Thread: Method of Characteristics

1. Method of Characteristics

Hi All,

Can any one confirm that $\displaystyle \int\frac{1}{x+y}dx=ln(x+y)+C$ since the top is the derivative of the bottom? It is part of my bigger question for solving for u using method of characteristics
$\displaystyle \frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}=\frac{u}{x+y}$

Thanks

2. Is $\displaystyle y$ a function of $\displaystyle x$, or are we treating it as constant?

3. What you have is correct but not applicable here. Your method od charactistics are

$\displaystyle \dfrac{dx}{1} = \dfrac{dy}{-1} = \dfrac{(x+y)du}{u}$

From the first pair

$\displaystyle \dfrac{dx}{1} = \dfrac{dy}{-1} \;\; \Rightarrow\;\; x + y = c_1$

Then from the first and third

$\displaystyle \dfrac{dx}{1} = \dfrac{(x+y)du}{u}$

or

$\displaystyle \dfrac{dx}{c_1} = \dfrac{du}{u}\;\; \Rightarrow\;\; u e^{-\frac{x}{x+y} = c_2}$

Your general solution is $\displaystyle c_2 = f(c_1)$ or $\displaystyle u = e^{\frac{x}{x+y}}f(x+y)$

4. Danny and Prove it
Not sure I understand how you got the last term on the 2nd last line...? Can you clarify. FYI, my procedure is

dy/dx = -1, therefore y=-x+k, du/dx=u/(x+y) therefore $\displaystyle ln u=\int\frac{1}{x+y} dx = ln (x+y) +f(k)$.
Not sure what I have done wrong so far...
prove it, the above integration is based on assuming y is constant....not sure of this right

Thanks

5. Originally Posted by bugatti79
Danny and Prove it
Not sure I understand how you got the last term on the 2nd last line...? Can you clarify. FYI, my procedure is

dy/dx = -1, therefore y=-x+k, du/dx=u/(x+y) therefore $\displaystyle ln u=\int\frac{1}{x+y} dx = ln (x+y) +f(k)$.
Not sure what I have done wrong so far...
prove it, the above integration is based on assuming y is constant....not sure of this right

Thanks
But from what you have $\displaystyle x + y = k$. You must use this. Thus, $\displaystyle \dfrac{du}{dx} = \dfrac{u}{k}$.