Verify Solutions to First Order PDEs

In summary: Thanks, I see where I went wrong now. In my substitution I didn't account for the fact that y = c_1e^{-x} = e^{-x}\left(\frac{1}{x} - \frac{1}{y}\right). I should have replaced \frac{1}{y} with \frac{1}{e^{-x}\left(\frac{1}{x} - \frac{1}{y}\right)}. Thanks for your help.In summary, the general solutions for the first order PDEs z_x - yz_y = z and x^2z_x + y^2z_y = xy are z = e^xg(ye^x) and z = \
  • #1
Oxymoron
870
0
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]
 
Last edited:
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  • #2
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]
 
  • #3
Yes,it looks okay.Can u do the same with the second...?

Daniel.
 
  • #4
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?
 
Last edited:
  • #5
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.
 

Related to Verify Solutions to First Order PDEs

What is a first order partial differential equation (PDE)?

A first order partial differential equation (PDE) is a mathematical equation that involves partial derivatives of an unknown function with respect to one or more independent variables. It can be written in the form F(x,y,u,∂u/∂x,∂u/∂y) = 0, where u is the unknown function and F is a function of the independent variables and their partial derivatives.

What is the difference between verifying and solving a first order PDE?

Verifying a first order PDE involves checking whether a given function satisfies the equation, while solving a first order PDE involves finding the unknown function that satisfies the equation and any given boundary conditions.

What are the common methods used to verify solutions to first order PDEs?

The most commonly used methods to verify solutions to first order PDEs are substitution, differentiation, and integration. These methods involve substituting the given function into the PDE, differentiating or integrating both sides of the equation, and then comparing the results to see if they are equal.

Why is it important to verify solutions to first order PDEs?

Verifying solutions to first order PDEs is important because it ensures the accuracy of the solution and helps to identify any potential errors. It also allows for the validation of the solution and gives a better understanding of the underlying mathematics.

What are some common mistakes to avoid when verifying solutions to first order PDEs?

Some common mistakes to avoid when verifying solutions to first order PDEs include forgetting to consider boundary conditions, making arithmetic errors, and mistaking the independent and dependent variables. It is important to double-check all calculations and ensure that all variables are correctly identified and accounted for.

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