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Question Number 91272 by jagoll last updated on 29/Apr/20

$$(D^2+4)y=\sin 2x$$

Answered by niroj last updated on 29/Apr/20

$$({\mathrm{D}}^2+4)\mathrm{y}=\sin 2\mathrm{x}$$$$\mathrm{A}.\mathrm{E}.,{\mathrm{m}}^2+4=0$$$$\mathrm{m}=0\stackrel{-}{+}2\mathrm{i}$$$$\mathrm{CF}={\mathrm{C}}_1\cos 2\mathrm{x}+{\mathrm{C}}_2\sin 2\mathrm{x}$$$$\mathrm{PI}=\frac{\sin 2\mathrm{x}}{{\mathrm{D}}^2+4}=\frac{\mathrm{x}\sin 2\mathrm{x}}{2\mathrm{D}}$$$$=\frac{1}{2}\int \mathrm{xsin}2\mathrm{xdx}$$$$=\frac{1}{2}[-\frac{\mathrm{x}\cos 2\mathrm{x}}{2}-\int 1.\frac{(-\cos 2\mathrm{x})}{2}\mathrm{dx}]$$$$=\frac{1}{2}[\frac{-\mathrm{x}\cos 2\mathrm{x}}{2}+\frac{1}{2}.\frac{\sin 2\mathrm{x}}{2}]$$$$=\frac{-\mathrm{x}\cos 2\mathrm{x}}{4}+\frac{\sin 2\mathrm{x}}{8}$$$$\mathrm{y}=\mathrm{CF}+\mathrm{PI}$$$$\mathrm{y}={\mathrm{C}}_1\cos 2\mathrm{x}+{\mathrm{C}}_2\sin 2\mathrm{x}-\frac{\mathrm{x}\cos 2\mathrm{x}}{4}+\frac{\sin 2\mathrm{x}}{8}//.$$

Commented by jagoll last updated on 29/Apr/20

$$thankyou$$

Commented by niroj last updated on 29/Apr/20

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