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Question Number 92510 by john santu last updated on 07/May/20

$$\frac{1}{{\mathrm{D}}^2+2}\left(\sin 2\mathrm{x}\right)?$$

Commented by john santu last updated on 08/May/20

$$\mathrm{consider}{\mathrm{y}}_{\mathrm{p}}=x\sin 2x$$$$D\left(x\sin 2x\right)=\sin 2x+2x\cos 2x$$$$D^2\left(x\sin 2x\right)=2\cos 2x+2\cos 2x-4x\sin 2x$$

Answered by niroj last updated on 08/May/20

$$=\frac{1}{{\mathrm{D}}^2+2}\left(\sin 2\mathrm{x}\right)$$$$=\frac{1}{-4+2}\sin 2\mathrm{x}$$$$=-\frac{1}{2}\sin 2\mathrm{x}//.$$$$$$

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