Verify Solutions to First Order PDEs

[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

 

[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)

 

[dextercioby]

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

Daniel.

 

[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 don't 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 can't 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?

 

[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.