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Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 44 maths

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Answer: y=e^{\sin ^{2} x}

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

Given: \frac{d y}{d x}=y \sin 2 x ; y(0)=1


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

        Integrating both sides

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

        Now Given that \text { at } x=0 ; y=1[y(0)=1]

        \begin{aligned} &\log |1|=\frac{-\cos 2(0)}{2}+c \\\\ &\Rightarrow 0=\frac{-1}{2}+c \Rightarrow c=\frac{1}{2} \end{aligned}

        Put in (1) we get

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

        \begin{aligned} &{\left[1-\cos 2 x=2 \sin ^{2} x\right]} \\\\ &\Rightarrow \log y=\sin ^{2} x \\\\ &{\left[\log _{\varepsilon} a=x \Rightarrow x=e^{x}\right]} \\\\ &\Rightarrow y=e^{\sin ^{2} x} \end{aligned}


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