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

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
  1. $x-2 y-\log (x-y+2)+2=0$
  2. $x-y-\log (x-y+2)+1=0$
  3. $2 x+y-\log (x-y+2)-1=0$
  4. $2 x-y-\log (x-y+2)+1=0$

Solution

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

Asked in: MHT CET 2022 (08 Aug Shift 1)

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