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  • Please Solve R.D. Sharma class 12 Chapter 21 Differential Equations Exercise 21.8 Question 2 Maths textbook Solution.

Please Solve R.D. Sharma class 12 Chapter 21  Differential Equations Exercise 21.8 Question 2 Maths textbook Solution.

Answers (1)

Answer: \cot \frac{x-y}{2}=y+C_{1}

Gievn:-\frac{dy}{dx}\cot( {x-y})=1

To find    :-  solve the differential equation.

Hint          :-  We will reduce the differential equation to variable separable form by substitution .

Solution  ;-  We have,

                    \frac{dy}{dx}\cot( {x-y})=1                                   .....(i)

Let x-y=v

Differentiating with respect to x,  we get,

            \frac{d}{dx}\left ( x-y \right )=\frac{d}{dx}\left ( v \right )

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

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

Putting the value of \frac{dy}{dx} in Equ. (i), we get

                \begin{aligned} &\left(1-\frac{d v}{d x}\right) \cos v=1 \; \; \; \; \; \; \; \; \; \; \; \; \; \quad(\because \mathrm{x}-\mathrm{y}=\mathrm{v}) \\ &\Rightarrow \quad 1-\frac{d v}{d x}=\frac{1}{\cos v} \\ &\Rightarrow \quad 1-\frac{d v}{d x}=\operatorname{sec} \mathrm{v} \\ &\Rightarrow \frac{d v}{d x}=1-\operatorname{sec} \mathrm{v} \end{aligned}

Now, taking like variable in same side, we get,

                    \begin{aligned} &\Rightarrow \frac{d v}{1-\sec v}=d x \\ &\Rightarrow \mathrm{dx}=\frac{d v}{1-\sec v} \\ &\Rightarrow \mathrm{dx}=\left(\frac{\cos v}{\cos v-1}\right) \mathrm{d} \mathrm{v} \end{aligned}

Integrating in both side, we get,

                    \begin{aligned} &\int d x=\int-\left(\frac{\cos v}{1-\cos v}\right) d v \\ &\Rightarrow \mathrm{x}=-\int \frac{\cos v}{1-\operatorname{Cos} v} \times \frac{1+\cos v}{1+\cos v} d v \\ &\Rightarrow \mathrm{x}=-\int \frac{\cos v+\cos ^{2} v}{1-\cos ^{2} v} d v \\ &\Rightarrow \mathrm{x}=-\int\left(\frac{\cos v+\cos ^{2} v}{\sin ^{2} v}\right) d v \\ &\Rightarrow \mathrm{x}=-\int\left(\frac{\cos v}{\sin ^{2} v}+\frac{\cos ^{2} v}{\sin ^{2} v}\right) d v \\ &\Rightarrow \mathrm{x}=-\int\left(\cot v \operatorname{Cosec} v+\cot ^{2}\right) d v \end{aligned}

                   \begin{aligned} &\Rightarrow \mathrm{x}=-\int\left(\cot v \operatorname{cosec} v+\operatorname{cosec}^{2} v-1\right) d v \\ &\Rightarrow \mathrm{x}=-(-\operatorname{Cosec} v-\cot v-v)+C \\ &\Rightarrow \mathrm{x}=\operatorname{Cosec} \mathrm{v}+\cot \mathrm{v}+\mathrm{v}+\mathrm{C}_{1} \\ &\Rightarrow \mathrm{x}=\frac{1}{\sin \mathrm{v}}+\frac{\operatorname{cosv}}{\sin \mathrm{v}}+\mathrm{v}+\mathrm{C}_{1} \\ &\Rightarrow \mathrm{x}-\mathrm{v}-\mathrm{C}_{1}=\frac{1+\operatorname{cosv}}{\sin \mathrm{v}} \end{aligned}

                  \begin{aligned} &\Rightarrow \mathrm{x}-\mathrm{v}-\mathrm{C}_{1}=\frac{2 \operatorname{los}^{2} \frac{\mathrm{v}}{2}}{2 \cdot 8 \mathrm{~s}_{2}^{\mathrm{T}} \sin \frac{\mathrm{q}}{2}} \\ &\Rightarrow \mathrm{x}-\mathrm{v}-\mathrm{C}_{1}=\operatorname{cot} \frac{\mathrm{v}}{2} \end{aligned}\text { (put } \left.\mathrm{C}=\mathrm{C}_{1} \text { and using } 1+\operatorname{Cos} 2 \mathrm{x}=2 \operatorname{Cos}^{2} \mathrm{x} ; \sin \mathrm{x}=2 \sin \frac{\mathrm{x}}{2} \cos \frac{\mathrm{x}}{2}\right)

Putting v=x-y

                \Rightarrow x-/x+/y+C_{1}=\cot \left ( \frac{x-y}{2} \right )

                \Rightarrow \cot \left ( \frac{x-y}{2} \right )=y+C_{1}

                                                (This is the required solution)

                  

 

 

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