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?
On
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 $$y(x)=\frac{1}{x^k}\,\left(\frac{a}{k+1}\,x^{k+1}+b\right)=\frac{a}{k+1}\,x+b\,\left(\frac{1}{x^k}\right)\,.$$
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 : $$xv''+v'=0$$ $$(v'x)=K_1$$ $$v'=\frac {K_1}x$$ $$v=K_1\int \frac {dx}x+K_2$$ $$v=K_1\ln(x)+K_2$$ $$y(x)=K_1x\ln(x)+K_2x$$