Question

given that y1(x) = x^-1 is one solution of the differential equation 2x^2 dy^2/dx^2 + 3xdy/dx - y=0, x>0 find a second linearly independent solution of the equation

Answer 1

Look for a solution in the form $y=y_1z=\frac{z}{x}$, then $y'=\frac{z'}{x}-\frac{z}{x^2}$ and $y''=z''\left(\frac{1}{x}\right)+2z'\left(\frac{1}{x}\right)'+z\left(\frac{1}{x}\right)''=\frac{z''}{x}-\frac{2z'}{x^2}+\frac{2z}{x^3}$

Then $0=2x^2\left(\frac{z''}{x}-\frac{2z'}{x^2}+\frac{2z}{x^3}\right)+3x\left(\frac{z'}{x}-\frac{z}{x^2}\right)-\frac{z}{x}=2z''x-z'$

Let $z'=u$, then $2u'x-u=0$ or $\frac{2u'ux-u^2x'}{x^2}=0$.

Since $2u'u=(u^2)'$, we have $0=\frac{2u'ux-u^2x'}{x^2}=\frac{(u^2)'x-u^2x'}{x^2}=\left(\frac{u^2}{x}\right)'$, so $\frac{u^2}{x}=C$

Then $u=\sqrt{Cx}=\sqrt{|C|}\sqrt{x}$ if $C\ge 0$, or $u=\sqrt{Cx}=\sqrt{|C|}\sqrt{-x}$ if $C\le 0$.

Denote $\sqrt{|C|}$ by $D$, then $u=D\sqrt{x}$ or $u=D\sqrt{-x}$

Replace $u$ by $z'$: $z'=D\sqrt{x}$ or $z'=D\sqrt{-x}$.

Then $z=\frac{2}{3}D\sqrt{x^3}+A$ or $z=-\frac{2}{3}D\sqrt{-x^3}+A$. In every case $z=B\sqrt{|x|^3}+A$.

Since $y=\frac{z}{x}$, we have $y=Bs(x)\sqrt{|x|}+\frac{A}{x}=E\sqrt{|x|}+\frac{A}{x}$, where $s(x)=\begin{cases} 1,&\text{if $x>0$}\\ 0,&\text{if $x=0$}\\ -1,&\text{if $x<0$} \end{cases}$

So $\sqrt{|x|}$ is the second solution.

Answer: $\sqrt{|x|}$