}{x^2}$$

$$\frac{dz}{dx} = \frac{y}{x}$$

Solving these two differential equations...

(1)
$$\int\frac{dy}{y^2} = \int\frac{dx}{x^2}$$
$$-\frac{1}{y} = -\frac{1}{x} + c_1$$
$$c_1 = \frac{1}{x} - \frac{1}{y}$$

(2)
$$\int dz = \int\frac{ydx}{x}$$
$$z = y\ln x + c_2$$
$$c_2 = z - y\ln x$$

Hence the general solution is $c_2 = g\left(\frac{1}{x}-\frac{1}{y}\right)$.

$$z - y\ln x = g\left(\frac{1}{x}-\frac{1}{y}\right)$$

$$z = g\left(\frac{1}{x}-\frac{1}{y}\right) + y\ln x$$

However when I check this I dont get equality...

$$z_x = g'\left(\frac{1}{x}-\frac{1}{y}\right)\left(-\frac{1}{x^2}\right) + \frac{y}{x}$$

$$z_y = g'\left(\frac{1}{x}-\frac{1}{y}\right)\left(\frac{1}{y^2}\right) + \ln x$$

And

$$x^2z_x - y^2z_y = -g'\left(\frac{1}{x}-\frac{1}{y}\right)+ \frac{y}{x} - g'\left(\frac{1}{x}-\frac{1}{y}\right) - y^2 \ln x$$
$$\neq g\left(\frac{1}{x}-\frac{1}{y}\right) + y\ln x$$

And I cant see that I made an error in obtaining $z = g\left(\frac{1}{x}-\frac{1}{y}\right)$

Can anyone see where I went wrong?

Last edited: May 18, 2005
5. May 18, 2005

### saltydog

Hello Oxymoron,

I also get:

$$c_1 = \frac{1}{x} - \frac{1}{y}$$

However at that point, I am to let:

$$w= \frac{1}{x} - \frac{1}{y}\quad\text{ and }\quad r=y$$

Letting:

$$V(w,r)\equiv z(x,y)$$

and substituting into the original equation, I end up with:

$$r^2V_r-\frac{r^2}{wr+1}=0$$

Solving for V(w,r) I get:

$$V(w,r)=\frac{1}{w}ln(wr+1)+F(w)$$

substituting back x and y I get:

$$z(x,y)=\frac{xy}{y-x}ln[\frac{y-x}{x}+1]+F[\frac{1}{x}-\frac{1}{y}]$$

That is, I'm not familiar with your use of (2) above. However backsubstitution of this does satisfy the PDE.