solve

x(d2y/dx2)-(2x-1)(dy/dx)+(x-1)y=0

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)

by Level 11 User (81.5k points)
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