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

Find a function f(x) satisfying the following equation. ∫_a ^( b) [(1/2){f(x)}^2 −(√({f(x)}^2 +{(d/dx)(f(x))}^2 ))]dx=0 b>0,a>0 , b≠a.

$${Find}\:{a}\:{function}\:{f}\left({x}\right)\:{satisfying} \\ $$ $${the}\:{following}\:{equation}. \\ $$ $$\int_{{a}} ^{\:{b}} \left[\frac{\mathrm{1}}{\mathrm{2}}\left\{{f}\left({x}\right)\right\}^{\mathrm{2}} −\sqrt{\left\{{f}\left({x}\right)\right\}^{\mathrm{2}} +\left\{\frac{{d}}{{dx}}\left({f}\left({x}\right)\right)\right\}^{\mathrm{2}} }\right]{dx}=\mathrm{0} \\ $$ $${b}>\mathrm{0},{a}>\mathrm{0}\:,\:{b}\neq{a}.\:\: \\ $$

Commented by123456 last updated on 21/Aug/15

(1/2)f^2 −(√(f^2 +(f′)^2 ))=0 f^2 =2(√(f^2 +(f′)^2 )) f^4 =4f^2 +4(f′)^2

$$\frac{\mathrm{1}}{\mathrm{2}}{f}^{\mathrm{2}} −\sqrt{{f}^{\mathrm{2}} +\left({f}'\right)^{\mathrm{2}} }=\mathrm{0} \\ $$ $${f}^{\mathrm{2}} =\mathrm{2}\sqrt{{f}^{\mathrm{2}} +\left({f}'\right)^{\mathrm{2}} } \\ $$ $${f}^{\mathrm{4}} =\mathrm{4}{f}^{\mathrm{2}} +\mathrm{4}\left({f}'\right)^{\mathrm{2}} \\ $$

Commented by112358 last updated on 25/Aug/15

f^4 =4f^2 +4(f^′ )^2 f^′ =±(1/2)(√(f^4 −4f^2 )) 2f^′ =±f(√(f^2 −4)) 2(df/dx)=±f(√(f^2 −4)) Integration by separation of variables ⇒2∫(df/(f(√(f^2 −4))))=±∫dx=K±x let f(x)=2secu⇒df=2secu×tanu du ∴f(√(f^2 −4))=2secu(√(4sec^2 u−4)) =(2secu)(2(√(sec^2 u−1))) ∵tan^2 u=sec^2 u−1 ⇒f(√(f^2 −4))=4secu(√(tan^2 u))=4secutanu (if tanu>0) ∴2∫((2secutanu)/(4secutanu))du=K±x ∫du=K±x⇒u=K±x ∵f(x)=2secu⇒u=cos^(−1) ((2/(f(x)))) ∴cos^(−1) ((2/(f(x))))=K±x (2/(f(x)))=cos(K±x) f(x)=2sec(K±x)

$${f}^{\mathrm{4}} =\mathrm{4}{f}^{\mathrm{2}} +\mathrm{4}\left({f}^{'} \right)^{\mathrm{2}} \\ $$ $${f}^{'} =\pm\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{f}^{\mathrm{4}} −\mathrm{4}{f}^{\mathrm{2}} } \\ $$ $$\mathrm{2}{f}^{'} =\pm{f}\sqrt{{f}^{\mathrm{2}} −\mathrm{4}} \\ $$ $$\mathrm{2}\frac{{df}}{{dx}}=\pm{f}\sqrt{{f}^{\mathrm{2}} −\mathrm{4}} \\ $$ $${Integration}\:{by}\:{separation}\:{of}\:{variables} \\ $$ $$\Rightarrow\mathrm{2}\int\frac{{df}}{{f}\sqrt{{f}^{\mathrm{2}} −\mathrm{4}}}=\pm\int{dx}={K}\pm{x} \\ $$ $${let}\:{f}\left({x}\right)=\mathrm{2}{secu}\Rightarrow{df}=\mathrm{2}{secu}×{tanu}\:{du} \\ $$ $$\therefore{f}\sqrt{{f}^{\mathrm{2}} −\mathrm{4}}=\mathrm{2}{secu}\sqrt{\mathrm{4}{sec}^{\mathrm{2}} {u}−\mathrm{4}} \\ $$ $$=\left(\mathrm{2}{secu}\right)\left(\mathrm{2}\sqrt{{sec}^{\mathrm{2}} {u}−\mathrm{1}}\right) \\ $$ $$\because{tan}^{\mathrm{2}} {u}={sec}^{\mathrm{2}} {u}−\mathrm{1} \\ $$ $$\Rightarrow{f}\sqrt{{f}^{\mathrm{2}} −\mathrm{4}}=\mathrm{4}{secu}\sqrt{{tan}^{\mathrm{2}} {u}}=\mathrm{4}{secutanu}\:\:\:\left({if}\:{tanu}>\mathrm{0}\right) \\ $$ $$\therefore\mathrm{2}\int\frac{\mathrm{2}{secutanu}}{\mathrm{4}{secutanu}}{du}={K}\pm{x} \\ $$ $$\int{du}={K}\pm{x}\Rightarrow{u}={K}\pm{x} \\ $$ $$\because{f}\left({x}\right)=\mathrm{2}{secu}\Rightarrow{u}={cos}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{f}\left({x}\right)}\right) \\ $$ $$\therefore{cos}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{f}\left({x}\right)}\right)={K}\pm{x} \\ $$ $$\frac{\mathrm{2}}{{f}\left({x}\right)}={cos}\left({K}\pm{x}\right) \\ $$ $${f}\left({x}\right)=\mathrm{2}{sec}\left({K}\pm{x}\right)\:\: \\ $$

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