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Please Solve R.D.Sharma class 12 Chapter 21 Differential Equations Exercise 21.8 Question 5 Maths textbook Solution.

Answers (1)

Answer: y-\tan ^{-1}\left ( x+y \right )=c

Given:\left ( x+y \right )^{2}\frac{dy}{dx}=1

Hint : -  Use variable separable form by substitution.

Solution : -  We have,

                    \left ( x+y \right )^{2}\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{dy}{dx}=\frac{dv}{dx}

                \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1                                                .....(ii)

Now, substituting equation (ii) in equation (i), we get

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

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

                \Rightarrow \frac{d v}{d x}-1=\frac{1}{v^{2}} \\

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

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

Taking like variables on same side, we get,

             \Rightarrow \frac{v^{2}}{1+v^{2}}dv=dx

Integrating on both the sides, we get,

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

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

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

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

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

                \Rightarrow v-\tan ^{-1} v=x+c \quad\left(\because \int\left(\frac{1}{x^{2}+1}\right) d x=\tan ^{-1} x\right)

Putting v=x+y, we have

             \Rightarrow \left ( x+y \right )-\tan ^{-1}\left ( x+y \right )=x+c

              \Rightarrow y-\tan ^{-1}\left ( x+y \right )=c

                            (This is the required solution).



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