Answer with Step-by-step explanation:
We are given that
[tex]ln\frac{2X-1}{X-1}=t[/tex]
DE:[tex]\frac{dX}{dt}=\frac{X-1}{1-2X}[/tex]
When given expression is a solution of given DE then it satisfied the DE.
Differentiate w.r.t t
[tex]\frac{1}{\frac{2X-1}{X-1}}\times \frac{2X'(X-1)-X'(2X-1)}{(X-1)^2}=1[/tex]
By using the formula:[tex]\frac{d(lnx)}{dx}=\frac{1}{x}[/tex]
[tex]\frac{d(\frac{u}{v})}{dx}=\frac{u'v-v'u}{v^2}[/tex]
[tex]\frac{2X'X-2X'-2X'X+X'}{(X-1)}\times \frac{1}{2X-1}=1[/tex]
[tex]\frac{-X'}{2X-1}=X-1[/tex]
[tex]X'=-\frac{X-1}{2X-1}[/tex]
[tex]\frac{dX}{dt}=\frac{X-1}{1-2X}[/tex]
LHS=RHS
Hence, given expression is an implicit solution of given DE.
