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

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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,


               (This is the required solution).



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