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Question Number 86142 by jagoll last updated on 27/Mar/20

y′ .sin t cos t = y + sin^3 t y((π/4)) = 0

$$\mathrm{y}'\:.\mathrm{sin}\:\mathrm{t}\:\mathrm{cos}\:\mathrm{t}\:=\:\mathrm{y}\:+\:\mathrm{sin}\:^{\mathrm{3}} \mathrm{t}\: \\ $$$$\mathrm{y}\left(\frac{\pi}{\mathrm{4}}\right)\:=\:\mathrm{0}\: \\ $$

Answered by Kunal12588 last updated on 27/Mar/20

(dy/dt)=(y/(sin t cos t))+sin t tan t ⇒(dy/dt)+(−(1/(sin t cos t)))y=sin t tan t P = −(1/(sin t cos t)) ; Q = sin t tan t I.F = e^(∫P dt) = e^(−∫(dt/(sin t cos t))) = e^(−∫ ((sin^2 t+cos^2 t)/(sin t cos t))dt) =e^(−∫tan t dt −∫cot t dt ) =e^(−ln ∣sec t∣−ln∣sin t∣ ) =e^(−ln ∣tan t∣) =e^(ln ∣cot t∣) =cot t y(I.F)=∫Q(I.F) dt ⇒y cot t = ∫sin t dt ⇒y cot t = −cos t + a ⇒y = −sin t + a tan t → general solution If question means when t = (π/4), y=0 . Then , 0=−(1/(√2))+a ⇒a=(1/(√2)) ∴ y = −sin t + (1/(√2)) tan t → particular solution

$$\frac{{dy}}{{dt}}=\frac{{y}}{\mathrm{sin}\:{t}\:\mathrm{cos}\:{t}}+\mathrm{sin}\:{t}\:\mathrm{tan}\:{t} \\ $$$$\Rightarrow\frac{{dy}}{{dt}}+\left(−\frac{\mathrm{1}}{\mathrm{sin}\:{t}\:\mathrm{cos}\:{t}}\right){y}=\mathrm{sin}\:{t}\:\mathrm{tan}\:{t} \\ $$$${P}\:=\:−\frac{\mathrm{1}}{\mathrm{sin}\:{t}\:\mathrm{cos}\:{t}}\:;\:{Q}\:=\:\mathrm{sin}\:{t}\:\mathrm{tan}\:{t} \\ $$$${I}.{F}\:=\:{e}^{\int{P}\:{dt}} \:=\:{e}^{−\int\frac{{dt}}{\mathrm{sin}\:{t}\:\mathrm{cos}\:{t}}} \:=\:{e}^{−\int\:\frac{\mathrm{sin}^{\mathrm{2}} \:{t}+\mathrm{cos}^{\mathrm{2}} \:{t}}{\mathrm{sin}\:{t}\:\mathrm{cos}\:{t}}{dt}} \\ $$$$={e}^{−\int\mathrm{tan}\:{t}\:{dt}\:−\int\mathrm{cot}\:{t}\:{dt}\:} ={e}^{−\mathrm{ln}\:\mid\mathrm{sec}\:{t}\mid−\mathrm{ln}\mid\mathrm{sin}\:{t}\mid\:} \\ $$$$={e}^{−\mathrm{ln}\:\mid\mathrm{tan}\:{t}\mid} ={e}^{\mathrm{ln}\:\mid\mathrm{cot}\:{t}\mid} =\mathrm{cot}\:{t} \\ $$$${y}\left({I}.{F}\right)=\int{Q}\left({I}.{F}\right)\:{dt} \\ $$$$\Rightarrow{y}\:\mathrm{cot}\:{t}\:=\:\int\mathrm{sin}\:{t}\:{dt} \\ $$$$\Rightarrow{y}\:\mathrm{cot}\:{t}\:=\:−\mathrm{cos}\:{t}\:+\:{a} \\ $$$$\Rightarrow{y}\:=\:−\mathrm{sin}\:{t}\:+\:{a}\:\mathrm{tan}\:{t}\:\rightarrow\:{general}\:{solution} \\ $$$${If}\:{question}\:{means}\:{when}\:{t}\:=\:\frac{\pi}{\mathrm{4}},\:{y}=\mathrm{0}\:.\:{Then}\:, \\ $$$$\mathrm{0}=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+{a} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\therefore\:{y}\:=\:−\mathrm{sin}\:{t}\:+\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{tan}\:{t}\:\rightarrow\:{particular}\:{solution} \\ $$

Commented by jagoll last updated on 27/Mar/20

thank you sir

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

Answered by TANMAY PANACEA. last updated on 27/Mar/20

(dy/dt)+((−y)/(sintcost))=((sin^2 t)/(cost)) e^(∫(dt/(−sintcost))) e^(∫−2cosec2t dt) ∫cosec2t=((lntant)/2) e^(∫−2cosec2t ) dt=e^(−2×((lntant)/2)) =(1/(tant))=cott cott×(dy/dt)+((−y×cott)/(sint×cost))=((sin^2 t)/(cost))×cott cott×(dy/dt)+y×(−cosec^2 t)=sint (d/dt)(y.cott)=sint ycott=−cost+c 0×1=−(1/(√2))+c ycott=−cost+(1/(√2))

$$\frac{{dy}}{{dt}}+\frac{−{y}}{{sintcost}}=\frac{{sin}^{\mathrm{2}} {t}}{{cost}} \\ $$$${e}^{\int\frac{{dt}}{−{sintcost}}} \\ $$$${e}^{\int−\mathrm{2}{cosec}\mathrm{2}{t}\:{dt}} \\ $$$$\int{cosec}\mathrm{2}{t}=\frac{{lntant}}{\mathrm{2}} \\ $$$${e}^{\int−\mathrm{2}{cosec}\mathrm{2}{t}\:} {dt}={e}^{−\mathrm{2}×\frac{{lntant}}{\mathrm{2}}} =\frac{\mathrm{1}}{{tant}}={cott} \\ $$$${cott}×\frac{{dy}}{{dt}}+\frac{−{y}×{cott}}{{sint}×{cost}}=\frac{{sin}^{\mathrm{2}} {t}}{{cost}}×{cott} \\ $$$${cott}×\frac{{dy}}{{dt}}+{y}×\left(−{cosec}^{\mathrm{2}} {t}\right)={sint} \\ $$$$\frac{{d}}{{dt}}\left({y}.{cott}\right)={sint} \\ $$$${ycott}=−{cost}+{c} \\ $$$$\mathrm{0}×\mathrm{1}=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+{c} \\ $$$${ycott}=−{cost}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$

Commented by jagoll last updated on 27/Mar/20

what is cott sir?

$$\mathrm{what}\:\mathrm{is}\:\mathrm{cott}\:\mathrm{sir}? \\ $$

Commented by TANMAY PANACEA. last updated on 27/Mar/20

cotθ → cot(t)

$${cot}\theta\:\rightarrow\:\:\:{cot}\left({t}\right) \\ $$

Commented by jagoll last updated on 27/Mar/20

ol it cot (t) ..thank you sir

$$\mathrm{ol}\:\mathrm{it}\:\mathrm{cot}\:\left(\mathrm{t}\right)\:..\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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