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Question Number 91825 by john santu last updated on 03/May/20

(cos x) (dy/dx)−y(sin x) = cot (x)

$$\left(\mathrm{cos}\:{x}\right)\:\frac{{dy}}{{dx}}−{y}\left(\mathrm{sin}\:{x}\right)\:=\:\mathrm{cot}\:\left({x}\right) \\ $$

Commented by Tony Lin last updated on 03/May/20

(ycosx)′=cotx ycosx=∫cotxdx=ln∣sinx∣+c y=((ln∣sinx∣+c)/(cosx))

$$\left({ycosx}\right)'={cotx} \\ $$$${ycosx}=\int{cotxdx}={ln}\mid{sinx}\mid+{c} \\ $$$${y}=\frac{{ln}\mid{sinx}\mid+{c}}{{cosx}} \\ $$

Commented by john santu last updated on 03/May/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by mathmax by abdo last updated on 04/May/20

cosx y^′ −sinx y =cotanx (he) →cosx y^′ −sinx y =0 ⇒cosx y^′ =sinx y ⇒ (y^′ /y) =((sinx)/(cosx)) ⇒ln∣y∣ =−ln(cosx)+c ⇒y =(k/(cosx)) let use mvc method →y^′ =(k^′ /(cosx))+k((sinx)/(cos^2 x)) (e)⇒k^′ +k ((sinx)/(cosx))−((ksinx)/(cosx)) =(1/(tanx)) ⇒k^′ =((cosx)/(sinx)) ⇒ k =∫ ((cosx)/(sinx))dx +c ⇒k(x)=ln(sinx) +c ⇒ the general solution is y(x)=(1/(cosx))(ln(sinx) +c) =((ln(sinx))/(cosx)) +(c/(cosx))

$${cosx}\:{y}^{'} −{sinx}\:{y}\:={cotanx} \\ $$$$\left({he}\right)\:\rightarrow{cosx}\:{y}^{'} \:−{sinx}\:{y}\:=\mathrm{0}\:\Rightarrow{cosx}\:{y}^{'} \:={sinx}\:{y}\:\Rightarrow \\ $$$$\frac{{y}^{'} }{{y}}\:=\frac{{sinx}}{{cosx}}\:\Rightarrow{ln}\mid{y}\mid\:=−{ln}\left({cosx}\right)+{c}\:\Rightarrow{y}\:=\frac{{k}}{{cosx}} \\ $$$${let}\:{use}\:{mvc}\:{method}\:\:\rightarrow{y}^{'} \:=\frac{{k}^{'} }{{cosx}}+{k}\frac{{sinx}}{{cos}^{\mathrm{2}} {x}} \\ $$$$\left({e}\right)\Rightarrow{k}^{'} +{k}\:\frac{{sinx}}{{cosx}}−\frac{{ksinx}}{{cosx}}\:=\frac{\mathrm{1}}{{tanx}}\:\Rightarrow{k}^{'} \:=\frac{{cosx}}{{sinx}}\:\Rightarrow \\ $$$${k}\:=\int\:\:\frac{{cosx}}{{sinx}}{dx}\:+{c}\:\Rightarrow{k}\left({x}\right)={ln}\left({sinx}\right)\:+{c}\:\Rightarrow \\ $$$${the}\:{general}\:{solution}\:{is}\:{y}\left({x}\right)=\frac{\mathrm{1}}{{cosx}}\left({ln}\left({sinx}\right)\:+{c}\right) \\ $$$$=\frac{{ln}\left({sinx}\right)}{{cosx}}\:+\frac{{c}}{{cosx}} \\ $$

Answered by john santu last updated on 03/May/20

Commented by john santu last updated on 03/May/20

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Answered by niroj last updated on 03/May/20

(dy/dx) −(tan x) y=cosec x IF= e^( ∫Pdx) = e^(−∫tan xdx) = e^(−log sec x) =e^(log (1/(sec x))) IF= cos x y.cos x = ∫cos x.cosec x dx+c y cos x= ∫cos x.(1/(sin x))dx+c y cos x= ∫cot x dx+c y cos x= log sin x+c y = (1/(cos x))(log sin x +C)//.

$$\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\:−\left(\mathrm{tan}\:\mathrm{x}\right)\:\mathrm{y}=\mathrm{cosec}\:\mathrm{x} \\ $$$$\:\:\:\mathrm{IF}=\:\mathrm{e}^{\:\int\mathrm{Pdx}} =\:\mathrm{e}^{−\int\mathrm{tan}\:\mathrm{xdx}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{e}^{−\mathrm{log}\:\mathrm{sec}\:\mathrm{x}} =\mathrm{e}^{\mathrm{log}\:\frac{\mathrm{1}}{\mathrm{sec}\:\mathrm{x}}} \\ $$$$\:\:\mathrm{IF}=\:\mathrm{cos}\:\mathrm{x} \\ $$$$\:\:\mathrm{y}.\mathrm{cos}\:\mathrm{x}\:=\:\int\mathrm{cos}\:\mathrm{x}.\mathrm{cosec}\:\mathrm{x}\:\mathrm{dx}+\mathrm{c} \\ $$$$\:\:\mathrm{y}\:\mathrm{cos}\:\mathrm{x}=\:\int\mathrm{cos}\:\mathrm{x}.\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}\mathrm{dx}+\mathrm{c} \\ $$$$\:\:\:\mathrm{y}\:\mathrm{cos}\:\mathrm{x}=\:\int\mathrm{cot}\:\mathrm{x}\:\mathrm{dx}+\mathrm{c} \\ $$$$\:\:\:\mathrm{y}\:\mathrm{cos}\:\mathrm{x}=\:\mathrm{log}\:\mathrm{sin}\:\mathrm{x}+\mathrm{c} \\ $$$$\:\:\:\:\:\:\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{x}}\left(\mathrm{log}\:\mathrm{sin}\:\mathrm{x}\:+\mathrm{C}\right)//. \\ $$$$ \\ $$

Commented by john santu last updated on 03/May/20

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