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Question Number 121960 by bemath last updated on 13/Nov/20

(dy/dx) + 3y tan x = (1/2)sin 2x

$$\:\:\frac{{dy}}{{dx}}\:+\:\mathrm{3}{y}\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\: \\ $$

Answered by liberty last updated on 13/Nov/20

y′ + 3y tan x = sin x cos x IF u = e^(∫ 3 tan x dx) = e^(−3∫ ((d(cos x))/(cos x))) = e^(ln (sec^3 x )) u = sec^3 x General solution y = ((∫ sin x cos x sec^3 x dx + C)/(sec^3 x )) y = C.cos^3 x + cos^3 x ∫ ((sin x)/(cos^2 x)) dx y= C.cos^3 x−cos^3 x ∫ ((d(cos x))/(cos^2 x)) y= C.cos^3 x + cos^2 x

$$\:\mathrm{y}'\:+\:\mathrm{3y}\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x} \\ $$$$\:\mathrm{IF}\:\mathrm{u}\:=\:\mathrm{e}^{\int\:\mathrm{3}\:\mathrm{tan}\:\mathrm{x}\:\mathrm{dx}} \:=\:\mathrm{e}^{−\mathrm{3}\int\:\frac{\mathrm{d}\left(\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{cos}\:\mathrm{x}}} =\:\mathrm{e}^{\mathrm{ln}\:\left(\mathrm{sec}\:^{\mathrm{3}} \mathrm{x}\:\right)} \\ $$$$\:\mathrm{u}\:=\:\mathrm{sec}\:^{\mathrm{3}} \mathrm{x}\: \\ $$$$\mathrm{General}\:\mathrm{solution}\: \\ $$$$\mathrm{y}\:=\:\frac{\int\:\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{sec}\:^{\mathrm{3}} \mathrm{x}\:\mathrm{dx}\:+\:\mathrm{C}}{\mathrm{sec}\:^{\mathrm{3}} \mathrm{x}\:} \\ $$$$\mathrm{y}\:=\:\mathrm{C}.\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:+\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:\int\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx} \\ $$$$\mathrm{y}=\:\mathrm{C}.\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}−\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:\int\:\frac{\mathrm{d}\left(\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$\mathrm{y}=\:\mathrm{C}.\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:+\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\: \\ $$

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