How to integrate this partial differential equation

**JulieK** · December 28th 2012, 07:06 AM

I have the following equation

$\frac{\partial}{\partial y}\left(m\frac{dy}{dx}\right)=0$

where $y$ is a function of $x$ and $m$ is a function of $y$ . If I integrate this equation first with respect to $y$ should I get a function of $x$ as the constant of integration (say $C\left(x\right)$ ) or it is just a constant? If it is a function, how can I then find its form (e.g. polynomial, etc.)? Should I use boundary conditions or I can decide about the form from inspecting the type of the equation.

**Prove It** · December 28th 2012, 04:56 PM

Originally Posted by JulieK

I have the following equation

$\frac{\partial}{\partial y}\left(m\frac{dy}{dx}\right)=0$

where $y$ is a function of $x$ and $m$ is a function of $y$ . If I integrate this equation first with respect to $y$ should I get a function of $x$ as the constant of integration (say $C\left(x\right)$ ) or it is just a constant? If it is a function, how can I then find its form (e.g. polynomial, etc.)? Should I use boundary conditions or I can decide about the form from inspecting the type of the equation.

$\displaystyle \begin{align*} \frac{\partial }{\partial y} \left( m\, \frac{dy}{dx} \right) &= 0 \\ m\,\frac{dy}{dx} &= \int{0\, dy} \\ m\,\frac{dy}{dx} &= f(x) \\ \frac{dy}{dx} &= \frac{f(x)}{m} \\ y &= \int{\frac{f(x)}{m}\,dx} \\ y &= \frac{1}{m} \int{f(x)\,dx} + C \end{align*}$

Unless you have some boundary conditions now, this is the simplest it's going to get.

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