Integration Question: Deriving \int\frac{V_{x}}{V}=ln(V) | Homework Help

[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?

 

[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.

 

[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}.

 

[CompuChip]

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

An alternative is to note that 3V + V_x = 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]<br /> where the A_{i[/i] are to be determined (for example from initial or boundary values).}

 

[arildno]

Note:

You can rewrite this as:
\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).