Verify Solutions to First Order PDEs

[Oxymoron]

Just need some verification.

Question 1

Find the general solutions of the following first order PDE

[tex]z_x - yz_y = z[/tex]

Question 2

Find the general solution of the following first order PDE

[tex]x^2z_x+y^2z_y = xy[/tex]

 

[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.

[tex]\frac{dy}{dx} = -y[/tex]
[tex]\frac{dz}{dx} = z[/tex]

Solving these gives me

[tex]y = c_1e^{-x}[/tex]
[tex]z = c_2e^x[/tex]

In proper form...

[tex]c_1 = ye^x[/tex]
[tex]c_2 = ze^{-x}[/tex]

So the general solution is

[tex]c_2 = g(c_1)[/tex]


[tex]z = e^xg(ye^x)[/tex]

Checking this solution is correct, I computed

[tex]z_x = e^xg(ye^x) + e^xg'(ye^x)ye^x = e^xg(ye^x) + ye^{2x}g'(ye^x) [/tex]

[tex]z_y = e^xg'(ye^x)e^x = e^{2x}g'(ye^x)[/tex]

And substitute these into the initial PDE we get

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

Note that I had earlier defined [itex]z = e^xg(ye^x)[/itex]

 

[dextercioby]

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

Daniel.

 

[Oxymoron]

Characteristic Equations are

[tex]\frac{dy}{dx} = \frac{y^2}{x^2}[/tex]

[tex]\frac{dz}{dx} = \frac{y}{x}[/tex]

Solving these two differential equations...

(1)
[tex]\int\frac{dy}{y^2} = \int\frac{dx}{x^2}[/tex]
[tex]-\frac{1}{y} = -\frac{1}{x} + c_1[/tex]
[tex]c_1 = \frac{1}{x} - \frac{1}{y}[/tex]

(2)
[tex]\int dz = \int\frac{ydx}{x}[/tex]
[tex]z = y\ln x + c_2[/tex]
[tex]c_2 = z - y\ln x[/tex]

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

[tex]z - y\ln x = g\left(\frac{1}{x}-\frac{1}{y}\right)[/tex]

[tex]z = g\left(\frac{1}{x}-\frac{1}{y}\right) + y\ln x[/tex]

However when I check this I don't get equality...

[tex]z_x = g'\left(\frac{1}{x}-\frac{1}{y}\right)\left(-\frac{1}{x^2}\right) + \frac{y}{x}[/tex]

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

And

[tex]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[/tex]
[tex]\neq g\left(\frac{1}{x}-\frac{1}{y}\right) + y\ln x[/tex]

And I can't see that I made an error in obtaining [itex]z = g\left(\frac{1}{x}-\frac{1}{y}\right)[/itex]

Can anyone see where I went wrong?

 

[saltydog]

Hello Oxymoron,

I also get:

[tex]c_1 = \frac{1}{x} - \frac{1}{y}[/tex]

However at that point, I am to let:

[tex]w= \frac{1}{x} - \frac{1}{y}\quad\text{ and }\quad r=y[/tex]

Letting:

[tex]V(w,r)\equiv z(x,y)[/tex]

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

[tex]r^2V_r-\frac{r^2}{wr+1}=0[/tex]

Solving for V(w,r) I get:

[tex]V(w,r)=\frac{1}{w}ln(wr+1)+F(w)[/tex]

substituting back x and y I get:

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

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