# Differential Equations - am I doing this right

1. Sep 27, 2009

### protonchain

1. The problem statement, all variables and given/known data

What is $$y(x)$$ [$$y$$ as a function of $$x$$] given that

2. Relevant equations

$$\frac{dy}{dx} = 3\frac{y}{x}$$

where you are given the boundary condition that $$y = y_0$$ at $$x = x_0$$ where $$y_0$$ and $$x_0$$ are constants.

3. The attempt at a solution

Separating variables $$\Rightarrow \frac{dy}{y} = \frac{3 dx}{x}$$

Integrating $$\Rightarrow \int{\frac{dy}{y}} = \int{\frac{3 dx}{x}$$

Integrals give $$\Rightarrow ln(y) = 3 ln(x) + C$$

Given the boundary conditions then $$\Rightarrow ln(y_0) = 3 ln(x_0) + C$$

Therefore $$\Rightarrow C = ln(y_0) - 3ln(x_0) = ln(\frac{y_0}{x_0^3})$$

Plugging back into $$ln(y) = 3 ln(x) + C$$ gives
$$ln(y) = 3 ln(x) + ln(\frac{y_0}{x_0^3})$$

Therefore $$\Rightarrow y = exp\left(ln(x^3) + ln(\frac{y_0}{x_0^3})\right)$$

$$\Rightarrow y = exp\left(ln(x^3 \bullet \frac{y_0}{x_0^3})\right) \Rightarrow y = y_0(\frac{x}{x_0})^3$$

2. Sep 27, 2009

### jamesrc

Looks good to me.

One way you can check your work is to find dy/dx from your final equation, and verify that you get your initial equation...

3. Sep 27, 2009

### protonchain

That's the part where I got worried about whether I did it right or not.

I think I was getting stucking trying to prove that it was correct, and I just did it.

Thank you for the assurance and the help! :) Really appreciate it