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Question Number 2054 by Yozzi last updated on 01/Nov/15

$$Findallrealsolutionsytotheequation$$$$sin\left(\frac{dy}{dx}\right)+sin\left(\frac{d^{2}y}{{dx}^2}\right)=0.$$$$Foreachsolutiondeterminethevalueof$$$$Q=max\left(\underset{r=1}{\sum ^n}sin\left(\frac{d^{r}y}{{dx}^r}\right)\right),givingthevalue\left(s\right)$$$$ofxforwhichQarises.$$

Commented by 123456 last updated on 01/Nov/15

$$y=k$$

Commented by 123456 last updated on 01/Nov/15

$$\mid \frac{dy}{dx}\mid \ll 1,\mid \frac{d^{2}y}{{dx}^2}\mid \ll 1,\mid \theta \mid \ll 1\Rightarrow \sin \theta \approx \theta$$$$\frac{dy}{dx}+\frac{d^{2}y}{{dx}^2}=0$$$$\lambda +{\lambda }^2=\lambda (1+\lambda )=0$$$$y={\mathrm{C}}_1+{\mathrm{C}}_2{e}^{-x}$$$$-----------$$$$y={\mathrm{C}}_1+{\mathrm{C}}_2{e}^{-x}$$$$\frac{dy}{dx}=-{\mathrm{C}}_2{e}^{-x}$$$$\frac{d^{2}y}{{dx}^2}={\mathrm{C}}_2{e}^{-x}$$$$\sin \frac{dy}{dx}+\sin \frac{d^{2}y}{{dx}^2}=0$$$$-\sin {\mathrm{C}}_2{e}^{-x}+\sin {\mathrm{C}}_2{e}^{-x}=0\left(\mathrm{T}\right)$$$$\mathrm{so}$$$$y={\mathrm{C}}_1+{\mathrm{C}}_2{e}^{-x}$$$$\mathrm{is}\mathrm{a}\mathrm{solution},\mathrm{but}\mathrm{not}\mathrm{sure}\mathrm{if}\mathrm{its}\mathrm{the}$$$$\mathrm{full}\mathrm{solution}.$$

Commented by prakash jain last updated on 01/Nov/15

$$$$$$Q=max\left(\underset{r=1}{\sum ^n}sin\left(\frac{d^{r}y}{{dx}^r}\right)\right)=\underset{r=1}{\sum ^n}\sin \frac{d^{r}y}{{dx}^r}$$$$maxseemstobeunnecessaryfunctionsince$$$$thereisonlyoneparameter\mathrm{sum}.$$

Commented by Yozzi last updated on 01/Nov/15

$$Sorryforthemiscommunication.$$$$Iwasjustthinkingthenabout$$$$thevalueofthesum\underset{r=1}{\sum ^n}sin\left(\frac{d^{r}y}{{dx}^r}\right)$$$$assumimgweuseoneofthesolutions.$$$$So,ify=A+{Be}^{-x}+2n\pi xuseitto$$$$deduceQ.$$

Commented by prakash jain last updated on 01/Nov/15

$$ok.understood.$$

Answered by prakash jain last updated on 01/Nov/15

$$u=\frac{dy}{dx},v=\frac{d^{2}y}{{dx}^2},\sin u=-\sin v$$$$v=2n\pi -u,v=(2n+1)\pi +u$$$$\frac{d^{2}y}{{dx}^2}=(2n+1)\pi +\frac{dy}{dx}\dots .\left(1\right)$$$$y=c_1+c_2{e}^x-(2n+1)\pi x\dots ..\left(\mathrm{solution}A\right)$$$$y^{\prime }=c_2{e}^x-(2n+1)\pi ,y^{″}=c_2{e}^x\Rightarrow \sin \left(y^{″}\right)=-\sin \left(y^{\prime }\right)$$$$\frac{d^{2}y}{{dx}^2}=2n\pi -\frac{dy}{dx}\dots \left(2\right)$$$$\mathrm{Solution}\mathrm{for}\left(2\right)$$$$y=c_1+c_2{e}^{-x}+2n\pi x\dots ..\left(\mathrm{solution}\mathrm{B}\right)$$$$y^{\prime }=-c_2{e}^{-x}+2n\pi ,y^{″}=c_2{e}^{-x}\Rightarrow \sin \left(y^{″}\right)=-\sin \left(y^{\prime }\right)$$

Commented by Yozzi last updated on 01/Nov/15

$$Lety=A+{Be}^{-x}+2n\pi x\left(SolutionB\right)$$$$y^{{}^{\prime }}=-{Be}^{-x}+2n\pi$$$$y^{{}^{″}}={Be}^{-x}$$$$y^{{}^{‴}}=-{Be}^{-x}$$$$y^4={Be}^{-x}$$$$\dots .$$$$y^n=\{\begin{array}{l}-{Be}^{-x}foroddn>1\\ {Be}^{-x}forevenn⩾2\end{array}$$$$\therefore \underset{r=1}{\sum ^n}sin\left(y^{\left(r\right)}\right)=sin(2n\pi -{Be}^{-x})+sin\left({Be}^{-x}\right)+sin(-{Be}^{-x})+sin\left({Be}^{-x}\right)+\dots +sin\left(y^{\left(n\right)}\right)$$$$\underset{r=1}{\sum ^n}sin\left(y^{\left(r\right)}\right)=\{\begin{array}{l}-{sinBe}^{-x}oddn\\ 0evenn\end{array}$$$$Q=max(-{sinBe}^{-x})=1$$$$when{Be}^{-x}=\frac{3\pi }{2}\Rightarrow e^{-x}=\frac{3\pi }{2B}$$$$IfB>0\Rightarrow -x=ln\left(\frac{3\pi }{2B}\right)$$$$x=ln\left(\frac{2B}{3\pi }\right).$$$$IfB<0,letB=-T,T>0$$$$\therefore \underset{r=1}{\sum ^n}sin\left(y^{\left(n\right)}\right)=sin\left({Te}^{-x}\right)$$$$Q=1\Rightarrow x=ln\left(\frac{2T}{\pi }\right).$$$$$$

Commented by Yozzi last updated on 01/Nov/15

$$y=A+{Be}^{-x}+2n\pi x$$$$\Rightarrow y^{{}^{\prime }}=-{Be}^{-x}+2n\pi$$$$y^{{}^{″}}={Be}^{-x}$$$$\therefore y^{{}^{″}}+y^{{}^{\prime }}=2n\pi$$$$$$$$y=A+{Be}^x+(2n+1)\pi x$$$$y^{{}^{\prime }}={Be}^x+(2n+1)\pi$$$$y^{{}^{″}}={Be}^x$$$$y^{{}^{″}}-y^{{}^{\prime }}=-(2n+1)\pi$$$$$$

Commented by prakash jain last updated on 01/Nov/15

$$\mathrm{Mistake}\mathrm{is}\mathrm{sign}\mathrm{of}\mathrm{solution}\left(\mathrm{A}\right).Corrected.$$

Commented by Yozzi last updated on 02/Nov/15

$$Awesome.$$

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