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idku
 one year ago
DE
idku
 one year ago
DE

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idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \displaystyle y'+y=y^2\) \(\large \displaystyle y^{2}y'+y^{1}=1\) \(\large \displaystyle v'+v=1\) \(\large \displaystyle e^xv'+e^xv=e^x\) \(\large \displaystyle ve^x=e^x+C\) \(\large \displaystyle v=1~+C/e^x\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \displaystyle y=\frac{1}{1+\frac{C}{e^x}}\) \(\large \displaystyle y=\frac{1}{\frac{C+e^x}{e^x}}\) \(\large \displaystyle y=\frac{e^x}{C+e^x}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \displaystyle \frac{1}{2}y'+y=y^3\) \(\large \displaystyle \frac{1}{2}y'y^{3}+y^{2}=1\) \(\large \displaystyle v'+v=1\) \(\large \displaystyle v=(e^x+c)/e^x\) \(\large \displaystyle y=\pm\sqrt{e^x/(e^x+c)}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0So, I can conclude, that: \(\large \displaystyle \frac{1}{n}y'+y=y^{n+1}\) has a solution of: \(\large \displaystyle y^{n}=e^x/(e^x+C)\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0for all integer n (besides 0) (perhaps not only integer)
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