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Question Number 115000 by mnjuly1970 last updated on 22/Sep/20

....nice math... if y =(cos(2x))^(−(1/2)) then prove :: y+y^(′′) = 3y^5 ...m.n.july.1970...

$$\:\:\:\:\:\:\:\:\:\:\:\:....{nice}\:\:\:{math}... \\ $$$$ \\ $$$$\:{if}\:\:{y}\:=\left({cos}\left(\mathrm{2}{x}\right)\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{then} \\ $$$${prove}\:::\:\:\:{y}+{y}^{''} =\:\mathrm{3}{y}^{\mathrm{5}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.{july}.\mathrm{1970}... \\ $$

Answered by Dwaipayan Shikari last updated on 22/Sep/20

y=(1/( (√(cos2x)))) y′=cos^(−(3/2)) 2x.sin2x y′′=3cos^(−(5/2)) 2x.sin^2 2x+2cos^(−(1/2)) 2x y+y′′=3(cos^(−(5/2)) 2x(1−cos^2 2x)+cos^(−(1/2)) 2x) y+y′′=3cos^(−(5/2)) 2x−3cos^(−(1/2)) 2x+3cos^(−(1/2)) 2x y+y′′=3y^5

$${y}=\frac{\mathrm{1}}{\:\sqrt{{cos}\mathrm{2}{x}}} \\ $$$${y}'={cos}^{−\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{2}{x}.{sin}\mathrm{2}{x} \\ $$$${y}''=\mathrm{3}{cos}^{−\frac{\mathrm{5}}{\mathrm{2}}} \mathrm{2}{x}.{sin}^{\mathrm{2}} \mathrm{2}{x}+\mathrm{2}{cos}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{2}{x} \\ $$$${y}+{y}''=\mathrm{3}\left({cos}^{−\frac{\mathrm{5}}{\mathrm{2}}} \mathrm{2}{x}\left(\mathrm{1}−{cos}^{\mathrm{2}} \mathrm{2}{x}\right)+{cos}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{2}{x}\right) \\ $$$${y}+{y}''=\mathrm{3}{cos}^{−\frac{\mathrm{5}}{\mathrm{2}}} \mathrm{2}{x}−\mathrm{3}{cos}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{2}{x}+\mathrm{3}{cos}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{2}{x} \\ $$$${y}+{y}''=\mathrm{3}{y}^{\mathrm{5}} \\ $$

Commented by mnjuly1970 last updated on 22/Sep/20

very nice very nice ...thank you sir...

$${very}\:{nice}\:{very}\:{nice}\:...{thank} \\ $$$${you}\:{sir}... \\ $$

Commented by Canovas last updated on 29/Sep/20

Correct

$${Correct} \\ $$

Answered by $@y@m last updated on 23/Sep/20

y=(1/( (√(cos 2x)))) y^2 =sec2 x ...(1) 2yy′=2sec2xtan 2x yy′=sec2xtan 2x yy′=y^2 tan2 x y′=ytan 2x ...(2) y′′=2ysec^2 2x+tan2 x.y′ y′′=2y.y^4 +tan 2x.ytan 2x {using(1) &(2) y′′=2y^5 +ytan^2 2x y′′=2y^5 +y(sec^2 2x−1) y′′=2y^5 +y(y^4 −1) y+y′′=3y^5

$${y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}\:\:\: \\ $$$${y}^{\mathrm{2}} =\mathrm{sec2}\:{x}\:...\left(\mathrm{1}\right) \\ $$$$\mathrm{2}{yy}'=\mathrm{2sec2}{x}\mathrm{tan}\:\mathrm{2}{x} \\ $$$${yy}'=\mathrm{sec2}{x}\mathrm{tan}\:\mathrm{2}{x} \\ $$$${yy}'={y}^{\mathrm{2}} \mathrm{tan2}\:{x} \\ $$$${y}'={y}\mathrm{tan}\:\mathrm{2}{x}\:...\left(\mathrm{2}\right) \\ $$$${y}''=\mathrm{2}{y}\mathrm{sec}\:^{\mathrm{2}} \mathrm{2}{x}+\mathrm{tan2}\:{x}.{y}' \\ $$$${y}''=\mathrm{2}{y}.{y}^{\mathrm{4}} +\mathrm{tan}\:\mathrm{2}{x}.{y}\mathrm{tan}\:\mathrm{2}{x}\:\:\left\{{using}\left(\mathrm{1}\right)\:\&\left(\mathrm{2}\right)\right. \\ $$$${y}''=\mathrm{2}{y}^{\mathrm{5}} +{y}\mathrm{tan}\:^{\mathrm{2}} \mathrm{2}{x} \\ $$$${y}''=\mathrm{2}{y}^{\mathrm{5}} +{y}\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{2}{x}−\mathrm{1}\right) \\ $$$${y}''=\mathrm{2}{y}^{\mathrm{5}} +{y}\left({y}^{\mathrm{4}} −\mathrm{1}\right) \\ $$$${y}+{y}''=\mathrm{3}{y}^{\mathrm{5}} \\ $$

Commented by mnjuly1970 last updated on 23/Sep/20

very nice awesom solution

$${very}\:{nice}\:{awesom}\:{solution} \\ $$$$ \\ $$

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