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Question Number 93828 by i jagooll last updated on 15/May/20
sin x ((dy/dx))−y = 2sin x
$$\mathrm{sin}\:\mathrm{x}\:\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)−\mathrm{y}\:=\:\mathrm{2sin}\:\mathrm{x}\: \\ $$
Commented by i jagooll last updated on 15/May/20
Commented by mathmax by abdo last updated on 15/May/20
sinx y^′ −y =2sinx (he)→sinxy^′ −y =0 ⇒sinx y^′ =y ⇒(y^′ /y)=(1/(sinx)) ⇒ ln∣y∣ =∫ (dx/(sinx)) +c but ∫ (dx/(sinx)) =_(tan((x/2))=t) ∫ ((2dt)/((1+t^2 )×((2t)/(1+t^2 )))) =∫ (dt/t) =ln∣t∣ +α =ln∣tan((x/2))∣ +α ⇒ln∣y∣=ln∣tan((x/2))∣ +c ⇒ y(x) =k tan((x/2)) let use mvc method ⇒ y^′ =k^′ tan((x/2)) +(k/2)(1+tan^2 ((x/2))) (e) ⇒k^′ sinxtan((x/2)) +(k/2)sinx(1+tan^2 ((x/2)))−ktan((x/2)) =2sinx ⇒ k^′ sinx tan((x/2)) +(k/2)×((2tan((x/2)))/(1+tan^2 ((x/2))))×(1+tan^2 ((x/2)))−ktan((x/2))=2sinx ⇒k^′ tan((x/2)) =2 ⇒k^′ =(2/(tan((x/2)))) ⇒k =2 ∫ (dx/(tan((x/2)))) =_((x/2)=u) 2∫ ((2du)/(tanu)) =4 ∫ ((cosu)/(sinu)) du =4ln∣sin((x/2))∣ +c ⇒ y(x) =(4 ln∣sin((x/2))+k)tan((x/2)) =4ln∣sin((x/2))∣ +k tan((x/2))
$${sinx}\:{y}^{'} −{y}\:=\mathrm{2}{sinx}\:\:\: \\ $$$$\left({he}\right)\rightarrow{sinxy}^{'} −{y}\:=\mathrm{0}\:\Rightarrow{sinx}\:{y}^{'} \:={y}\:\Rightarrow\frac{{y}^{'} }{{y}}=\frac{\mathrm{1}}{{sinx}}\:\Rightarrow \\ $$$${ln}\mid{y}\mid\:=\int\:\frac{{dx}}{{sinx}}\:+{c}\:\:\:{but}\:\int\:\frac{{dx}}{{sinx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)×\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$=\int\:\frac{{dt}}{{t}}\:={ln}\mid{t}\mid\:+\alpha\:={ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)\mid\:+\alpha\:\Rightarrow{ln}\mid{y}\mid={ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)\mid\:+{c}\:\Rightarrow \\ $$$${y}\left({x}\right)\:={k}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:\:\:\:\:{let}\:{use}\:{mvc}\:{method}\:\Rightarrow \\ $$$${y}^{'} \:={k}^{'} \:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:+\frac{{k}}{\mathrm{2}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\:\left({e}\right)\:\Rightarrow{k}^{'} \:{sinxtan}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$+\frac{{k}}{\mathrm{2}}{sinx}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)−{ktan}\left(\frac{{x}}{\mathrm{2}}\right)\:=\mathrm{2}{sinx}\:\Rightarrow \\ $$$${k}^{'} \:{sinx}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:+\frac{{k}}{\mathrm{2}}×\frac{\mathrm{2}{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}×\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)−{ktan}\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{2}{sinx} \\ $$$$\Rightarrow{k}^{'} \:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:=\mathrm{2}\:\Rightarrow{k}^{'} \:=\frac{\mathrm{2}}{{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\:\Rightarrow{k}\:=\mathrm{2}\:\int\:\:\frac{{dx}}{{tan}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$=_{\frac{{x}}{\mathrm{2}}={u}} \:\:\mathrm{2}\int\:\:\frac{\mathrm{2}{du}}{{tanu}}\:=\mathrm{4}\:\int\:\:\frac{{cosu}}{{sinu}}\:{du}\:=\mathrm{4}{ln}\mid{sin}\left(\frac{{x}}{\mathrm{2}}\right)\mid\:+{c}\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\left(\mathrm{4}\:{ln}\mid{sin}\left(\frac{{x}}{\mathrm{2}}\right)+{k}\right){tan}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$=\mathrm{4}{ln}\mid{sin}\left(\frac{{x}}{\mathrm{2}}\right)\mid\:+{k}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$
Answered by niroj last updated on 15/May/20
sin x ((dy/dx))−y= 2sin x (dy/dx)− (1/(sin x))y= 2 P = − (1/(sin x)) , Q= 2 IF= e^(∫Pdx) −∫(1/(sin x))dx=−∫cosec xdx IF= e^(−log (cosec x−cot x)) IF= e^(log (cosec x−cot x)^(−1) ) IF= (1/(cosec x−cot x)) y× (1/(cosec x−cot x))= ∫ (1/(cosecx−cot x)).2 dx+C (y/(cosec x−cot x)) = ∫(2/((1/(sinx)) −((cos x)/(sinx))))dx+C =2∫ ((sin xdx)/(1−cos x)) +C let, 1−cos x=t sinxdx=dt 2∫ (1/t)dt+C=2log t+C 2log (1−cosx)+C Complete solution: (y/(cosec x−cot x))=log (1−cos x)^2 +C y= (cosec x−cot x){log(1−cos x)^2 +C}
$$\:\:\:\mathrm{sin}\:\mathrm{x}\:\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)−\mathrm{y}=\:\mathrm{2sin}\:\mathrm{x} \\ $$$$\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}−\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}\mathrm{y}=\:\mathrm{2} \\ $$$$\:\:\mathrm{P}\:=\:−\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}\:,\:\mathrm{Q}=\:\mathrm{2} \\ $$$$\:\:\:\mathrm{IF}=\:\mathrm{e}^{\int\mathrm{Pdx}} \\ $$$$\:\:\:\:\:\:−\int\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}\mathrm{dx}=−\int\mathrm{cosec}\:\mathrm{xdx} \\ $$$$\:\:\:\mathrm{IF}=\:\mathrm{e}^{−\mathrm{log}\:\left(\mathrm{cosec}\:\mathrm{x}−\mathrm{cot}\:\mathrm{x}\right)} \\ $$$$\:\:\mathrm{IF}=\:\mathrm{e}^{\mathrm{log}\:\left(\mathrm{cosec}\:\mathrm{x}−\mathrm{cot}\:\mathrm{x}\right)^{−\mathrm{1}} } \\ $$$$\:\:\mathrm{IF}=\:\frac{\mathrm{1}}{\mathrm{cosec}\:\mathrm{x}−\mathrm{cot}\:\mathrm{x}} \\ $$$$\:\:\:\:\mathrm{y}×\:\frac{\mathrm{1}}{\mathrm{cosec}\:\mathrm{x}−\mathrm{cot}\:\mathrm{x}}=\:\int\:\frac{\mathrm{1}}{\mathrm{cosecx}−\mathrm{cot}\:\mathrm{x}}.\mathrm{2}\:\mathrm{dx}+\mathrm{C} \\ $$$$\:\:\frac{\mathrm{y}}{\mathrm{cosec}\:\mathrm{x}−\mathrm{cot}\:\mathrm{x}}\:=\:\int\frac{\mathrm{2}}{\frac{\mathrm{1}}{\mathrm{sinx}}\:−\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{sinx}}}\mathrm{dx}+\mathrm{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int\:\:\frac{\mathrm{sin}\:\mathrm{xdx}}{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}\:+\mathrm{C} \\ $$$$\:\:\mathrm{let},\:\mathrm{1}−\mathrm{cos}\:\mathrm{x}=\mathrm{t} \\ $$$$\:\:\:\:\:\:\:\mathrm{sinxdx}=\mathrm{dt} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}\int\:\frac{\mathrm{1}}{\mathrm{t}}\mathrm{dt}+\mathrm{C}=\mathrm{2log}\:\mathrm{t}+\mathrm{C} \\ $$$$\:\:\:\:\:\mathrm{2log}\:\left(\mathrm{1}−\mathrm{cosx}\right)+\mathrm{C} \\ $$$$\:\mathrm{Complete}\:\mathrm{solution}: \\ $$$$\:\:\frac{\mathrm{y}}{\mathrm{cosec}\:\mathrm{x}−\mathrm{cot}\:\mathrm{x}}=\mathrm{log}\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} +\mathrm{C} \\ $$$$\:\:\mathrm{y}=\:\left(\mathrm{cosec}\:\mathrm{x}−\mathrm{cot}\:\mathrm{x}\right)\left\{\mathrm{log}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} +\mathrm{C}\right\} \\ $$$$\: \\ $$$$ \\ $$
Commented by i jagooll last updated on 15/May/20
cooll man ������

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