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I'm currently trying to solve the following ODE:

$$x^4yy''+x^4(y')^2+3x^3yy'-1=0$$

I've tried the substitution $$\upsilon=\frac{y}{x}$$which didn't simplify the whole lot. Then I tried rewriting it by division by $y^2$ which yields:

$$x^4\frac{y''}{y}+x^4\left(\frac{y'}{y}\right)^2+3x^3\frac{y'}{y}-\frac{1}{y^2}=0.$$

Now if it wasn't for the $1/y^2$-term the substitution $v=y'/y$ might be a very logical choice. Unfortinately the last term makes this substitution cumbersome.

Now I was wondering if there was any logical substitution that I could make to simplify/solve this differential equation. I'm pretty stuck on this one.

Edit. The substitution $\,\upsilon=yy'\,$ is starting to look promising ...

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1 Answer 1

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$$ x^4yy''+x^4(y')^2+3x^3yy'-1=0, $$ or $$ x^4\big(yy''+(y')^2\big)+3x^3yy'-1=0, $$ or $$ x^4\big(yy'\big)'+3x^3yy'-1=0, $$ and setting $z=yy'$ we obtain $$ x^3z'+3x^2z-\frac 1x=0, $$ or $$ (x^3z)'=(\log x)' $$ or $$ x^3z-\log x=c, $$ for some constant $c$. Hence $$ yy'=z=cx^{-3}+x^{-3}\log x $$ or $$ \frac{1}{2}y^2=-\frac{c}{2x^2}-\frac{\log x+1}{2x^2}=-\frac{\log x}{2x^2}+\frac{c'}{x^2}+c'' $$ and thus for suitable constants $c_1,c_2$ we have the expression $$ y=\pm\left(c_1+\frac{c_2}{x^2}-\frac{\log x}{x^2}\right)^{1/2}\,. $$

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