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

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

Commented by mohammad17 last updated on 25/Sep/20

$$y^{{}^{″}}+lny=0\Rightarrow y^{{}^{″}}=-lny$$$$$$$$\int d\left(y^{{}^{\prime }}\right)=-\int lnydy$$$$$$$$y^{{}^{\prime }}=-ylny+y+C_1$$$$$$$$\int d\left(y\right)=\int (C_1+y-ylny)dy$$$$$$$$y=C_{1}y+\frac{1}{2}{y}^2-\frac{1}{2}{y}^{2}lny+\frac{1}{4}{y}^2+C_2$$$$$$$$y=\frac{3}{4}{y}^2-\frac{1}{2}{y}^{2}lny+C_{1}y+C_2$$$$$$$$y=y^2\left(\frac{3-2lny}{4}\right)+C_{1}y+C_2$$$$$$$$⟨m.t⟩$$

Commented by Olaf last updated on 26/Sep/20

$$sorrymister.$$$$\int d\left(y^{\prime }\right)=-\int lnydx....!$$$$andnot-\int lnydy$$$$Andasyoucansee,$$$$ifyouderivateyourfinalresult$$$$toverifytheinitialequation$$$$unafornately{it}^{\prime }swrong.$$

Answered by Olaf last updated on 24/Sep/20

$$y^{″}+\ln y=0$$$$y^{″}{y}^{\prime }+y^{\prime }\ln y=0$$$$\int y^{″}{y}^{\prime }dx+\int y^{\prime }\ln ydx=0$$$$\frac{1}{2}{y}^{\prime 2}+y\ln y-\int y\frac{y^{\prime }}{y}={\mathrm{C}}_1$$$$\frac{1}{2}{y}^{\prime 2}+y\ln y-y={\mathrm{C}}_1$$$$y^{\prime 2}=2y(1-\ln y)+{\mathrm{C}}_2$$$$\frac{y^{\prime }}{\sqrt{2y(1-\ln y)+{\mathrm{C}}_2}}=1$$$$\int \frac{y^{\prime }}{\sqrt{2y(1-\ln y)+{\mathrm{C}}_2}}dx=x+{\mathrm{C}}_3$$$$$$

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

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

Commented by Olaf last updated on 26/Sep/20

$$nosir.$$$$onlyapproximatesolutions$$

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