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Question Number 115298 by Dwaipayan Shikari last updated on 24/Sep/20

(d^2 y/dx^2 )+log(y)=0

$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }+\mathrm{log}\left(\mathrm{y}\right)=\mathrm{0} \\ $$

Commented by mohammad17 last updated on 25/Sep/20

y^(′′) +lny=0⇒y^(′′) =−lny ∫d(y^′ )=−∫lnydy y^′ =−ylny+y+C_1 ∫d(y)=∫(C_1 +y−ylny)dy y=C_1 y+(1/2)y^2 −(1/2)y^2 lny+(1/4)y^2 +C_2 y=(3/4)y^2 −(1/2)y^2 lny+C_1 y+C_2 y=y^2 (((3−2lny)/4))+C_1 y+C_2 ⟨m.t⟩

$${y}^{''} +{lny}=\mathrm{0}\Rightarrow{y}^{''} =−{lny} \\ $$$$ \\ $$$$\int{d}\left({y}^{'} \right)=−\int{lnydy} \\ $$$$ \\ $$$${y}^{'} =−{ylny}+{y}+{C}_{\mathrm{1}} \\ $$$$ \\ $$$$\int{d}\left({y}\right)=\int\left({C}_{\mathrm{1}} +{y}−{ylny}\right){dy} \\ $$$$ \\ $$$${y}={C}_{\mathrm{1}} {y}+\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} {lny}+\frac{\mathrm{1}}{\mathrm{4}}{y}^{\mathrm{2}} +{C}_{\mathrm{2}} \\ $$$$ \\ $$$${y}=\frac{\mathrm{3}}{\mathrm{4}}{y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} {lny}+{C}_{\mathrm{1}} {y}+{C}_{\mathrm{2}} \\ $$$$ \\ $$$${y}={y}^{\mathrm{2}} \left(\frac{\mathrm{3}−\mathrm{2}{lny}}{\mathrm{4}}\right)+{C}_{\mathrm{1}} {y}+{C}_{\mathrm{2}} \\ $$$$ \\ $$$$\langle{m}.{t}\rangle \\ $$

Commented by Olaf last updated on 26/Sep/20

sorry mister. ∫d(y′) = −∫lnydx ....! and not −∫lnydy And as you can see, if you derivate your final result to verify the initial equation unafornately it′s wrong.

$${sorry}\:{mister}. \\ $$$$\int{d}\left({y}'\right)\:=\:−\int{lnydx}\:....! \\ $$$${and}\:{not}\:−\int{lnydy} \\ $$$${And}\:{as}\:{you}\:{can}\:{see}, \\ $$$${if}\:{you}\:{derivate}\:{your}\:{final}\:{result} \\ $$$${to}\:{verify}\:{the}\:{initial}\:{equation} \\ $$$${unafornately}\:{it}'{s}\:{wrong}. \\ $$

Answered by Olaf last updated on 24/Sep/20

y′′+lny = 0 y′′y′+y′lny = 0 ∫y′′y′dx+∫y′lnydx = 0 (1/2)y′^2 +ylny−∫y((y′)/y) = C_1 (1/2)y′^2 +ylny−y = C_1 y′^2 = 2y(1−lny)+ C_2 ((y′)/( (√(2y(1−lny)+C_2 )))) = 1 ∫((y′)/( (√(2y(1−lny)+C_2 ))))dx = x+C_3

$${y}''+\mathrm{ln}{y}\:=\:\mathrm{0} \\ $$$${y}''{y}'+{y}'\mathrm{ln}{y}\:=\:\mathrm{0} \\ $$$$\int{y}''{y}'{dx}+\int{y}'\mathrm{ln}{ydx}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{y}'^{\mathrm{2}} +{y}\mathrm{ln}{y}−\int{y}\frac{{y}'}{{y}}\:=\:\mathrm{C}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{y}'^{\mathrm{2}} +{y}\mathrm{ln}{y}−{y}\:=\:\mathrm{C}_{\mathrm{1}} \\ $$$${y}'^{\mathrm{2}} \:=\:\mathrm{2}{y}\left(\mathrm{1}−\mathrm{ln}{y}\right)+\:\mathrm{C}_{\mathrm{2}} \\ $$$$\frac{{y}'}{\:\sqrt{\mathrm{2}{y}\left(\mathrm{1}−\mathrm{ln}{y}\right)+\mathrm{C}_{\mathrm{2}} }}\:=\:\mathrm{1} \\ $$$$\int\frac{{y}'}{\:\sqrt{\mathrm{2}{y}\left(\mathrm{1}−\mathrm{ln}{y}\right)+\mathrm{C}_{\mathrm{2}} }}{dx}\:=\:{x}+\mathrm{C}_{\mathrm{3}} \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 25/Sep/20

Is there any definite solution?

$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{definite}\:\mathrm{solution}? \\ $$

Commented by Olaf last updated on 26/Sep/20

no sir. only approximate solutions

$${no}\:{sir}. \\ $$$${only}\:{approximate}\:{solutions} \\ $$

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