Question

A. Solve the differential equation ="y'=2x \sqrt{1-y^2} " alt="y'=2x \sqrt{1-y^2} " align="absmiddle" class="latex-formula">.
b. Explain why the initial value problem $y'=2x \sqrt{1-y^2}$ with y(0) = 3 does not have a solution.

Answer 1

$y' = \frac{dy}{dx}$

seperable differential equations will have the form
$\frac{dy}{dx} = F(x) G(y)$

what you do from here is isolate all the y terms on one side and all the X terms on the other
$\frac{dy}{G(y)} = F(x) dx$
just divided G(y) to both sides and multiply dx to both sides

then integrate both sides
$\int \frac{1}{G(y)} dy = \int F(x) dx$

once you integrate, you will have a constant. use the initial value condition to solve for the constant, then try to isolate x or y if the question asks for it


In your problem,
$G(y) = \sqrt{1-y^2} F(x) = 2x$

so all you need to integrate is
$\int \frac{1}{\sqrt{1-y^2}} dy = \int 2x dx$