Question

Find a second solution y2 for (x-1)y"-xy'+y=0; y1=e^x that isn't a constant multiple of the solution y1.

Answer 1

@SithsAndGiggles

Answer 2

Using reduction of order, we want to find a second solution \(y_2=uy_1\), where \(u,y_1,y_2\) are all functions of \(x\).
\[y_2=uy_1\\
{y_2}'=u'y_1+u{y_1}'\\
{y_2}''=u''y_1+2u'{y_1}'+u{y_1}''\]
With \(y_1=e^x\), you have
\[y_2=ue^x\\
{y_2}'=(u'+u)e^x\\
{y_2}''=(u''+2u'+u)e^x\]
Subbing into the ODE, you have
\[\begin{align*}
0&=(x-1)y''-xy'+y\\\\
0&=\bigg((x-1)(u''+2u'+u)-x(u'+u)+u\bigg)e^x\\\\
0&=(x-1)u''+(2x-2)u'+(x-1)u-x(u'+u)+u\\\\
0&=(x-1)u''+(x-2)u'\end{align*}\]
Substituting \(t=u'\) gives you an equation linear in \(t\):
\[0=(x-1)t'+(x-2)t\]

Answer 3

So (x-1)t'=-(x-2)t?

Answer 4

Right, you can solve this by separation of variables, or find an integrating factor. Your choice.

Answer 5

But I'm not getting the answer in the book. \[t'=-\frac{ x-2 }{ x-1 }t\]

Answer 6

I got \[t=C(x-1)e^ {-x}\]

Answer 7

And \[u'=Ce^ {-x}(x-1)\]

Answer 8

\[\frac{ u }{ c }=-xe^ {-x}+C\]

Answer 9

The answer in the book is y2=x.

Answer 10

\[t'=-\frac{ x-2 }{ x-1 }t~~\implies~~\dfrac{dt}{t}=-\frac{x-2}{x-1}\,dx\]
Integrate both sides and back-substitute,
\[\begin{align*}
\int\dfrac{dt}{t}&=-\int\frac{x-2}{x-1}\,dx\\\\
\ln t&=-\int\frac{x-1-1}{x-1}\,dx\\\\
&=\int\left(\frac{1}{x-1}-1\right)\,dx\\\\
&=\ln(x-1)-x+C_1\\\\
t&=C_1(x-1)e^{-x}\\\\
u'&=C_1(x-1)e^{-x}\\\\
u&=-C_1xe^{-x}+C_2\\\\
ue^x&=-C_1x+C_2e^x\\\\
\implies y_2&=x\end{align*}\]

Answer 11

I see my mistake now. :) Thanks a lot!