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anonymous
 4 years ago
Main dish
\[(1x^2) y'x^2y = (1+x)\sqrt{1x^2}\]
anonymous
 4 years ago
Main dish \[(1x^2) y'x^2y = (1+x)\sqrt{1x^2}\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y'  \frac{x^2}{1x^2}y = \frac{(1+x)\sqrt{1x^2}}{1x^2}\]\[y'  \frac{x^2}{1x^2}y = \frac{\sqrt{1x^2}}{1x}\] Not yet finished.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\alpha =\exp[\int\frac{x^2}{1x^2}dx=e^{\frac{1}{2}ln\frac{x+1}{x1}+x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{e^x}{\frac{x+1}{x1}^\frac{1}{2}}=\frac{\sqrt{x1}e^x}{\sqrt{1+x}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y= \frac{\sqrt{1+x}}{\sqrt{x1}e^x}\int \frac{\sqrt{x1}e^x}{\sqrt{1+x}}(\frac{\sqrt{1x^2}}{1x})dx\]\[=\frac{\sqrt{1+x}}{\sqrt{x1}e^x}\int \frac{\sqrt{(1x)(1x)}e^x}{1x}dx\]\[=\frac{\sqrt{1+x}}{\sqrt{x1}e^x}\int e^xdx\]\[=\frac{\sqrt{1+x}}{\sqrt{x1}e^x}( e^x+C)\]\[=\frac{\sqrt{1+x}}{\sqrt{x1}}+\frac{C\sqrt{1+x}}{\sqrt{x1}e^x}\] Oh.. I don't know how to do it :(
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