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calculus-1-if-y-1-sin-x-cos-x-1-3-then-3y-12-y-2-y-2-




Question Number 132104 by mnjuly1970 last updated on 11/Feb/21
               ... calculus (1)...    if  y=((1/(sin(x)+cos(x))))^(1/3)   then :      3y′′−12(y′)^2 −y^2 = ???       .........
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{calculus}\:\left(\mathrm{1}\right)… \\ $$$$\:\:{if}\:\:{y}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{sin}\left({x}\right)+{cos}\left({x}\right)}}\:\:{then}\:: \\ $$$$\:\:\:\:\mathrm{3}{y}''−\mathrm{12}\left({y}'\right)^{\mathrm{2}} −{y}^{\mathrm{2}} =\:??? \\ $$$$\:\:\:\:\:……… \\ $$
Answered by mnjuly1970 last updated on 11/Feb/21
     solution::     (1/y^3 )=sin(x)+cos(x)   differentiation both sides:   ((−3y′y^2 )/y^6 )=cos(x)−sin(x)       simplification::  ((3y′)/y^4 )=sin(x)−cos(x)          differentiation again:: ((3y′′y^4 −12(y′)^2 y^3 )/y^8 ) =cos(x)+sin(x)=(1/y^3 )      simplification:: ((3y′′y−12(y′)^2 )/y^5 )=(1/y^3 )        ∴  3y′′y−12(y′)^2 −y^2 =0
$$\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\frac{\mathrm{1}}{{y}^{\mathrm{3}} }={sin}\left({x}\right)+{cos}\left({x}\right) \\ $$$$\:{differentiation}\:{both}\:{sides}:\:\:\:\frac{−\mathrm{3}{y}'{y}^{\mathrm{2}} }{{y}^{\mathrm{6}} }={cos}\left({x}\right)−{sin}\left({x}\right) \\ $$$$\:\:\:\:\:{simplification}::\:\:\frac{\mathrm{3}{y}'}{{y}^{\mathrm{4}} }={sin}\left({x}\right)−{cos}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:{differentiation}\:{again}::\:\frac{\mathrm{3}{y}''{y}^{\mathrm{4}} −\mathrm{12}\left({y}'\right)^{\mathrm{2}} {y}^{\mathrm{3}} }{{y}^{\mathrm{8}} }\:={cos}\left({x}\right)+{sin}\left({x}\right)=\frac{\mathrm{1}}{{y}^{\mathrm{3}} }\:\:\:\: \\ $$$${simplification}::\:\frac{\mathrm{3}{y}''{y}−\mathrm{12}\left({y}'\right)^{\mathrm{2}} }{{y}^{\mathrm{5}} }=\frac{\mathrm{1}}{{y}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\therefore\:\:\mathrm{3}{y}''{y}−\mathrm{12}\left({y}'\right)^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\: \\ $$

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