Solve the following DE using Laplace Transform method
(2t-1)^2*x”+(2t-1)x’-x=0
x(0)=0 x'(0)=0
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applying LT I got another variable coefficients second order DE possibly harder than the given one
The main theorem is the one used to transform t²x” and tx’
http://www.solitaryroad.com/c914.html
L[t^m Y⁽ⁿ⁾(t)] = (-1)^m d^m/(ds)^m L[Y⁽ⁿ⁾(t)]
your equation
(2t – 1)²x” + (2t – 1)x’ – x = 0
becomes
L[(2t – 1)²x”] + L[(2t – 1) x’] – L[x] = 0
expanding X(s) is X
s²X – 4L[tx”] + 4L[t²x”] -sX + 2L[tx’] – X = 0……….(*)
then, applying the theorem above we have
L[tx”] = -(d/ds)L[x”] = -(d/ds)(s²X) = -2sX – s² dx/ds
and
L[tx’] = -(d/ds)L[x’] = -(d/ds)(sX)= -X – sdX/ds
substitute in (*)
s²X -4(-2sX- s² dx/ds) + 4(2X + s dX/ds – 2s dX/ds – s² d²X/ds²) -sX + 2(-X – s dX/ds) – X = 0
order and rearrange
4s² d²X/ds² + dX/ds (6s – 4s²) – X (s² + 7s + 5) = 0
or
4s² X” + (6s – 4s²)X’ – (s² + 7s + 5)X = 0
???
On the other side it’s easy to find a trivial solution x(t) = 0
As far as I remember this is unique, for the fundamental theorem about DE