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  • Please Solve R.D. Sharma class 12 Chapter 21 Differential Equations Exercise 21.8 Question 1 Maths textbook Solution.

Please Solve R.D. Sharma class 12 Chapter 21  Differential Equations Exercise 21.8 Question 1 Maths textbook Solution.

Answers (1)

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

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

To find    :-  solve the differential equation.

Hint          :-  differential equation of the form

                      \frac{dy}{dx}=\left ( ax+by+c \right )can be reduced to variable separable form by the substitution ax+by+c=v

Solution:-we have

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

         Let\: x+y+1=v

         Differentiating with respect to x, we get

\begin{aligned} &\Rightarrow 1+\frac{d y}{d x}=\frac{d v}{d x} \\ &\Rightarrow \frac{d y}{d x}=\frac{d v}{d x}-1 \end{aligned}                                                                    ....(ii)

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

            \begin{aligned} & \frac{d v}{d x}-1=(x+y+1)^{2} \\ \Rightarrow & \frac{d v}{d x}-1=v^{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad(\because x+y+1=v) \\ \Rightarrow & \frac{d v}{d x}=v^{2}+1 \\ \Rightarrow & \frac{d v}{v^{2}+1}=d x \end{aligned}( taking like variable on same side)

Integrating on both sides, we get,

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

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

Putting the value of V, we get,

                      \Rightarrow \tan ^{-1}\left ( x+y+1 \right )=x+c            ,                    which is the required solution.

 

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