Question

Main dish
\[(1-x^2) y'-x^2y = (1+x)\sqrt{1-x^2}\]

Answer 1

\[y' - \frac{x^2}{1-x^2}y = \frac{(1+x)\sqrt{1-x^2}}{1-x^2}\]\[y' - \frac{x^2}{1-x^2}y = \frac{\sqrt{1-x^2}}{1-x}\]
Not yet finished.

Answer 2

\[\alpha =\exp[-\int\frac{x^2}{1-x^2}dx=e^{-\frac{1}{2}ln|\frac{x+1}{x-1}|+x}\]

Answer 3

\[=\frac{e^x}{|\frac{x+1}{x-1}|^\frac{1}{2}}=\frac{\sqrt{x-1}e^x}{\sqrt{1+x}}\]

Answer 4

\[y= \frac{\sqrt{1+x}}{\sqrt{x-1}e^x}\int \frac{\sqrt{|x-1|}e^x}{\sqrt{1+x}}(\frac{\sqrt{1-x^2}}{1-x})dx\]\[=\frac{\sqrt{1+x}}{\sqrt{x-1}e^x}\int \frac{\sqrt{(|1-x|)(1-x)}e^x}{1-x}dx\]\[=\frac{\sqrt{1+x}}{\sqrt{x-1}e^x}\int e^xdx\]\[=\frac{\sqrt{1+x}}{\sqrt{x-1}e^x}( e^x+C)\]\[=\frac{\sqrt{1+x}}{\sqrt{x-1}}+\frac{C\sqrt{1+x}}{\sqrt{x-1}e^x}\]
Oh.. I don't know how to do it :(