Find a second solution of the given differential equation. $$
xy''+y'=0; y_1=ln(x)
$$
I solved this all the way to the end and found my second solution to be $y_2=-1$, but the book says it is $y_2=1$. I am checking my algebra and the method I used was to get it to $u''+u'(\frac{2+ln(x)}{xln(x)})=0$. I then made the substitution $w=u'$. Is this correct? I feel as though I was supposed to use $w=u'(\frac{2+ln(x)}{xln(x)})$ but I'm not sure. Is there a faster way to do this problem?
 A: The given equation can be written as $$\frac{\mathrm{d}\,(xy')}{\mathrm{d}\,x}=0$$
Hence, you are looking for functions $y$ such that $$xy'=c_1$$ Where $c_1$ is a constant.
A: The way in which the problem is presented shows that we should use reduction of order method and solve the second solution with the help of the first one so we assume the second solution as:
$$y_2(x)=u(x)\ln(x)$$
$$y_2'(x)=u'(x)\ln(x)+\frac{u(x)}{x}$$
$$y_2''(x)=u''(x)\ln(x)+\frac{u'(x)}{x}+\frac{u'(x)x-u(x)}{x^2}$$
Substituting $y_2(x)$ in the main differential equation gives us:
$$u''(x)x\ln x+u'(x)(2+\ln x)=0$$
$$\frac{u''(x)}{u'(x)}=-\frac{2+\ln x}{x\ln x}=-\frac{1}{x}-\frac{2}{x\ln x}$$
Hint: because you are dealing with indefinite integrals you should a constant $c_n$ when computing the integral so:
Assume that $w=u''$
$$\frac{w'(x)}{w(x)}=-\frac{1}{x}-\frac{2}{x\ln x}$$
$$\ln w(x)=-\ln x-2\int x^{-1}{\ln x}^{-1}dx+c_1\qquad(c_1\,is\,constant)$$
To calculate $\int x^{-1}{\ln x}^{-1}dx$ we use the method change of variables so assume that $v=\ln x\Rightarrow dv=x^{-1}dx$ then $\int x^{-1}{\ln x}^{-1}dx=\int\frac{dv}{v}=\ln v=\ln(\ln x)$ and also there is no difference. We can wirite $c_1$ as $\ln c_1$ (both of them are arbitrary constants) so we will have:
$$\ln w(x)+\ln x=-2\ln(\ln x)+\ln c_1$$
$$\ln w(x)x=\ln(\ln x)^{-2}+\ln c_1$$
$$\ln w(x)x=\ln c_1(\ln x)^{-2}$$
$$xw(x)=c_1(\ln x)^{-2}$$
$$w(x)=c_1x^{-1}(\ln x)^{-2}$$
$$u'(x)=c_1x^{-1}(\ln x)^{-2}$$
Hint: you are dealing with an indefinite integral and you should put a constant $c_2$ in your calculations
$$u(x)=c_1(\int x^{-1}(\ln x)^{-2}dx+c_2$$
In order to calculate $\int x^{-1}(\ln x)^{-2}dx$ we use change of variables method so assume that $v=\ln x\Rightarrow dv=x^{-1}dx$ then $\int x^{-1}(\ln x)^{-2}dx=\int v^{-2}dv=-v^{-1}=-(\ln x)^{-1}$ so we will have:
$$u(x)=c_1(-(\ln x)^{-1}+c_2)=-c_1(\ln x)^{-1}+c_1c_2$$
There is nodifference. We can write $-c_1$ as $c_1$ and $c_1c_2$ as $c_2$ because all of them are arbitrary constants. so:
$$u(x)=c_1(\ln x)^{-1}+c_2$$
$$y_2(x)=u(x)y_1(x)=c_1+c_2\ln x$$
So in general form the second solution will be $y_2(x)=c_1+c_2\ln x$ and because we have no initial values we can't go further and specify the unique solution. If we had initial values, we could calculate $c_1$ and $c_2$ so both $y_2=1$ and $y_2=-1$ can be the answer. For example:  
Initial values: $y_2(1)=y_2(2)=1\Rightarrow\,c_1=1\;c_2=0\Rightarrow \, y_2=1$
Initial values: $y_2(1)=y_2(2)=-1\Rightarrow\,c_1=-1\;c_2=0\Rightarrow \, y_2=-1$
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From the formula that you provided in the comments:
$$y_2(x)=y_1(x)\int\frac{e^{-\int p(x)dx}}{y_1(x)^{2}}dx$$
we have $y''+\frac{1}{x}y'=0\Rightarrow p(x)=\frac{1}{x}$ so:
$$y_2(x)=\ln x\int\frac{e^{-\int\frac{dx}{x}}}{(\ln x)^2}dx=\ln x\int\frac{e^{\ln\frac{1}{x}}}{(ln x)^2}dx=-1$$
In fact when using the reduction of order method we say that if $y_1(x)$ and $y_2(x)$ are two special solutions of a homogenous second-order differential equation, any linear equation of them will be the general solution of the equation so the general solution will be:
$$y(x)=c_1y_1(x)+c_2y_2(x)=c_1(-1)+c_2(\ln x)$$
substituting $-c_1$ by $c_1$ (because they are both arbitrary constants) we can write:
$$y(x)=c_1+c_2(\ln x)$$
so there's no difference. either $-1$ or $1$ can be the second solution of the equation
A: Put $ y^{'} =p \tag{1}$
$$ p + x \, p^{'} =0 $$
$$ xp = c_1$$
$$ \dfrac{dx}{x}=  \dfrac{d y}{c_1}   $$
$$ y/c_1 = log ( x/c_2) $$
Back substituting 
$$ x p = x \dfrac{dy}{dx} = $$ const. 
p is the negative constant, from (1).
Since p= any const., $ y_2 = x\cdot $ that constant, not a trivial solution.
