# Seperable diff eq, differing intial-conditions

1. Jun 27, 2011

### Asphyxiated

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

Solve:

$$\frac {dy}{dx} = 2x \sqrt{1-y^{2}}$$

then find a solution for:

$$y(0)=0$$

and can you find a solution for:

$$y(0)=2$$

2. Relevant equations

3. The attempt at a solution

Just want to know if this is right.

First the equation can be rearranged to:

$$\frac {dy}{\sqrt{1-y^{2}}}= 2x dx$$

$$\int \frac {1}{\sqrt{1-y^{2}}} dy = \int 2x dx$$

$$sin^{-1}(y) = x^{2}+c$$

$$y=sin(x^{2}+c)$$

for y(0)=0

$$0=sin(c)$$

which is valid when:

$$c = 0 \;\; or \;\; c=k \pi, \;\; \forall \; k \; \in \; Z$$

and for y(0)=2

$$2=sin(c)$$

$$c = sin^{-1}(2) = \frac {\pi}{2}-\frac {ln(4\sqrt{3}+7)}{2} i$$

which is non-real. I assume that they are looking for me to realize that the answer isn't real? I am use to dealing with complex numbers from electrical engineering so this isn't that strange to me. Although I am not sure what this really means in this context. (the context being abstract math-land.)

2. Jun 27, 2011

### rock.freak667

I think they just wanted you to realize that for sinc>1 yields no real solutions for c.

3. Jun 27, 2011

### micromass

Staff Emeritus
Hi Asphyxiated!

That seems very good.

Regarding the y(0)=2 case. When solve ODE's like this, it is assumed that we work in the real numbers (unless specified otherwise). So only the real solutions count. So fro y(0)=2 there are no solutions.