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Question Number 93451 by mhmd last updated on 13/May/20

Solve by using change the conistant megbod   4y^(′′) +y=((x^2 −1)/(x(√x)))?

$${Solve}\:{by}\:{using}\:{change}\:{the}\:{conistant}\:{megbod}\: \\ $$$$\mathrm{4}{y}^{''} +{y}=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}\sqrt{{x}}}? \\ $$

Commented by i jagooll last updated on 13/May/20

what is conistant megbod?

$$\mathrm{what}\:\mathrm{is}\:\mathrm{conistant}\:\mathrm{megbod}? \\ $$

Commented by mhmd last updated on 13/May/20

sory method not megbod

$${sory}\:{method}\:{not}\:{megbod} \\ $$

Commented by john santu last updated on 13/May/20

characteristic eq : 4r^2 +1 = 0  r = ± (1/2) i . homogenous solution  y_h = C_1 e^((1/2)i) +C_2 e^(−(1/2)i)   y_h  = C_1 cos ((1/2)x)+C_2 sin ((1/2)x)  Particular solution   y_p  =(((√x)−x^(−3/2) )/(4D^2 +1))

$$\mathrm{characteristic}\:\mathrm{eq}\::\:\mathrm{4r}^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{r}\:=\:\pm\:\frac{\mathrm{1}}{\mathrm{2}}\:{i}\:.\:\mathrm{homogenous}\:\mathrm{solution} \\ $$$$\mathrm{y}_{\mathrm{h}} =\:\mathrm{C}_{\mathrm{1}} \mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}{i}} +\mathrm{C}_{\mathrm{2}} \mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{i}} \\ $$$$\mathrm{y}_{\mathrm{h}} \:=\:\mathrm{C}_{\mathrm{1}} \mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{2}}{x}\right)+\mathrm{C}_{\mathrm{2}} \mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}{x}\right) \\ $$$$\mathrm{Particular}\:\mathrm{solution}\: \\ $$$$\mathrm{y}_{\mathrm{p}} \:=\frac{\sqrt{{x}}−{x}^{−\mathrm{3}/\mathrm{2}} }{\mathrm{4D}^{\mathrm{2}} +\mathrm{1}}\: \\ $$

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