# Question of integration

1. Feb 23, 2010

### BustedBreaks

So in doing a homework problem I have convinced my self that $$\int\frac{V_{x}}{V}=ln(V)$$ which I vaguely remember learning in class, but I'm having trouble deriving it. Can someone help me out?

2. Feb 23, 2010

### CompuChip

You mean, you want to show that
$$\frac{d}{dx} \ln(V(x)) = \frac{dV/dx}{x}$$ ?

Well, this follows from the chain rule and the fact that $\frac{d}{du} \ln(u) = \frac{1}{u}$.
The latter can be derived, for example by differentiating $$x = e^{\ln(x)}$$ w.r.t. x using the chain rule, and solving for d ln(x)/dx.

3. Feb 23, 2010

### BustedBreaks

To give a bit more context.

I was trying to solve the partial differential equation: $$3U_{y}+U_{xy}=0$$ with the hint, let $$V=U_y$$

substituting we have $$3V+V_{x}=0$$
then $$-3=\frac{V_{x}}{V}$$

I didn't really know how to continue from here so I just played around and figured out that $$V=e^{3x}$$ and $$U(x,y)=ye^{-3x}$$

I'm trying to figure out what to do when I come to $$-3=\frac{V_{x}}{V}$$.

4. Feb 23, 2010

### CompuChip

Ah, then indeed you can use the integral you gave.

An alternative is to note that 3V + Vx = 0 is a linear ordinary differential equation. In that case, you can usually stick in $V(x) = e^{\lambda x}$ as a "trial" solution. The differential equation then turns into a polynomial equation for $\lambda$ with solutions $\lambda_1, \cdots, \lambda_n$, where n is the order of the ODE (in this case, n = 1). The general solution to the ODE is then
$$V(x) = A_1 e^{\lambda_1 x} + \cdots + A_n e^{\lambda_n x}[/itex] where the Ai[/i] are to be determined (for example from initial or boundary values). 5. Feb 23, 2010 ### arildno Note: You can rewrite this as: [tex]\frac{\partial}{\partial{y}}(3U+U_{x})=0\to{3}U+U_{x}=F(x)$$
where F is some arbitrary function of x.

This can be re-written as:
$$\frac{\partial}{\partial{x}}(e^{3x}U)=e^{3x}F(x)$$

whereby the solution is given as:
$$U(x,y)=e^{-3x}\int(e^{3x}F(x))dx+G(y)e^{-3x}$$

where G(y) is some arbitrary function in y.

Obviously, the first term can be replaced by some arbitrary function H(x).

Last edited: Feb 23, 2010