DE

Question

DE

idku · Answer

$\large \displaystyle -y'+y=y^2$

$\large \displaystyle -y^{-2}y'+y^{-1}=1$

$\large \displaystyle v'+v=1$

$\large \displaystyle e^xv'+e^xv=e^x$

$\large \displaystyle ve^x=e^x+C$

$\large \displaystyle v=1~+C/e^x$

idku · Answer

$\large \displaystyle y=\frac{1}{1+\frac{C}{e^x}}$

$\large \displaystyle y=\frac{1}{\frac{C+e^x}{e^x}}$

$\large \displaystyle y=\frac{e^x}{C+e^x}$

idku · Answer

$\large \displaystyle -\frac{1}{2}y'+y=y^3$

$\large \displaystyle -\frac{1}{2}y'y^{-3}+y^{-2}=1$

$\large \displaystyle v'+v=1$

$\large \displaystyle v=(e^x+c)/e^x$

$\large \displaystyle y=\pm\sqrt{e^x/(e^x+c)}$

idku · Answer

So, 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 · Answer

for all integer n (besides 0)

(perhaps not only integer)