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4x-2-y-y-0-x-gt-0-




Question Number 103037 by bramlex last updated on 12/Jul/20
4x^2 y′′ +y = 0 , x > 0
$$\mathrm{4x}^{\mathrm{2}} \mathrm{y}''\:+\mathrm{y}\:=\:\mathrm{0}\:,\:\mathrm{x}\:>\:\mathrm{0} \\ $$
Answered by maths mind last updated on 12/Jul/20
let y=x^t   ⇒4x^2 (t(t−1))x^(t−2) +x^t =0  ⇔(4t^2 −4t+1)x^t =0  ⇔(2t−1)^2 x^t =0⇒t=(1/2)  let y=(√x) z  ⇒y′=z′(√x)+(z/(2(√x))),y′′=((z′)/( (√x)))+z′′(√x)−(z/(4x(√x)))  ⇔4x(√x)z′+4x^2 (√x)z′′−z(√x)+z(√x)=0  ⇒z′+xz′′=0  ⇒((z′′)/(z′))=−(1/x)⇒ln(z′)=−ln(x)+c  ⇒z′=(k/x)⇒z=kln(x)+c  ⇒y=(kln(x)+c)(√x)
$${let}\:{y}={x}^{{t}} \\ $$$$\Rightarrow\mathrm{4}{x}^{\mathrm{2}} \left({t}\left({t}−\mathrm{1}\right)\right){x}^{{t}−\mathrm{2}} +{x}^{{t}} =\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{4}{t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{1}\right){x}^{{t}} =\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{2}{t}−\mathrm{1}\right)^{\mathrm{2}} {x}^{{t}} =\mathrm{0}\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${let}\:{y}=\sqrt{{x}}\:{z} \\ $$$$\Rightarrow{y}'={z}'\sqrt{{x}}+\frac{{z}}{\mathrm{2}\sqrt{{x}}},{y}''=\frac{{z}'}{\:\sqrt{{x}}}+{z}''\sqrt{{x}}−\frac{{z}}{\mathrm{4}{x}\sqrt{{x}}} \\ $$$$\Leftrightarrow\mathrm{4}{x}\sqrt{{x}}{z}'+\mathrm{4}{x}^{\mathrm{2}} \sqrt{{x}}{z}''−{z}\sqrt{{x}}+{z}\sqrt{{x}}=\mathrm{0} \\ $$$$\Rightarrow{z}'+{xz}''=\mathrm{0} \\ $$$$\Rightarrow\frac{{z}''}{{z}'}=−\frac{\mathrm{1}}{{x}}\Rightarrow{ln}\left({z}'\right)=−{ln}\left({x}\right)+{c} \\ $$$$\Rightarrow{z}'=\frac{{k}}{{x}}\Rightarrow{z}={kln}\left({x}\right)+{c} \\ $$$$\Rightarrow{y}=\left({kln}\left({x}\right)+{c}\right)\sqrt{{x}} \\ $$

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