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Question Number 24233 by Nayon.Sm last updated on 14/Nov/17

$$if\frac{d^{2}y}{{dx}^2}=ksinytheny=?$$

Answered by ajfour last updated on 15/Nov/17

$$lety\to xandx\to t$$$$\Rightarrow then\frac{d^{2}x}{{dt}^2}=k\sin x$$$$\frac{vdv}{dx}=k\sin x$$$$\frac{v^2}{2}=-k\cos x+C$$$$\Rightarrow v=\frac{dx}{dt}=\pm \sqrt{2}\left(\sqrt{C-k\cos x}\right)$$$$\Rightarrow \int \frac{dx}{\sqrt{C-k\cos x}}=\pm \sqrt{2}t+c_1$$$$notalwayseasilyintegrable..$$

Answered by abwayh last updated on 16/Nov/17

$$\mathrm{just}\mathrm{try};$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\mathrm{ksin}\mathrm{y}$$$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{ksin}\mathrm{y}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\int \mathrm{ksin}\mathrm{y}\mathrm{dx}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{kxsin}\mathrm{y}+\mathrm{c}\left(\mathrm{y}\right);\mathrm{c}\left(\mathrm{y}\right)\mathrm{is}\mathrm{cons}.\mathrm{respect}\mathrm{to}\mathrm{y}$$$$\mathrm{dy}=\{\mathrm{kxsin}\mathrm{y}+\mathrm{c}\left(\mathrm{y}\right)\}\mathrm{dx}$$$$\mathrm{y}=\int \mathrm{kxsin}\mathrm{y}\mathrm{dx}+\int \mathrm{c}\left(\mathrm{y}\right)\mathrm{dx}$$$$\mathrm{y}=\frac{1}{2}{\mathrm{kx}}^2\sin \mathrm{y}+\mathrm{c}\left(\mathrm{y}\right)\mathrm{x}+\mathrm{c}1\left(\mathrm{y}\right)......;\mathrm{c},\mathrm{c}1\mathrm{cons}.$$

Commented by mrW1 last updated on 16/Nov/17

$$yisafunctionofx,\sin \left(y\right)isalso$$$$afunctionofx,i.e.\sin \left(y\right)=f\left(x\right)$$$$therefore$$$$\int \sin \left(y\right)dx=\int f\left(x\right)dx\ne xf\left(x\right)!!!$$$$i.e.\int \sin \left(y\right)dx\ne x\sin \left(y\right)!!!$$

Commented by abwayh last updated on 16/Nov/17

$$\mathrm{sorry}.\mathrm{sir};\mathrm{I}\mathrm{mean}\mathrm{for}\mathrm{i}.\mathrm{e}.\mathrm{take}\mathrm{c}\left(\mathrm{y}\right),{\mathrm{c}}_1\left(\mathrm{y}\right)\mathrm{as}\mathrm{functions}\mathrm{of}\mathrm{y}$$$$\mathrm{put}(\mathrm{c}=\sin \mathrm{y};{\mathrm{c}}_1=\cos \mathrm{y})(\mathrm{or}\mathrm{any}\mathrm{func}.\mathrm{for}\mathrm{y})$$$$\mathrm{then}:$$$$\mathrm{y}=\frac{1}{2}{\mathrm{kx}}^2\sin \mathrm{y}+\mathrm{xsin}\mathrm{y}+\cos \mathrm{y}$$$$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{kxsin}\mathrm{y}+\sin \mathrm{y}$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\mathrm{ksin}\mathrm{y}$$$$$$

Commented by mrW1 last updated on 17/Nov/17

$${that}^{\prime }stotallywrongwhatyoudo,sir.$$$$forexample:$$$$ify=\frac{1}{2}{kx}^2\sin y+x\sin y+\cos y$$$$then$$$$\frac{dy}{dx}=kx\sin y+\frac{1}{2}{kx}^2\cos y\frac{dy}{dx}+\sin y+\cos y\frac{dy}{dx}-\sin y\frac{dy}{dx}$$$$\ne kx\sin y+\sin y$$$$$$$$AsIhavesaid,yisafunctionofx,$$$${it}^{\prime }snotaconstantwrtx!therefore$$$$\int k\sin ydx\ne kx\sin y+c\left(y\right)$$

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