$\text{ The solution of the differential equation }(2x-4y+3)\frac{dy}{dx}+(x-2y+1)=0\text{ is }$$\log \left[4\right(\mathrm{x}-2\mathrm{y})+5]=\lambda (\mathrm{x}-2\mathrm{y})+\mathrm{c}\text{ then }\lambda \text{ is }$

$\begin{array}{l}\frac{dy}{dx}=\frac{-(x-2y+1)}{2(x-2y)+3}\\ \text{ Put }x-2y=t\\ 1-2\frac{dy}{dx}=\frac{dt}{dx}\\ \Rightarrow \frac{1}{2}\left(1-\frac{dt}{dx}\right)=\frac{dy}{dx}\\ \Rightarrow \frac{1}{2}\left(1-\frac{dt}{dx}\right)=-\frac{(t+1)}{2t+3}\\ \Rightarrow 1-\frac{dt}{dx}=\frac{-2t-2}{2t+3}\\ \frac{dt}{dx}=1+\frac{2t-2}{2t+3}=\frac{4t+5}{2t+3}\\ \Rightarrow \frac{1}{2}\int \left(1+\frac{1}{4t+5}\right)dt=\int dx\end{array}$

$\Rightarrow \frac{1}{4}\log \left|4\right(x-2y)+5|=x+2y+c/4$

$\begin{array}{l}\Rightarrow \log \left|4\right(x-2y)+5|=4(x+2y)+c\\ \Rightarrow \lambda =4\end{array}$