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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.8 Question 8 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.8 Question 8 Maths Textbook Solution.

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Answer: y-x+log|\sin \left ( x+y \right )+\cos \left ( x+y \right )|=c

Given:\frac{dy}{dx}=\tan (x+y)

Hint : -  first, we will separate variables and then solve.

Solution : -  We have,

                    \frac{dy}{dx}=\tan (x+y)

                   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

Putting x+y=v and \frac{dy}{dx}=\frac{dv}{dx}-1 in the given differential equation, we get,

                \therefore \frac{dy}{dx}=\frac{dv}{dx}-1=\tan v

            \Rightarrow \frac{dv}{dx}=1+\tan v

            Taking like variables on same side, we get,

            \begin{aligned} &\Rightarrow \frac{d v}{1+\tan v}=\mathrm{d} \mathrm{x} \\ &\Rightarrow\left(\frac{1}{1+\frac{\sin v}{\cos v}}\right) d v=d x \\ &\Rightarrow\left(\frac{\cos v}{\cos v+\sin v}\right) d v=d x \end{aligned}

Multiplying 2 on both the sides, we get,

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

Integrating on both sides, we get,

                \begin{aligned} &\Rightarrow \int\left(1+\frac{\cos v-\sin v}{\cos v+\sin v}\right) \mathrm{d} \mathrm{v}=2 \mathrm{~d} \mathrm{x} \\ &\Rightarrow \int \mathrm{d} \mathrm{v}+\int \frac{\cos v-\sin v}{\cos v+\sin v} \mathrm{~d} \mathrm{v}=2 \mathrm{~d} \mathrm{x} \\ &\Rightarrow \mathrm{v}+\log |\operatorname{cos} \mathrm{v}+\sin \mathrm{v}|=2 \mathrm{x}+\mathrm{c} \end{aligned}

Putting v = x + y, we get,

            \begin{aligned} &\Rightarrow(x+y)+\log \mid \operatorname{cos}(x+y)+\sin (x+y)=2 x+c \\ &\Rightarrow y-x+\log |\operatorname{cos}(x+y)+\sin (x+y)|=c \end{aligned}

(This is the required solution).

Posted by

infoexpert21

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