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Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 25 maths textbook solution.

Answers (1)

Answer : x=\tan y+C \sqrt{\tan y}

Hint : To solve this equation we use \tan \; x and convert it to \sin x \; and \cos\; x .

Give : (x+\tan y) d y=\sin 2 y d x

Solution : (x+\tan y) d y=\sin 2 y d x

        \begin{aligned} &=\frac{d x}{d y}=\frac{(x+\tan y)}{\sin 2 y} \\ \end{aligned}

       \begin{aligned} &=\frac{d x}{d y}=\frac{x}{\sin 2 y}+\frac{\tan y}{\sin 2 y} \\ \end{aligned}

       \begin{aligned} &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{\sin y}{2 \cos y} \\ \end{aligned}

       \begin{aligned} &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{\frac{\sin y}{\cos y}}{2 \sin y \cos y} \\ \end{aligned}

      \begin{aligned} &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{1}{2 \cos ^{2} y} \\ \end{aligned}

      \begin{aligned} &=\frac{d x}{d y}=x \operatorname{cosec} 2 y+\frac{1}{2} \sec ^{2} y \end{aligned}


        \begin{aligned} &R=\operatorname{cosec} 2 y, S=\frac{1}{2} \sec ^{2} y \\ \end{aligned}

        \begin{aligned} &I f=e^{\int R d y} \\ \end{aligned}

       \begin{aligned} &=e^{\int(-\operatorname{cosec} 2 y) d y} \\ \end{aligned}

       \begin{aligned} &=e^{-\log |\operatorname{cosec} 2 y-\cot 2 y|} \\ \end{aligned}

       \begin{aligned} &=\operatorname{cosec} 2 y-\cot 2 y \\ \end{aligned}

        \begin{aligned} &=\frac{1}{\sin 2 y}-\frac{\cos 2 y}{\sin 2 y} \\ \end{aligned}

        \begin{aligned} &=\frac{1-\cos 2 y}{\sin 2 y} \\ \end{aligned}

        \begin{aligned} &=\frac{2 \sin ^{2} y}{2 \sin y \cos y} \\ \end{aligned}

       \begin{aligned} &=\frac{\sin y}{\cos y} \\ \end{aligned}

       \begin{aligned} &=\tan y \\ \end{aligned}

       \begin{aligned} &=e^{-\log |\tan y|} \\ \end{aligned}

       \begin{aligned} &=e^{\log |\cot y|} \\ &=\cot y \end{aligned}

        \begin{aligned} &=x I f=\int S I f+C \\ &=x \cot y=\int \frac{1}{2} \sec ^{2} y \cot y d y+C \\ &=x \cot y=\int \frac{1}{2 \cos ^{2} y} \frac{\cos y}{\sin y} d y+C \\ &=x \cot y=\int \frac{1}{2 \cos y} \frac{1}{\sin y} d y+C \end{aligned}

        \begin{aligned} &=x \cot y=\int \frac{1}{\sin 2 y} d y+C \\ &=x \cot y=\int \operatorname{cosec} 2 y d y+C \\ &=x \cot y=\frac{1}{2} \log |\operatorname{cosec} 2 y-\cot 2 y|+C \end{aligned}

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