Answer to Question #177502 in Differential Equations for utkarsh

Question

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

Answer 1

$y'+\dfrac{x}{1-x^2}y-x\sqrt{y}=0$

Let $y=u*v$

$u'v+v'u+\dfrac{x}{1-x^2}uv-x\sqrt{uv}=0$

$v(u'+\dfrac{x}{1-x^2}u)+v'u-x\sqrt{uv}=0$

Let first part equal to 0. Then:

$\begin{cases} u'+\dfrac{x}{1-x^2}u=0\\ v'u-x\sqrt{uv}=0 \end{cases}$

$\begin{cases} u'=-\dfrac{x}{1-x^2}u\\ v'u=x\sqrt{uv} \end{cases}$

$\begin{cases} \dfrac{du}{u}=-\dfrac{xdx}{1-x^2}\\ \dfrac{dv}{dx}u=x\sqrt{uv} \end{cases}$

$\begin{cases} \ln{u}=\ln{\dfrac{1}{\sqrt{1-x^2}}}\\ \dfrac{dv}{\sqrt{v}}=\dfrac{xdx}{\sqrt{u}} \end{cases}$

$\begin{cases} u=\dfrac{1}{\sqrt{1-x^2}}\\ \dfrac{dv}{\sqrt{v}}=x\sqrt[4]{1-x^2}dx \end{cases}$

$\begin{cases} u=\dfrac{1}{\sqrt{1-x^2}}\\ 2\sqrt{v}=-\dfrac{2}{5}(1-x^2)^{5/4}+C \end{cases}$

$\begin{cases} u=\dfrac{1}{\sqrt{1-x^2}}\\ 2\sqrt{v}=-\dfrac{2}{5}(1-x^2)^{5/4}+C \end{cases}$

$\begin{cases} u=\dfrac{1}{\sqrt{1-x^2}}\\ v=(\dfrac{1}{5}(1-x^2)^{5/4}-\dfrac{C}{2})^2 \end{cases}$

$y=\dfrac{1}{\sqrt{1-x^2}}*(\dfrac{1}{5}(1-x^2)^{5/4}-\dfrac{C}{2})^2$

$y(0)=1=>\\ 1=\dfrac{1}{\sqrt{1-0^2}}*(\dfrac{1}{5}(1-0^2)^{5/4}-\dfrac{C}{2})^2$

$1=\dfrac{1}{\sqrt{1}}*(\dfrac{1}{5}(1)^{5/4}-\dfrac{C}{2})^2\\ 1=(\dfrac{1}{5}-\dfrac{C}{2})^2=>\\ C=-\dfrac{8}{5} \space or\space С=\dfrac{12}{5}$

$y_1=\dfrac{1}{25\sqrt{1-x^2}}*((1-x^2)^{5/4}+4)^2\\ y_2=\dfrac{1}{25\sqrt{1-x^2}}*((1-x^2)^{5/4}-6)^2$