Question

The solution of the differential equation $\frac{dy}{dx}=\frac{x-y+3}{2(x-y)+5}$ is

• A. $2(x-y)+\log (x-y)=x+c$
• B. $2(x-y)-\log (x-y+2)=x+c$
• C. $2(x-y)+\log (x-y+2)=x+c$
• D. None of these

Answer 1

The correct option is C

$2(x-y)+\log (x-y+2)=x+c$

Explanation for the correct option

Given: $\frac{dy}{dx}=\frac{x-y+3}{2(x-y)+5}$

Let $v=x-y$

$\frac{dv}{dx}=1-\frac{dy}{dx}$

Now ,

$$\begin{gathered} \frac{dy}{dx}=\frac{x-y+3}{2(x-y)+5} \\ \Rightarrow 1-\frac{dv}{dx}=\frac{v+3}{2v+5} \\ \Rightarrow \frac{dv}{dx}=1-\frac{v+3}{2v+5} \\ \Rightarrow \frac{dv}{dx}=\frac{2v+5-v-3}{2v+5} \\ \Rightarrow \frac{dv}{dx}=\frac{v+2}{2v+5} \\ \Rightarrow \frac{dv}{dx}=\frac{v+2}{2v+5} \\ \Rightarrow \frac{2v+5}{v+2}dv=dx \end{gathered}$$

Integrating both sides

$$\begin{gathered} \int \frac{2v+5}{v+2}dv=\int dx \\ \Rightarrow \int 2+\frac{1}{v+2}dv=\int dx \\ \Rightarrow 2v+\log (v+2)=x+C \end{gathered}$$

Put $v=x-y$

Therefore $2\left(x-y\right)+\log (x-y+2)=x+C$

Hence option C is correct.