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Question Number 86120 by jagoll last updated on 27/Mar/20
(dy/dx) + ((sin 2y)/x) = x^3 cos^2 y
$$\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\frac{\mathrm{sin}\:\mathrm{2y}}{\mathrm{x}}\:=\:\mathrm{x}^{\mathrm{3}} \:\mathrm{cos}\:^{\mathrm{2}} \:\mathrm{y} \\ $$
Commented by Prithwish Sen 1 last updated on 27/Mar/20
sir it is sin2x not sin2y .
$${sir}\:{it}\:{is}\:{sin}\mathrm{2}{x}\:{not}\:{sin}\mathrm{2}{y}\:. \\ $$
Commented by jagoll last updated on 27/Mar/20
o sorry . i′m wrong write the equation. i will correct
$$\mathrm{o}\:\mathrm{sorry}\:.\:\mathrm{i}'\mathrm{m}\:\mathrm{wrong}\:\mathrm{write}\:\mathrm{the}\: \\ $$$$\mathrm{equation}.\:\mathrm{i}\:\mathrm{will}\:\mathrm{correct} \\ $$
Commented by jagoll last updated on 27/Mar/20
sir. john. soory. you answer not correct
$$\mathrm{sir}.\:\mathrm{john}.\:\mathrm{soory}.\:\mathrm{you}\:\mathrm{answer}\: \\ $$$$\mathrm{not}\:\mathrm{correct} \\ $$
Commented by john santu last updated on 27/Mar/20
ok
$${ok} \\ $$
Commented by niroj last updated on 27/Mar/20
(dy/dx)+ ((sin2y)/x)= x^3 cos^2 y divide both side by cos^2 y (1/(cos^2 y ))(dy/dx)+ ((2siny.cos y)/(x.cos^2 y))=((x^3 cos^2 y)/(cos^2 y)) sec^2 y (dy/dx) + (2/x)tany=x^3 multiply both side by x^2 x^2 sec^2 y (dy/dx) + x^2 .(2/x)tany =x^3 .x^2 x^2 sec^2 y (dy/dx) + 2x tany=x^5 (d/dx) (x^2 tan y)=x^5 On integratin both side by ∫dx. ∫ (d/dx)(x^2 tan y).dx=∫x^5 dx x^2 tan y= (x^6 /6)+C//.
$$\:\:\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}+\:\frac{\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}}=\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} \:\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\mathrm{y}} \\ $$$$\:\:\mathrm{divide}\:\mathrm{both}\:\mathrm{side}\:\mathrm{by}\:\mathrm{cos}^{\mathrm{2}} \mathrm{y}\:\:\: \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{y}\:}\frac{\mathrm{dy}}{\mathrm{dx}}+\:\frac{\mathrm{2siny}.\mathrm{cos}\:\mathrm{y}}{\mathrm{x}.\mathrm{cos}^{\mathrm{2}} \mathrm{y}}=\frac{\mathrm{x}^{\mathrm{3}} \mathrm{cos}^{\mathrm{2}} \mathrm{y}}{\mathrm{cos}^{\mathrm{2}} \mathrm{y}} \\ $$$$\:\:\:\mathrm{sec}^{\mathrm{2}} \mathrm{y}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\frac{\mathrm{2}}{\mathrm{x}}\mathrm{tany}=\mathrm{x}^{\mathrm{3}} \\ $$$$\:\:\mathrm{multiply}\:\mathrm{both}\:\mathrm{side}\:\mathrm{by}\:\mathrm{x}^{\mathrm{2}} \\ $$$$\:\:\mathrm{x}^{\mathrm{2}} \:\mathrm{sec}^{\mathrm{2}} \mathrm{y}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{x}^{\mathrm{2}} .\frac{\mathrm{2}}{\mathrm{x}}\mathrm{tany}\:=\mathrm{x}^{\mathrm{3}} .\mathrm{x}^{\mathrm{2}} \\ $$$$\:\:\mathrm{x}^{\mathrm{2}} \mathrm{sec}^{\mathrm{2}} \mathrm{y}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{2x}\:\mathrm{tany}=\mathrm{x}^{\mathrm{5}} \\ $$$$\:\:\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left(\mathrm{x}^{\mathrm{2}} \mathrm{tan}\:\mathrm{y}\right)=\mathrm{x}^{\mathrm{5}} \\ $$$$\:\:\mathrm{On}\:\mathrm{integratin}\:\mathrm{both}\:\mathrm{side}\:\mathrm{by}\:\int\mathrm{dx}. \\ $$$$\:\:\:\:\:\int\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{2}} \:\mathrm{tan}\:\mathrm{y}\right).\mathrm{dx}=\int\mathrm{x}^{\mathrm{5}} \mathrm{dx} \\ $$$$\:\:\mathrm{x}^{\mathrm{2}} \mathrm{tan}\:\mathrm{y}=\:\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{6}}+\mathrm{C}//. \\ $$$$\:\: \\ $$$$\:\: \\ $$
Answered by mind is power last updated on 27/Mar/20
⇒(1/(cos^2 (y)))(dy/dx)+((2tg(y))/x)=x^3 u=tg(y)⇒(du/dx)=(dy/dx)(1+tg^2 (y)) ⇒u′+((2u)/x)=x^3 u=(1/x^2 ) homegenius ⇒(k/x^2 )=u⇒((k′)/x^2 )−((2k)/x^3 )+((2k)/x^3 )=x^3 k′=x^5 ⇒k=(x^6 /6)+c (x^4 /6)+(c/x^2 )=u y=tan^(−1) ((x^4 /6)+(c/x^2 ))
$$\Rightarrow\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({y}\right)}\frac{{dy}}{{dx}}+\frac{\mathrm{2}{tg}\left({y}\right)}{{x}}={x}^{\mathrm{3}} \\ $$$${u}={tg}\left({y}\right)\Rightarrow\frac{{du}}{{dx}}=\frac{{dy}}{{dx}}\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({y}\right)\right) \\ $$$$\Rightarrow{u}'+\frac{\mathrm{2}{u}}{{x}}={x}^{\mathrm{3}} \\ $$$${u}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{homegenius} \\ $$$$\Rightarrow\frac{{k}}{{x}^{\mathrm{2}} }={u}\Rightarrow\frac{{k}'}{{x}^{\mathrm{2}} }−\frac{\mathrm{2}{k}}{{x}^{\mathrm{3}} }+\frac{\mathrm{2}{k}}{{x}^{\mathrm{3}} }={x}^{\mathrm{3}} \\ $$$${k}'={x}^{\mathrm{5}} \Rightarrow{k}=\frac{{x}^{\mathrm{6}} }{\mathrm{6}}+{c} \\ $$$$\frac{{x}^{\mathrm{4}} }{\mathrm{6}}+\frac{{c}}{{x}^{\mathrm{2}} }={u} \\ $$$${y}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{4}} }{\mathrm{6}}+\frac{{c}}{{x}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$$ \\ $$

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