#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.8 Question 11 Maths Textbook Solution.

Answer:  $-e^{-\left ( x+y \right )}=x+c$

Given: $\frac{dy}{dx}+1=e^{x+y}$

Hint: Differential equation of the form $\frac{dy}{dx}=\int \left ( ax+by+c \right )$ can be reduced to variable separable form by substitution     $ax+by+c=v$

Solution:  We have,

$\frac{dy}{dx}+1=e^{x+y}$                .......(i)

Let $x+y=v$

Differentiating with respect to x, we get,

$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\mathrm{y})=\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}} \\$

$\Rightarrow 1+\frac{d y}{d x}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}} \quad \cdots \text { (ii) } \\$

Substituting (ii) in equation (i), we get,

$\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}=\mathrm{e}^{\mathrm{x}+\mathrm{y}} \\$

$\Rightarrow \frac{\mathrm{d} \mathrm{v}}{\mathrm{d} x}=\mathrm{e}^{\mathrm{v}} \quad[\because \mathrm{x}+\mathrm{y}=\mathrm{v}]$

Taking like variables on the same side, we get,

$\Rightarrow \frac{\mathrm{d} v}{\mathrm{e}^{\mathrm{v}}}=\mathrm{d} \mathrm{x} \\$

$\Rightarrow \mathrm{e}^{-\mathrm{v}} \mathrm{d} \mathrm{v}=\mathrm{d} \mathrm{x} \\$

Integrating on both sides, we get,

$\Rightarrow \int \mathrm{e}^{-\mathrm{v}} \mathrm{d} \mathrm{v}=\int \mathrm{d} \mathrm{x} \\$

$\Rightarrow-\mathrm{e}^{-\mathrm{v}}=\mathrm{x}+\mathrm{c} \\$

Putting $v=x+y$, we get,

$\Rightarrow-\mathrm{e}^{-(x+y)}=\mathrm{x}+\mathrm{c}$

(This is the required solution).