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

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

Answers (1)

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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