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  • Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 48 maths

Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 48 maths

Answers (1)

Answer: 2 y \sin y=2 x^{2} \log x+x^{2}-1

Hint: Separate the terms of x and y and then integrate them.

Given: \frac{d y}{d x}=\frac{2 x(\log x+1)}{\sin y+y \cos y}, y=0 \text { when } x=1

Solution:

        \begin{aligned} &\frac{d y}{d x}=\frac{2 x(\log x+1)}{\sin y+y \cos y} \\\\ &\Rightarrow(\sin y+y \cos y) d y=2 x(\log x+1) d x \end{aligned}

          Integrating both sides

        \begin{aligned} &\Rightarrow \int(\sin y+y \cos y) d y=\int 2 x(\log x+1) d x \\\\ &\Rightarrow \int \sin y d y+\int y \cos y d y=\int 2 x \log x d x+\int 2 x d x \end{aligned}                    [?Integration by parts]

        \begin{aligned} &\Rightarrow-\cos y+\left[y \sin y-\int \sin y d y\right]=2\left[\log x \cdot \frac{x^{2}}{2}-\int \frac{1}{x} \frac{x^{2}}{2} d x\right] \\\\ &\Rightarrow-\cos y+y \sin y+\cos y=x^{2} \log x-\frac{x^{2}}{2}+c \\\\ &\Rightarrow y \sin y=x^{2} \log x+\frac{x^{2}}{2}+c \end{aligned}   ...........(1)

        Given that y=0 \text { when } x=1

        \begin{aligned} &\Rightarrow 0 \sin 0=1 \cdot \log 1+\frac{1}{2}+c \\\\ &\Rightarrow 0=0+\frac{1}{2}+c \\\\ &\Rightarrow c=-\frac{1}{2} \end{aligned}

        Put in (1)

        \begin{aligned} &\Rightarrow y \sin y=x^{2} \log x+\frac{x^{2}}{2}-\frac{1}{2} \\\\ &\Rightarrow 2 y \sin y=2 x^{2} \log x+x^{2}-1 \end{aligned}

 

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