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Question Number 116375 by mnjuly1970 last updated on 03/Oct/20

$$...nicecalculus...$$$$ordinarydifferential$$$$equation(o.d.e)$$$$$$$$y\frac{d^{2}y}{{dx}^2}-{\left(\frac{dy}{dx}\right)}^2=y^2\left(lny\right)...$$$$find:generalsolution$$$$..m.n.1970..$$

Answered by Olaf last updated on 03/Oct/20

$$\frac{y\left(\frac{d^{2}y}{{dx}^2}\right)-\left(\frac{dy}{dx}\right)\left(\frac{dy}{dx}\right)}{y^2}=\ln y$$$$\frac{d}{dx}\left(\frac{\left(\frac{dy}{dx}\right)}{y}\right)=\ln y$$$$y=e^u$$$$\frac{d}{dx}\left(\frac{\left(\frac{du}{dx}\right)e^u}{e^u}\right)=u$$$$\frac{d^{2}u}{{dx}^2}=u$$$$\frac{d^{2}u}{{dx}^2}-u=0$$$$r^2-1=0$$$$r=\pm 1$$$$u=\mathrm{A}{e}^x+\mathrm{B}{e}^{-x}$$$$y=e^u=e^{\mathrm{A}{e}^x+\mathrm{B}{e}^{-x}}$$$$$$$$$$

Answered by mnjuly1970 last updated on 04/Oct/20

$$\frac{dy}{dx}=p\Rightarrow \frac{d^{2}y}{{dx}^2}=\frac{dp}{dx}=p\frac{dp}{dy}$$$$p\frac{dp}{dy}-y^{-1}{p}^2=yln\left(y\right)$$$$p^2=u\left(bernoulli\right)\Rightarrow 2p\frac{dp}{dy}=\frac{du}{dy}$$$$\frac{du}{dy}-2\frac{u}{y}=2yln\left(y\right)$$$$\mu =e^{\int \frac{-2}{y}dy}=\frac{1}{y^2}(I.F)$$$$u=y^2(\int \frac{1}{y^2}2yln\left(y\right)dy+k)=y^2{\left(lny\right)}^2+{ky}^2$$$$p=y\sqrt{{\left(lny\right)}^2+k}$$$$\int \frac{dy}{y\sqrt{{\left(lny\right)}^2+k}}=x+c_1$$$$t=ln\left(y\right)\Rightarrow dt=\frac{1}{y}dy$$$$\int \frac{dt}{\sqrt{t^2+k}}=x+c_1\Rightarrow t=\sqrt{k}sinh\left(r\right)$$$$dt=\sqrt{k}cosh\left(r\right)dr$$$$\frac{1}{\sqrt{k}}\int \sqrt{k}\frac{cosh\left(r\right)}{cosh\left(r\right)}dr=x+c_1$$$$r=x+c_1\Rightarrow {sinh}^{-1}\left(\frac{t}{\sqrt{k}}\right)=x+c_1$$$$\left(\frac{t}{\sqrt{k}}\right)=sinh(x+c_1)$$$$t=\sqrt{k}sinh(x+c_1)\Rightarrow y=e^{\sqrt{k}}{}^{sinh(x+c_1)}$$$$$$$$m.n.197o$$$$$$$$$$$$$$

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