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  • Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 19

Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 19

Answers (1)

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

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

Given: \frac{d y}{d x}=\frac{e^{x}\left(\sin ^{2} x+\sin 2 x\right)}{y(2 \log y+1)}

Solution: \frac{d y}{d x}=\frac{e^{x}\left(\sin ^{2} x+\sin 2 x\right)}{y(2 \log y+1)}

        \begin{aligned} &\Rightarrow y(2 \log y+1) d y=e^{x}\left(\sin ^{2} x+\sin 2 x\right) d x \\\\ &\Rightarrow(2 y \log y+y) d y=\left(e^{x} \sin ^{2} x+e^{x} \sin 2 x\right) d x \\\\ &\Rightarrow 2 y \log y d y+y d y=e^{x} \sin ^{2} x d x+e^{x} \sin 2 x d x \end{aligned}

          Integrating both sides and using integrating by parts

        \Rightarrow 2\left[\log y \int y d y-\int\left\{\frac{d}{d y}(\log y) \int y d y\right\}\right] d y+\int y d y =\sin ^{2} x \int e^{x} d x-\int\left[\frac{d}{d x} \sin ^{2} x \int\left(e^{x} d x\right)\right] d x+\int e^{x} \sin 2 x d x

        \Rightarrow 2\left[\log y\left(\frac{y^{2}}{2}\right)-\int\left(\frac{1}{y}\right)\left(\frac{y^{2}}{2}\right) d y\right]+y d y =\sin ^{2} x e^{x}-\int\left[2 \sin x \cos x e^{x}\right] d x+\int e^{x} \sin 2 x+c

        \begin{aligned} &\Rightarrow y^{2} \log y-\int y d y+\int y d y=-\int e^{x} \sin 2 x d x+\int e^{x} \sin 2 x d x+e^{x} \sin ^{2} x \\\\ &y^{2} \log y=e^{x} \sin ^{2} x+c \end{aligned}

        

 

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