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Explain Solution R.D. Sharma Class 12 Chapter 21 Differential Equations Exercise 21.8 Question 10 Maths Textbook Solution.

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Answer: x=c\: e^{y}y-2

Given:\left ( x+y+1 \right )\frac{dy}{dx}=1

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,

                    \left ( x+y+1 \right )\frac{dy}{dx}=1                ........(i)

Let x+y=v

Differentiating with respect to x, we get,

                    \frac{d}{dx}\left ( x+y \right )\frac{dv}{dx}

                  \Rightarrow 1+\frac{d}{dx}=\frac{dv}{dx}

                \Rightarrow \frac{d}{dx}=\frac{dv}{dx}-1

Putting  dydx  = dvdx -1 and x + y = v in equation (i), we get,

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

Taking like variables on the same side, we get,

                        \Rightarrow \frac{v+1}{v+2} d v=\mathrm{d} \mathrm{x} \\

Integrating on both sides, we get,     

                        \Rightarrow \int \frac{v+1}{v+2} d v=\int \mathrm{d} \mathrm{x} \\

                        \Rightarrow \int\left(1-\frac{1}{v+2}\right) d v=\int \mathrm{d} \mathrm{x} \\

                        \Rightarrow(\mathrm{v}-\log |v+2|)=\mathrm{x}+\log \mathrm{c}_{1} \\

Putting v = x + y, we get,

                        \Rightarrow \mathrm{x}+\mathrm{y}-\log |\mathrm{x}+\mathrm{y}+2|=\mathrm{x}+\log \mathrm{c}_{1} \\

                        \Rightarrow \mathrm{x}+\mathrm{y}-\mathrm{x}=\log |\mathrm{x}+\mathrm{y}+2|+\log \mathrm{c}_{1} \\

                        \Rightarrow \mathrm{y}=\log \mathrm{c}_{1}|\mathrm{x}+\mathrm{y}+2| \\

                        \Rightarrow \mathrm{e}^{\mathrm{y}}=\mathrm{c}_{1}(\mathrm{x}+\mathrm{y}+2) \quad\left(\because \mathrm{c}=\frac{1}{\mathrm{c}_{1}}\right) \\

                        \Rightarrow \mathrm{x}=\mathrm{c} \mathrm{e}^{\mathrm{y}}-\mathrm{y}-2

(This is the required solution).





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