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y-2-x-y-y-0-y-




Question Number 198688 by tri26112004 last updated on 23/Oct/23
y′′ + (2/x).y′ + y = 0  y=¿
$${y}''\:+\:\frac{\mathrm{2}}{{x}}.{y}'\:+\:{y}\:=\:\mathrm{0} \\ $$$${y}=¿ \\ $$
Answered by witcher3 last updated on 23/Oct/23
xy=z  xy′+y=z′  2y′+xy′′=z′′  y′′+(2/x)y′+y=0⇔z′′+z=0  z=acos(x)+bsin(x)  y=((acos(x)+bsin(x))/x),(a,b)∈R^2
$$\mathrm{xy}=\mathrm{z} \\ $$$$\mathrm{xy}'+\mathrm{y}=\mathrm{z}' \\ $$$$\mathrm{2y}'+\mathrm{xy}''=\mathrm{z}'' \\ $$$$\mathrm{y}''+\frac{\mathrm{2}}{\mathrm{x}}\mathrm{y}'+\mathrm{y}=\mathrm{0}\Leftrightarrow\mathrm{z}''+\mathrm{z}=\mathrm{0} \\ $$$$\mathrm{z}=\mathrm{acos}\left(\mathrm{x}\right)+\mathrm{bsin}\left(\mathrm{x}\right) \\ $$$$\mathrm{y}=\frac{\mathrm{acos}\left(\mathrm{x}\right)+\mathrm{bsin}\left(\mathrm{x}\right)}{\mathrm{x}},\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$

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