Question

solve the differential equation dy/dx + (x/1-x^2)y = x(y)^1/2 , y(0)=1

Answer 1

This is the Bernoully equation

$y=uv,y'=u'v+uv',$ then $v(u'+\frac{x}{1-x^2}u)+(uv'-xu^\frac{1}{2}v^\frac{1}{2})=0$

$\int \frac{du}{u}=-\int \frac{x}{1-x^2}dx$ hence $u=(1-x^2)^\frac{1}{2},$ then

$(1-x^2)^\frac{1}{2}v'-x(1-x^2)^\frac{1}{4}v^\frac{1}{2}=0$ hence $\int \frac{dv}{v^\frac{1}{2}}=\int \frac{x}{(1-x^2)^\frac{1}{4}}dx$ hence

$v=(-\frac{(1-x^2)^\frac{3}{4}}{3}+C)^2$ then $y=(1-x^2)^\frac{1}{2}v$

as $y(0)=1$ hence $C=4/3$ then $y(x)=\frac{1}{9}(1-x^2)^\frac{1}{2}(4-(1-x^2)^\frac{3}{4})^2$