Finding a Second Solution to (x-1)y"-xy'+y=0

[Punchlinegirl]

Use reduction of order to find a second solution to the given differential equation
(x-1)y"-xy'+y=0 x>1 y_1=e^x


Putting it in standard form gives
$$y"-x/x-1 y' + 1/x-1 y =0$$
y(t)=v(t)e^x
y'(t)= v'(t)e^x +v(t)e^x
y"(t)= v"(t)+2v'(t)e^x +v(t)e^x
plugging into the initial equation:
$$v"(t)e^x+2v'(t)e^x+v(t)e^x-xv'(t)e^x/x-1 +xv(t)e^x/x-1 +v(t)e^x/x-1$$

I'm not sure how to simplify this further if it can even be done, or what I should do next. Can someone please help me out?

 

[Fermat]

Leave it in the form: (x-1)y"-xy'+y=0 x>1 y_1=e^x

then substitute in the y, y' and y'' values for the 2nd solution.

Simplify and you will find the v-term disappear. I just did :)