The particular solution of the differential equation \((2 x-2 y+3) \mathrm{d} x-(x-y+1) \mathrm{d} y=0\) when \(x=0, y=1\) is

Let \(x-y+1=v\)
\(\begin{aligned}
& \Rightarrow 1-\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d} v}{\mathrm{~d} x} \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=1-\frac{\mathrm{d} v}{\mathrm{~d} x} \\
& \text { now } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 x-2 y+3}{x-y+1} \\
& \Rightarrow 1-\frac{\mathrm{d} v}{\mathrm{~d} x}=\frac{2 v+1}{v} \\
& \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} x}=1-\frac{2 v+1}{v}=\frac{-(v+1)}{v} \\
& \Rightarrow \frac{v}{v+1} \mathrm{~d} v=-\mathrm{d} x \\
& \Rightarrow \int\left(1-\frac{1}{v+1}\right) \mathrm{d} v=-\int \mathrm{d} x \\
& \Rightarrow n-\log |n+1|=-x+c \\
& \Rightarrow(x-y+1)-\log (x-y+1+1)=-x+c \\
& \Rightarrow 2 x-y+1+\log (x-y+2)=c
\end{aligned}\)
\(\begin{aligned} & \text { for } x=0, y=1 ; c=0 \\ & \Rightarrow 2 x-y-\log (x-y+2)+1=0\end{aligned}\)