Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.8 Question 3 Maths textbook Solution.

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

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

Hint : - first, we will separate the variables and then solve .

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

Let $x-y=v$

Differentiating with respect to  x , we get,

\begin{aligned} &\frac{d}{d x}(x-y)=\frac{d v}{d x} \\ &\Rightarrow \quad 1-\frac{d y}{d x}=\frac{d v}{d x} \\ &\Rightarrow \frac{d y}{d x}=1-\frac{d v}{d x} \end{aligned}

Putting $\frac{dy}{dx}=1-\frac{dv}{dx}$ and $x-y=v$ in equation (i), we get,

\begin{aligned} &1-\frac{d v}{d x}=\frac{v+3}{2 v+5} \\ &\Rightarrow \frac{d v}{d x}=1-\left(\frac{v+3}{2 v+5}\right) \\ &\Rightarrow \frac{d v}{d x}=\frac{2 v+5-v-3}{2 v+5} \\ &\Rightarrow \frac{d v}{d x}=\frac{v+2}{2 v+5} \end{aligned}

Taking like variables in the same side,

$\Rightarrow \frac{2v+5}{v+2}dv=dx$

Now, integrating in both sides, we get,

$\Rightarrow\int \frac{2v+5}{v+2}dv=dx$

\begin{aligned} &\Rightarrow \int\left(\frac{2 v+4+1}{v+2}\right) \mathrm{d} \mathrm{v}=\int \mathrm{d} \mathrm{x} \\ &\Rightarrow \int\left(\frac{2(v+2)}{v+2}+\frac{1}{\mathrm{v}+2}\right) \mathrm{d} \mathrm{v}=\int \mathrm{d} \mathrm{x} \\ &\Rightarrow \quad \int\left(2+\frac{1}{v+2}\right) \mathrm{d} \mathrm{v}=\int \mathrm{d} \mathrm{x} \\ &\Rightarrow 2 \mathrm{v}+\log |\mathrm{v}+2|=\mathrm{x}+\mathrm{c} \end{aligned}

Putting $v=x-y$

$\Rightarrow 2\left ( x-y \right )+log |x-y+2|=x+c$

( this is the required solution).