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