Question

Dy/dx+{x/(1-x^2)}y=x√y,y(0)=1

Answer 1

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

Trivial solution $y=0$, but there is $y(0)=1$ . That's why we can't take it.

Further we consider that $y\neq0.$

$y'+\frac{x}{1-x^2}y=x\sqrt{y}\\ \frac{y'}{\sqrt{y}}+\frac{x\sqrt{y}}{1-x^2}=x$

Let $\sqrt{y}=z$, then $z'=(\sqrt{y})'=\frac{y'}{2\sqrt{y}}$ and $\frac{y'}{\sqrt{y}}=2z'$

$2z'+\frac{xz}{1-x^2}=x$

Let $z=uv$, then $z'=u'v+uv'$

$2u'v+2uv'+\frac{xuv}{1-x^2}=x\\ 2u'v+u(2v'+\frac{xv}{1-x^2})=x$ (1)

$2v'+\frac{xv}{1-x^2}=0\\ 2v'=\frac{xv}{x^2-1}\\ 2\frac{dv}{dx}=\frac{xv}{x^2-1}\\ 2\frac{dv}{v}=\frac{xdx}{x^2-1}\\ 2\int\frac{dv}{v}=\int\frac{xdx}{x^2-1}\\ 2log(|v|)=1/2log(|x^2-1|)\\ v=\sqrt[4]{|x^2-1|}$

Now (1) is $2u'\sqrt[4]{|x^2-1|}=x$

$2\frac{du}{dv}=\frac{x}{\sqrt[4]{|x^2-1|}}\\ 2du=\frac{xdx}{\sqrt[4]{|x^2-1|}}\\ 2\int du=\int \frac{xdx}{\sqrt[4]{|x^2-1|}}\\ 2\int du=\frac{1}{2}\int \frac{dx^2}{\sqrt[4]{|x^2-1|}}\\ u=\frac{1}{3}\sqrt[4]{|x^2-1|}^3+C$

Finally

$z=uv=\sqrt[4]{|x^2-1|}\frac{1}{3}(\sqrt[4]{|x^2-1|}^3+C)=\frac{1}{3}(|x^2-1|)+C\sqrt[4]{|x^2-1|}$

$\sqrt{y}=\frac{1}{3}(|x^2-1|)+C\sqrt[4]{|x^2-1|}\\ y=(\frac{1}{3}(|x^2-1|)+C\sqrt[4]{|x^2-1|})^2$

$y(0)=1, \implies \\ 1=(\frac{1}{3}|-1|+C\sqrt[4]{|-1|})^2\\ 1=(\frac{1}{3}+C)^2$

$C=\frac{2}{3}$ or $C=-\frac{4}{3}$

$y=(\frac{1}{3}(|x^2-1|)+\frac{2}{3}\sqrt[4]{|x^2-1|})^2$ or

$y=(\frac{1}{3}(|x^2-1|)-\frac{4}{3}\sqrt[4]{|x^2-1|})^2$