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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.8 Question 6 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.8 Question 6 Maths Textbook Solution.

Answers (1)

Answer:x=\tan \left ( x-2y \right )+c

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

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

Solution : -  We have,

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

Let x-2y=v

Differentiating with respect to x, we get,

                            \frac{d}{d x}(\mathrm{x}-2 \mathrm{y})=\frac{d v}{d x} \\

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

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

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

                \begin{aligned} & \cos ^{2}(x-2 y)=1-\left(1-\frac{d v}{d x}\right) \\ \Rightarrow & \cos ^{2} v=1-1+\frac{d v}{d x} \quad(\because \mathrm{v}=\mathrm{x}-2 \mathrm{y}) \\ \Rightarrow & \cos ^{2} v=\frac{d v}{d x} \end{aligned}

Taking like variables on same side, we get,

                 \Rightarrow dx=\frac{dv}{\cos ^{2}v}

                  \Rightarrow dx=\sec ^{2}vdv

Integrating on both the sides, we get,

                  \int dx=\int \sec ^{2}v\; dv

                \Rightarrow x=\tan \left ( x-2y \right )+c

(This is the required solution).

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