Solving a separable equation Solve the following separable equations subject to the given boundary conditions
a) $\frac{dy}{dx}$ = $x\sqrt{1−y^2}$, $y=0$ at $x=0$, 
b)$(ω^2+x^2)\frac{dy}{dx} = y$, $y=2$ at $x=0$ ($ω>0$ is a parameter).
Edit:
Sorry I wasn't clear, I am fine with everything up to and including integration, it's just the boundary conditions that I cannot seem to grasp.
 A: Hint: For the first one, rearrange as
$$\frac{\mathrm dy}{\mathrm dx} = x\sqrt{1 - y^2} \implies \frac{\mathrm dy}{\sqrt{1 - y^2}} = x\,\mathrm dx$$
For the second one, we have:
$$\left(\omega^2 + x^2\right)\frac{\mathrm dy}{\mathrm dx} = y \implies \frac{\mathrm dy}y = \frac{\mathrm dx}{\omega^2 + x^2}$$
After integrating you'll have to substitute the initial conditions in order to find the constants.
Example
After integrating the first one, you'll get:
$$\arcsin y = \frac{x^2}2 + C\tag1$$
Now, substitute $x=0$, $y=0$ to get
$$0 = 0 + C \implies C = 0$$
Once you know the constant, substitute back into $(1)$ and solve for $y$:
$$\arcsin y = \frac{x^2}2 + 0 \implies y = \sin\left(\frac{x^2}2\right)$$
A: Hint: $(a)$ They are called separable because
$$\frac{1}{\sqrt{1 - y^2}}dy = xdx$$
integrate both sides. Then use the boundary conditions to find your coefficients. 
Similar approach to $(b)$.
Spoiler: 

$$\int\frac{1}{\sqrt{1 - y^2}}dy = \int xdx \implies \arcsin y = \frac{x^2}{2} + C$$

Now as $x = 0$ and $y = 0$ we have that 

$$ 0 = 0 + C \implies C = 0 $$

Finally 

 $$\arcsin y = \frac{x^2}{2} + C = \frac{x^2}{2} \implies y = \sin \Big(\frac{x^2}{2}\Big)$$

A: a) $$ \int \frac{dy}{\sqrt{1-y^{2}}}=\int x dx$$
b) $$\int\frac{dy}{y} =\int \frac{dx}{\omega^{2}+x^{2}}$$
If you're studying separation of variables, I think you'd be able to evaluate these integrals.
