Question Number 91272 by jagoll last updated on 29/Apr/20
$$\left({D}^{\mathrm{2}} +\mathrm{4}\right){y}\:=\:\mathrm{sin}\:\mathrm{2}{x} \\ $$
Answered by niroj last updated on 29/Apr/20
![(D^2 +4)y= sin2x A.E., m^2 +4=0 m=0+^− 2i CF= C_1 cos2x+C_2 sin2x PI= ((sin2x)/(D^2 +4))= ((x sin2x)/(2D)) = (1/2)∫xsin2xdx = (1/2)[ −((x cos2x)/2)−∫1.(((−cos2x))/2)dx] = (1/2)[ ((−x cos2x)/2)+ (1/2).((sin2x)/2)] = ((−x cos2x)/4)+ ((sin2x)/8) y= CF+PI y= C_1 cos2x+C_2 sin2x−((x cos2x)/4) + ((sin2x)/8) //.](https://www.tinkutara.com/question/Q91286.png)
$$\:\left(\mathrm{D}^{\mathrm{2}} +\mathrm{4}\right)\mathrm{y}=\:\mathrm{sin2x} \\ $$$$\:\mathrm{A}.\mathrm{E}.,\:\:\mathrm{m}^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{m}=\mathrm{0}\overset{−} {+}\mathrm{2i} \\ $$$$\:\:\:\mathrm{CF}=\:\mathrm{C}_{\mathrm{1}} \mathrm{cos2x}+\mathrm{C}_{\mathrm{2}} \mathrm{sin2x} \\ $$$$\:\:\mathrm{PI}=\:\frac{\mathrm{sin2x}}{\mathrm{D}^{\mathrm{2}} +\mathrm{4}}=\:\frac{\mathrm{x}\:\mathrm{sin2x}}{\mathrm{2D}} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{xsin2xdx} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\:−\frac{\mathrm{x}\:\mathrm{cos2x}}{\mathrm{2}}−\int\mathrm{1}.\frac{\left(−\mathrm{cos2x}\right)}{\mathrm{2}}\mathrm{dx}\right] \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\:\frac{−\mathrm{x}\:\mathrm{cos2x}}{\mathrm{2}}+\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{sin2x}}{\mathrm{2}}\right] \\ $$$$\:\:=\:\frac{−\mathrm{x}\:\mathrm{cos2x}}{\mathrm{4}}+\:\frac{\mathrm{sin2x}}{\mathrm{8}} \\ $$$$\:\:\mathrm{y}=\:\mathrm{CF}+\mathrm{PI} \\ $$$$\:\:\mathrm{y}=\:\mathrm{C}_{\mathrm{1}} \mathrm{cos2x}+\mathrm{C}_{\mathrm{2}} \mathrm{sin2x}−\frac{\mathrm{x}\:\mathrm{cos2x}}{\mathrm{4}}\:+\:\frac{\mathrm{sin2x}}{\mathrm{8}}\://. \\ $$
Commented by jagoll last updated on 29/Apr/20
$${thank}\:{you} \\ $$
Commented by niroj last updated on 29/Apr/20
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