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

$x^2y'=x^2y^2-2$

$y'=y^2-\dfrac{2}{x^2}$

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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

2 Answers 2

up vote 4 down vote accepted
+25

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

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

$\therefore-\dfrac{u''}{u}+\dfrac{(u')^2}{u^2}=\dfrac{(u')^2}{u^2}-\dfrac{2}{x^2}$

$-\dfrac{u''}{u}=-\dfrac{2}{x^2}$

$x^2\dfrac{d^2u}{dx^2}-2u=0$

Let $t=\ln x$ ,

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

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

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

$\dfrac{d^2u}{dt^2}-\dfrac{du}{dt}-2u=0$

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

$(\lambda-2)(\lambda+1)=0$

$\lambda=2$ 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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