# DE with initial conditions

1. Jan 29, 2006

### Pengwuino

Ok I have a problem here. I have gotten to the unfortunate point where I feel like I'm nearly done, Mathematica has given me a solution that is correct… but of course, I can't figure out how I was suppose to get from point A to B. I currently have:

$$2x - 6y\sqrt {x^2 + 1} \frac{{dy}}{{dx}} = 0,y(0) = 3$$

So I did this…

$$\begin{array}{l} 2xdx = 6y(x^2 + 1)^{1/2} \frac{{dy}}{{dx}} \\ \int {2x(x^2 + 1)^{ - 1/2} dx = } \int {6ydy} \\ u = x^2 + 1 \\ dx = \frac{{du}}{{2x}} \\ 2u^{1/2} = 3y^2 + c \\ 2(x^2 + 1)^{1/2} = 3y^2 + c \\ \end{array}$$

Kinda not sure where to go from here…. Or if I did that right in the first place

The correct answer according to Mathematica is...

$$\frac{{\sqrt {{\rm 25 + 2}\sqrt {{\rm 1 + x}^{\rm 2} } } }}{{\sqrt 3 }}$$

So yup, put x=0 and 3 pops out.

Last edited: Jan 29, 2006
2. Jan 29, 2006

### d_leet

It looks right, now use your initial condition to solve for the constant of integration.

3. Jan 29, 2006

### Pengwuino

oh my god, sorry guys, i guess i had one of those "forgot basic algebra" moments. I tried to square root the left side of the equation before bringing c to the left...