The solution of the differential equation $\frac{d y}{d x}=\frac{x-2 y+1}{2 x-4 y}$ is

The solution of the differential equation $\frac{d y}{d x}=\frac{x-2 y+1}{2 x-4 y}$ is
  1. $(x-2 y)^2+2 x=c$
  2. $(x-2 y)^2+x=c$
  3. $(x-2 y)+2 x^2=c$
  4. $(x-2 y)+x^2=c$

Solution

Given that, $\frac{d y}{d x}=\frac{x-2 y+1}{2 x-4 y}$ Put $x-2 y=z \Rightarrow 1-2 \frac{d y}{d x}=\frac{d z}{d x}$ $ \begin{aligned} & \therefore \quad \frac{1}{2}\left[-\frac{d z}{d x}+1\right]=\frac{z+1}{2 z} \\ & \Rightarrow \quad-\frac{d z}{d x}+1=\frac{z+1}{z} \\ & \Rightarrow \quad \frac{d z}{d x}=-\frac{1}{z} \\ & \Rightarrow \quad z d z=-d x \\ & \Rightarrow \quad \frac{z^2}{2}=-x+c_1 \\ & \Rightarrow(x-2 y)^2+2 x=c \\ & \end{aligned} $

Asked in: AP EAMCET 2008

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