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Question Number 102298 by Ar Brandon last updated on 08/Jul/20

sinx∙(dy/dx)−ycosx=y^3 sin^2 xcosx

$$\mathrm{sinx}\centerdot\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{ycosx}=\mathrm{y}^{\mathrm{3}} \mathrm{sin}^{\mathrm{2}} \mathrm{xcosx} \\ $$

Answered by john santu last updated on 08/Jul/20

(dy/dx)−ycot x=sin xcos x .y^3 set v = y^(−2) ⇒(dv/dx) = −2y^(−3) (dy/dx) (dy/dx) = −(1/2)y^3 (dv/dx) ⇔ −(1/2)y^3 (dv/dx)−ycot x=sin xcos x y^3 (dv/dx) +2cot x.v = sin 2x IF u(x)=e^(∫2cot x dx) = e^(2ln(sin x)) =sin^2 x v = ((∫2sin^3 x d(sin x) +C)/(sin^2 x)) ((sin^2 x)/y^2 ) = (1/2)sin^4 x + C (1/y^2 ) = (1/2)sin^2 x + C. csc^2 x (JS ⊛)

$$\frac{{dy}}{{dx}}−{y}\mathrm{cot}\:{x}=\mathrm{sin}\:{x}\mathrm{cos}\:{x}\:.{y}^{\mathrm{3}} \\ $$$${set}\:{v}\:=\:{y}^{−\mathrm{2}} \:\Rightarrow\frac{{dv}}{{dx}}\:=\:−\mathrm{2}{y}^{−\mathrm{3}} \:\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{3}} \:\frac{{dv}}{{dx}} \\ $$$$\Leftrightarrow\:−\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{3}} \:\frac{{dv}}{{dx}}−{y}\mathrm{cot}\:{x}=\mathrm{sin}\:{x}\mathrm{cos}\:{x}\:{y}^{\mathrm{3}} \\ $$$$\frac{{dv}}{{dx}}\:+\mathrm{2cot}\:{x}.{v}\:=\:\mathrm{sin}\:\mathrm{2}{x} \\ $$$${IF}\:{u}\left({x}\right)={e}^{\int\mathrm{2cot}\:{x}\:{dx}} \:=\:{e}^{\mathrm{2ln}\left(\mathrm{sin}\:\mathrm{x}\right)} =\mathrm{sin}\:^{\mathrm{2}} {x} \\ $$$${v}\:=\:\frac{\int\mathrm{2sin}\:^{\mathrm{3}} {x}\:{d}\left(\mathrm{sin}\:{x}\right)\:+{C}}{\mathrm{sin}\:^{\mathrm{2}} {x}} \\ $$$$\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{4}} {x}\:+\:{C}\: \\ $$$$\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} {x}\:+\:{C}.\:\mathrm{csc}^{\mathrm{2}} {x}\: \\ $$$$\left({JS}\:\circledast\right) \\ $$$$ \\ $$

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