*Question: Solve: second order linear differential equation with variable coefficients.*

x∙(d^2 y)/(dx^2 )-(2x-1)∙dy/dx+(x-1)∙y=0

(d^2 y)/(dx^2 )-((2x-1))/x∙dy/dx+((x-1))/x∙y=0

Or,

(d^2 y)/(dx^2 )-p(x)∙dy/dx+q(x)∙y=0

By inspection, one solution is

y_1=Ae^x

Assume the 2nd solution, y_2 (x), is given by

y_2(x) =y_1 (x)*v(x)

y_2=Ae^x∙v

Wronskian

W(y_1,y_2 )=(■(y_1&y_2@y_1^'&y_2^' )) <--- matrix form of ([y1 y2] [y1' y2'])

W=y_1∙y_2^' - y_1^'∙y_2

W= Ae^x (Ae^x∙v+Ae^x ∙v^' ) - Ae^x∙Ae^x∙v

W=A^2 e^2x∙v + A^2 e^2x ∙v^' - A^2 e^2x∙v

W=A^2 e^2x v^'

Abel’s Theorem:

W(y_1,y_2 )=Ce^(-∫p(x) dx),

Using p(x)=-(2x-1)/x,

W(y_1,y_2 )=Ce^(-∫-2+1/x dx)

W(y_1,y_2 )=Ce^(2x-ln(x) )

W(y_1,y_2 )=Ce^2x∙(1/x)

Equating the two forms of the Wronskian, we get,

A^2 e^2x ∙v^' = Ce^2x∙(1/x)

After cancelling,

dv/dx=K/x

∫dv=K∫dx/x

v=K ln(x)

Hence the 2nd solution is,

y_2 (x)=AK.e^x.lnx)

The general solution then is

y(x)=y_1 (x)+y_2 (x)

**y(x)=A.e^x+ AK.e^x.lnx)**