Question

Find a second solution y2 for x^2*y"+xy'-y=0; y1=x that isn't a constant multiple of the solution y1. Choose K conveniently to simplify y2.

Answer 1

So I divided by x^2 and got...\[y''+\frac{ 1 }{ x }y'-\frac{ 1 }{ x^2 }y=0\]

Answer 2

Now it's in the form of y"+p(x)y'+q(x)y=0.

Answer 3

@SithsAndGiggles

Answer 4

I'm not sure if this is the right approach, but notice that you have two "hidden" derivatives:
\[\frac{d}{dx}[y']=y''\quad\text{and}\quad\frac{d}{dx}\left[\frac{1}{x}y\right]=\frac{1}{x}y'-\frac{1}{x^2}y\]
so the ODE is equivalent to
\[\frac{d}{dx}\left[y'+\frac{1}{x}y\right]=0\]
Integrating both sides yields
\[y'+\frac{1}{x}y=C_1\]
which is linear in \(y\). If you were to look for an integrating factor, you'd find that \(\mu(x)=x\):
\[\begin{align*}xy'+y&=C_1x\\\\
\frac{d}{dx}[xy]&=C_1x\\\\
xy&=\frac{1}{2}C_1x^2+C_2\\\\
y&=\frac{1}{2}C_1\underbrace{x}_{y_1}+C_2\underbrace{\frac{1}{x}}_{y_2}
\end{align*}\]

Answer 5

Another method (again, not sure if it's what you're looking for): the given ODE is of the Euler-Cauchy form,
\[x^2y''+xy'-y=0\]
Setting \(y=x^r\), you have the derivatives \(y'=rx^{r-1}\) and \(y''=r(r-1)x^{r-2}\). Subbing these in, you get
\[r(r-1)x^r+rx^r-x^r=0~~\implies~~r(r-1)+r-1=(r+1)(r-1)=0\]
which has roots \(r=\pm1\). Here you find that the fundamental solutions would agree with the ones found before, \(\left\{x,\dfrac{1}{x}\right\}\).

Answer 6

Another method, this time using reduction of order (probably what you wanted in the first place, now that I think about it).

Assume you have a solution of the form \(y=y_1v=xv\), then \(y'=xv'+v\) and \(y''=xv''+2v'\). Subbing into the ODE, you have
\[\begin{align*}x^2(xv''+2v')+x(xv'+v)-xv&=0\\
x^3v''+3x^2v'&=0\\
xv''+3v'&=0\end{align*}\]
Substituting \(t=v'\), you get the linear ODE,
\[xt'+3t=0\]

Answer 7

I love your first method of solving this problem. Thank you so much!