here is the problem i've been working on for acouple days and can't get right...
Given a solution y1=x of the DE x^2y"-xy'+y=0
Find solution by using the reduction of order method
I've worked through it using the substitution y=xu then found the first and second derivative of y and plugged them back into the orig. equation..and for the orig equation i divided everything through buy the leading coeff x^2 to modify the equation to y"-1/xy'+1/x^2y
when i plugged everything into that equation and simplified I got u"+u'+2u
I then let v=u' to get it to a first order equation
which gave me
v'+v+2u and this is where i'm stuck I don't know what to do since there is still a u in the equation
$\displaystyle y=ux$
$\displaystyle y'=u+u'x$
$\displaystyle y''=u'+u'+u''x$
$\displaystyle x^2(u'+u'+u''x)-x(u+u'x)+ux=0$
$\displaystyle 2x^2u'+x^3u''-xu-x^2u'+ux=0 \iff x^3u''+x^2u'=0$
Now we need to solve this with the sub v=u' v'=u''
$\displaystyle x^3\frac{dv}{dx}+x^2v=0 \iff \frac{dv}{v}=-\frac{dx}{x}$
$\displaystyle \ln|v|=-\ln|x|+\ln(C) \iff \ln|v|=\ln\left( \frac{C}{x}\right) \iff v=\frac{C}{x}$
but v=u' integrating again we get
$\displaystyle u=C\ln|x|+d$
Now if we choose C=1 and d=0 we get
$\displaystyle u=ln|x|$
so the 2nd solution is $\displaystyle y=x\ln|x|$