#### Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (viii)

Answer:  \begin{aligned} & x+y-1=C e^{-y} \\ & \end{aligned}

Give:  $(x+y) \frac{d y}{d x}=1$

Hint: Using integration by parts.

Explanation:  \begin{aligned} &(x+y) \frac{d y}{d x}=1 \\ & \end{aligned}

$=\frac{d x}{d y}=x+y \\$

$=\frac{d x}{d y}-x=y$

This is a first order linear differential equation of the form

\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=-1 \text { and } Q=y \end{aligned}

The integrating factor  $If$ of this differential equation is

\begin{aligned} &I f=e^{\int P d y}\\ &=e^{\int-1 d y}\\ &=e^{-\int d y}\\ &=e^{-y} \quad\left[\int d y=y+C\right] \end{aligned}

Hence, the solution of different equation is

\begin{aligned} &y I f=\int Q I f d x+C \\ &=y e^{-y}=\int y e^{-y} d y+C \ldots(i) \end{aligned}

Using integration by parts we have

\begin{aligned} &=\int y e^{-y} d y=\frac{y e^{-y}}{-1}-\int(1) e^{-y} d y+C \\ &=-y e^{-y}-\frac{e^{-y}}{-1}+C \\ &=-y e^{-y}+e^{-y}+C \end{aligned}

From i

$=x e^{-y}=-y e^{-y}+e^{-y}+C$

Divide by $e^{-y}$

\begin{aligned} &=x=-y+1+\frac{C}{e^{-y}} \\ &=x=-y+1+C e^{-y} \end{aligned}