#### Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (viii) maths textbook solution.

Answer : $x+y-1= Ce^{-y}$

Give : $(x+y)\frac{dy}{dx}=1$

Hint : Using integration by parts.

Explanation : $(x+y)\frac{dy}{dx}=1$

$=\frac{dx}{dy}=x+y$

$=\frac{dx}{dy}-x=y$

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\; \; \; \; \; \; ....(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}