# Simple ODE

1. Apr 5, 2006

### cscott

Can anyone walk me through solving this?

$$\frac{dy}{dx} = \frac{2y}{x}$$
Seperate veriables...
$$\frac{dy}{2y} = \frac{dx}{x}$$
...and integrate from here, but the answer is $y = cx^2$ and I don't see how they got that.

Last edited: Apr 5, 2006
2. Apr 5, 2006

### TD

Show us how you integrated, it shouldn't be so hard

3. Apr 5, 2006

### cscott

Ya, I know :uhh:

Can't I say that it's equal to
$$\int \frac{1}{2}y^{-1}dx =\int \frac{1}{x} dx$$
...isn't the integral of $\frac{1}{x} = ln (|x|)$?

4. Apr 5, 2006

### d_leet

Yes, that looks right so far.

5. Apr 5, 2006

### mathmike

the left side of your equation should be inegrated with respect to y not x

6. Apr 6, 2006

### cscott

Sorry, that was a typo.

So I should get 1/2 ln(|y|) + C' = ln(|x|) + C after I integrate? Hows that equal to y =cx^2? :\

7. Apr 6, 2006

### nocturnal

combine C' and C into one constant, D, and refer to the properties of natural log

8. Apr 6, 2006

### cscott

C - C' = D

$$\ln y = \ln x^2 + D$$

and I can just remove ln, even with D there?

Last edited: Apr 6, 2006
9. Apr 6, 2006

### nocturnal

no, use the properties of ln to simplify your equation.

hint:
$$\ln a - \ln b = \ \ ??$$

$$e^{\ln a} = \ \ ??$$

$$b \ln a = \ \ ??$$

10. Apr 6, 2006

### cscott

So,

$$\ln \frac{y}{x^2} = D$$
$$e^{\ln y/x^2} = e^D$$

But then wouldn't I end up with

$$y = e^Dx^2$$

11. Apr 6, 2006

### TD

Yes, but since e is a constant and D is too, e^d is just another constant! Name it c and you have what you want

12. Apr 6, 2006

### cscott

Oh, well that makes sense. Thanks everyone.