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Question Number 1581 by 112358 last updated on 21/Aug/15

$$Findafunctionf\left(x\right)satisfying$$$$thefollowingequation.$$$${\int }_a^b[\frac{1}{2}{\left\{f\left(x\right)\right\}}^2-\sqrt{{\left\{f\left(x\right)\right\}}^2+{\left\{\frac{d}{dx}\left(f\left(x\right)\right)\right\}}^2}]dx=0$$$$b>0,a>0,b\ne a.$$

Commented by 123456 last updated on 21/Aug/15

$$\frac{1}{2}{f}^2-\sqrt{f^2+{\left(f^{\prime }\right)}^2}=0$$$$f^2=2\sqrt{f^2+{\left(f^{\prime }\right)}^2}$$$$f^4=4f^2+4{\left(f^{\prime }\right)}^2$$

Commented by 112358 last updated on 25/Aug/15

$$f^4=4f^2+4{\left(f^{{}^{\prime }}\right)}^2$$$$f^{{}^{\prime }}=\pm \frac{1}{2}\sqrt{f^4-4f^2}$$$$2f^{{}^{\prime }}=\pm f\sqrt{f^2-4}$$$$2\frac{df}{dx}=\pm f\sqrt{f^2-4}$$$$Integrationbyseparationofvariables$$$$\Rightarrow 2\int \frac{df}{f\sqrt{f^2-4}}=\pm \int dx=K\pm x$$$$letf\left(x\right)=2secu\Rightarrow df=2secu\times tanudu$$$$\therefore f\sqrt{f^2-4}=2secu\sqrt{4{sec}^{2}u-4}$$$$=\left(2secu\right)\left(2\sqrt{{sec}^{2}u-1}\right)$$$$\because {tan}^{2}u={sec}^{2}u-1$$$$\Rightarrow f\sqrt{f^2-4}=4secu\sqrt{{tan}^{2}u}=4secutanu(iftanu>0)$$$$\therefore 2\int \frac{2secutanu}{4secutanu}du=K\pm x$$$$\int du=K\pm x\Rightarrow u=K\pm x$$$$\because f\left(x\right)=2secu\Rightarrow u={cos}^{-1}\left(\frac{2}{f\left(x\right)}\right)$$$$\therefore {cos}^{-1}\left(\frac{2}{f\left(x\right)}\right)=K\pm x$$$$\frac{2}{f\left(x\right)}=cos(K\pm x)$$$$f\left(x\right)=2sec(K\pm x)$$

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