Question

xy''−xf(x)y' + f(x)y = 0. solucion general

Answer 1

$\texttt{Sea }x\neq 0\texttt{ entonces hacemos: }\\ \\ \hspace*{3cm}y''-f(x)y'+\dfrac{f(x)}{x}y=0\\ \\ \\ \texttt{Luego un cambio de variable: }u(x)=\dfrac{y}{x}\to y=ux\\ \\ \\ (ux)''-f(x)(ux)'+\dfrac{f(x)}{x}(ux)=0\\ \\ \\ (u''x+2u')-f(x)(u'x+u-u)=0\\ \\ \\ u''x+\left[2-xf(x)\right]u'=0\\ \\ \\ u''+\dfrac{2-xf(x)}{x}u'=0\\ \\ \\ \texttt{Sea }u'=w(x)\texttt{ entonces: }\\ \\ \\ w'+\left(\dfrac{2}{x}-f(x)\right)w=0$


$\dfrac{dw}{dx}=\left(f(x)-\dfrac{2}{x}\right)w\\ \\ \\ \dfrac{dw}{w}=\left(f(x)-\dfrac{2}{x}\right)dx\\ \\ \\ \displaystyle \ln |w| =\int f(x)-\dfrac{2}{x}\; dx+C_1\\ \\ \\ \ln |w| =\int f(x)\,dx-2\ln x+C_1\\ \\ \\ w=\dfrac{C_2\cdot e^{\int f(x)\,dx}}{x^2}\\ \\ \\ u=C_2\int \dfrac{e^{\int f(x)\,dx}}{x^2}\, dx+C_3\\ \\ \\ \boxed{y=x\cdot C_2\int \dfrac{e^{\int f(x)\,dx}}{x^2}\, dx+x\cdot C_3}$


          $\texttt{Donde los }C_i\texttt{ son constantes y } \int f(x)\,dx \texttt{ se considera como}\\ \texttt{una funci\'on de solo la variable }x$