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Question Number 97799 by abdomathmax last updated on 09/Jun/20

solve y^′ cosx +y sinx =cosx +sinx

$$\mathrm{solve}\:\mathrm{y}^{'} \mathrm{cosx}\:+\mathrm{y}\:\mathrm{sinx}\:=\mathrm{cosx}\:+\mathrm{sinx} \\ $$

Commented by john santu last updated on 10/Jun/20

(dy/dx) + tan x .y = 1+tan x IF u(x)=e^(∫ tan x dx) = e^(ln(sec x)) = sec x ⇔sec x (dy/dx) + tan x sec x .y = sec x(1+tan x) ∫ (d/dx) (y.sec x) = ∫ sec x(1+tan x)dx y.sec x = ln∣sec x+tan x∣ + sec x + c y = cos x ln ∣sec x+tan x∣ + c .cos x + 1

$$\frac{{dy}}{{dx}}\:+\:\mathrm{tan}\:{x}\:.{y}\:=\:\mathrm{1}+\mathrm{tan}\:{x} \\ $$$$\mathrm{IF}\:{u}\left({x}\right)={e}^{\int\:\mathrm{tan}\:{x}\:{dx}} \:=\:{e}^{\mathrm{ln}\left(\mathrm{sec}\:{x}\right)} =\:\mathrm{sec}\:{x} \\ $$$$\Leftrightarrow\mathrm{sec}\:{x}\:\frac{{dy}}{{dx}}\:+\:\mathrm{tan}\:{x}\:\mathrm{sec}\:{x}\:.{y}\:=\:\mathrm{sec}\:{x}\left(\mathrm{1}+\mathrm{tan}\:{x}\right) \\ $$$$\int\:\frac{{d}}{{dx}}\:\left({y}.\mathrm{sec}\:{x}\right)\:=\:\int\:\mathrm{sec}\:{x}\left(\mathrm{1}+\mathrm{tan}\:{x}\right){dx} \\ $$$${y}.\mathrm{sec}\:{x}\:=\:\mathrm{ln}\mid\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\mid\:+\:\mathrm{sec}\:{x}\:+\:{c} \\ $$$${y}\:=\:\mathrm{cos}\:{x}\:\mathrm{ln}\:\mid\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\mid\:+\:\mathrm{c}\:.\mathrm{cos}\:{x}\:+\:\mathrm{1}\: \\ $$

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