1. The problem statement, all variables and given/known data Solve (x-1)y''-xy'+y=0 , given x>1 and y1=ex 2. Relevant equations 3. The attempt at a solution I've tried solving it by multiplying everythin for (1/x-1), so my equation looks like: y''-(x/x-1)y'+y=0 (because x-1 is unequal to 0) (1) so now the equation has the form of: y''+p(x)y'+y=0, with p(x)=(-x/x-1). Then, given the fact that y1 is a solution, then another solution has the form Y(x)=v(x)y1 Then: Y'(x)=v'(x)y1+v(x)y1' Y''(X)=v''(x)y1+v'(x)y1+v'(x)y1+v(x)y1' Now, after replacing the y factors in (1) with Y and its derivatives, I get: v''(x)y1+(2y1+p(x)y1=0, Which can me solved as a first order differential equation by doing v''(x)=z'(x) and v'(x)=z(x). Then, let μ(x)=e∫p(x)dx=e∫(-x/x-1)dx=e-x-ln(x-1)=e-x(1/x-1) Then, z(x)=(∫μ(x)g(x)dx+c1)/μ(x)=c1/(e-x/(x-1))=c1(ex(x-1)) Now, because z(x)=v'(x)→∫z(x)dx=v(x) Then I have v(x)=c1∫(ex(x-1)dx=c1((x-2)ex+c2) But, whenever I do Y(x)=v(x)y1, and replace it in (1), it doesn't satisfy the equation- Can somebody help?