Solve $x^2y''-xy'+y=0$ given that $y_1=x$ using reduction of order Hello so I need to solve $x^2y''-xy'+y=0$ given that $y_1=x$ using the idea that $y_2=v(x)x$. So I went through all the steps
$$
x^3v''+2x^2v'-xv-x^2v'+xv = x
$$
$$
 x^3v''+x^2v' = x
$$
$$
 xv'' + v' = \dfrac{1}{x}
$$
$$
\int\dfrac{d}{dx}(xv')dx = \int\dfrac{1}{x}dx
$$
$$
xv' = \log|x|+C_1
$$
$$
\int v'dx = \int \dfrac{\log(x)}{x}dx + \int \dfrac{C_1}{x}dx
$$
$$
v = \dfrac{1}{2}\log^2|x|+C_1\log|x|+C_2
$$
However I know that the true answer is $v = \log(x)$. What am I missing?
 A: Hint 
Your equation is not homogenous. 
It has an x thats not part of the equation in the title
And $y_1=x$ is not a solution of the inhomogenous equation but a solution of the homogenous equation
It should be :
$$xv''+v'=0$$
$$(v'x)=K_1$$
$$v'=\frac {K_1}x$$
$$v=K_1\int \frac {dx}x+K_2$$
$$v=K_1\ln(x)+K_2$$
$$y(x)=K_1x\ln(x)+K_2x$$
A: I shall provide an alternative solution by solving the differential equation
$$x^2\,y''(x)+k\,x\,y'(x)-k\,y(x)=0\,,$$
where $k$ is a given constant.  In this particular question, $k=-1$.  
The trick is to note that
$$\frac{\text{d}}{\text{d}x}\left(y'(x)+\frac{k}{x}\,y(x)\right)=y''(x)+\frac{k}{x}\,y'(x)-\frac{k}{x^2}\,y(x)=\frac{x^2\,y''(x)+k\,x\,y'(x)-k\,y(x)}{x^2}=0\,.$$
That is, for some constant $a$, we have
$$\frac{1}{x^k}\,\frac{\text{d}}{\text{d}x}\,\big(x^k\,y(x)\big)=y'(x)+\frac{k}{x}\,y(x)=a\,.$$
Therefore,
$$y(x)=\frac{1}{x^k}\,\int\,(a\,x^k)\,\text{d}x\,.$$
If $k=-1$, then we get
$$y(x)=x\,\int\,\left(\frac{a}{x}\right)\,\text{d}x=x\,\big(a\,\ln(x)+b\big)=a\,\big(x\,\ln(x)\big)+b\,x\,,$$
for some constant $b$.  If $k\neq -1$, then there exists a constant $b$ for which
$$y(x)=\frac{1}{x^k}\,\left(\frac{a}{k+1}\,x^{k+1}+b\right)=\frac{a}{k+1}\,x+b\,\left(\frac{1}{x^k}\right)\,.$$
