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I am new to the differential equation and I need some ideas how to solve this problem.



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Hint: try to solve y'=y*y. – pppqqq Jan 18 '13 at 17:04
Have you looked into Riccati Equations?[(1),(2),(3)] – torrho Jan 18 '13 at 20:54
Looking for an answer drawing from credible and/or official sources... Huh? What kind of source? – Did Jan 23 '13 at 7:44
up vote 4 down vote accepted

Let $y=-\dfrac{u'}{u}$ ,

Then $y'=-\dfrac{u''}{u}+\dfrac{(u')^2}{u^2}$




Let $t=\ln x$ ,

Then $\dfrac{du}{dx}=\dfrac{du}{dt}\dfrac{dt}{dx}=\dfrac{1}{x}\dfrac{du}{dt}$


$\therefore x^2\left(\dfrac{1}{x^2}\dfrac{d^2u}{dt^2}-\dfrac{1}{x^2}\dfrac{du}{dt}\right)-2u=0$


The auxiliary equation is $\lambda^2-\lambda-2=0$


$\lambda=2$ or $-1$

$\therefore u=C_1e^{2t}+C_2e^{-t}=C_1x^2+\dfrac{C_2}{x}$

Hence $y=-\dfrac{\left(C_1x^2+\dfrac{C_2}{x}\right)'}{C_1x^2+\dfrac{C_2}{x}}=-\dfrac{2C_1x-\dfrac{C_2}{x^2}}{C_1x^2+\dfrac{C_2}{x}}=\dfrac{C_2-2C_1x^3}{C_1x^4+C_2x}=\dfrac{\dfrac{C_2}{C_1}-2x^3}{x^4+\dfrac{C_2}{C_1}x}=\dfrac{C-2x^3}{x^4+Cx}$

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Let $z(x)=xy(x)$ then $xz'=x^2y'+xy=(x^2y^2-2)+(xy)=z^2+z-2$ hence $$ \frac3x=\frac{3z'}{z^2+z-2}=\frac{z'}{z-1}-\frac{z'}{z+2}, $$ which yields $$ \frac{z-1}{z+2}=cx^3, $$ for some $c$. Finally, on either one of the halflines bounded by $c^{-1/3}$ (or on the whole real line if $c=0$), $$ y(x)=\frac1x\,\frac{1+2cx^3}{1-cx^3}. $$

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