Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am asked to solve $$u_x+u_y=1$$ If is was homogeneous i.e., $u_x+u_y$ the answer would be $u(x,y)=f(y-x)$ where $f$ is an arbitrary function. I have found the following set of solutions: $$u(x,y)=\lambda x +(1-\lambda)y$$ where $\lambda$ is an arbitrary constant(real or imaginary). I just have no idea what method other then trial and error would have lead me here. Any ideas? Thanks!

share|cite|improve this question
This is a linear first order PDE. Googling will provide lots of write-ups on how to solve it! – Mariano Suárez-Alvarez Feb 14 '13 at 21:04
Method of characteristics is used on general first-order PDEs. – Ron Gordon Feb 14 '13 at 21:50
up vote 5 down vote accepted

You can observe that the only difference between homogeneous an inhomogeneous equations is $1$. So you can assume that particular solution is linear on both $x$ and $y$, or $u^p = ax + by$. $$ u_x^p + u_y^p = a + b = 1 $$ In your case $a = \lambda$ and $b = 1 - \lambda$. General solution of inhomogeneous PDE given is the sum of general solution of homogeneous PDE and particular solution of inhomogeneous PDE, so $$ u = f(x-y)+\lambda x + (1-\lambda)y $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.