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Question Number 112456 by bobhans last updated on 08/Sep/20

solve (dy/dx)−((4y)/x) = 1+(2/x)

$$\mathrm{solve}\:\frac{\mathrm{dy}}{\mathrm{dx}}−\frac{\mathrm{4y}}{\mathrm{x}}\:=\:\mathrm{1}+\frac{\mathrm{2}}{\mathrm{x}} \\ $$

Answered by john santu last updated on 08/Sep/20

let u = yx^(−4) →(du/dx) = −4yx^(−5) +x^(−4) (dy/dx) (dy/dx) = x^4 (du/dx)+4yx^(−1) ⇔ x^4 (du/dx)+((4y)/x)−((4y)/x) = 1+(2/x) ⇔ (du/dx) = (1/x^4 ) +(2/x^5 ) ; du = (x^(−4) +2x^(−5) )dx ∫ du = ∫ (x^(−4) +2x^(−5) ) dx ⇒ u = −(1/(3x^3 )) −(1/(2x^4 )) + c ⇒(y/x^4 ) = −(1/(3x^3 )) −(1/(2x^4 )) + C ⇒y = −(1/3)x −(1/2) + Cx^4

$${let}\:{u}\:=\:{yx}^{−\mathrm{4}} \:\rightarrow\frac{{du}}{{dx}}\:=\:−\mathrm{4}{yx}^{−\mathrm{5}} +{x}^{−\mathrm{4}} \:\frac{{dy}}{{dx}} \\ $$$$\frac{{dy}}{{dx}}\:=\:{x}^{\mathrm{4}} \:\frac{{du}}{{dx}}+\mathrm{4}{yx}^{−\mathrm{1}} \\ $$$$\Leftrightarrow\:{x}^{\mathrm{4}} \:\frac{{du}}{{dx}}+\frac{\mathrm{4}{y}}{{x}}−\frac{\mathrm{4}{y}}{{x}}\:=\:\mathrm{1}+\frac{\mathrm{2}}{{x}} \\ $$$$\Leftrightarrow\:\frac{{du}}{{dx}}\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:+\frac{\mathrm{2}}{{x}^{\mathrm{5}} }\:;\:{du}\:=\:\left({x}^{−\mathrm{4}} +\mathrm{2}{x}^{−\mathrm{5}} \right){dx} \\ $$$$\int\:{du}\:=\:\int\:\left({x}^{−\mathrm{4}} +\mathrm{2}{x}^{−\mathrm{5}} \right)\:{dx}\: \\ $$$$\Rightarrow\:{u}\:=\:−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{4}} }\:+\:{c}\: \\ $$$$\Rightarrow\frac{{y}}{{x}^{\mathrm{4}} }\:=\:−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{4}} }\:+\:{C} \\ $$$$\Rightarrow{y}\:=\:−\frac{\mathrm{1}}{\mathrm{3}}{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\:{Cx}^{\mathrm{4}} \\ $$

Answered by abdomsup last updated on 08/Sep/20

h→y^′ −(4/x) y =0 ⇒(y^′ /y)=(4/x) ⇒ ln∣y∣ =4ln∣x∣ +c ⇒y = k x^4 lagrange method y^′ =k^′ x^4 +4kx^3 e⇒k^′ x^4 +4k x^3 −4kx^3 =1+(2/x) ⇒ k^′ =x^(−4) +2x^(−5) ⇒ k =−(1/3)x^(−3) −(1/2)x^(−4) +λ ⇒ y =(−(1/3)x^(−3) −(1/2)x^(−4 ) +λ)x^4 =−(x/3)−(1/2) +λx^4

$${h}\rightarrow{y}^{'} −\frac{\mathrm{4}}{{x}}\:{y}\:=\mathrm{0}\:\Rightarrow\frac{{y}^{'} }{{y}}=\frac{\mathrm{4}}{{x}}\:\Rightarrow \\ $$$${ln}\mid{y}\mid\:=\mathrm{4}{ln}\mid{x}\mid\:+{c}\:\Rightarrow{y}\:=\:{k}\:{x}^{\mathrm{4}} \\ $$$${lagrange}\:{method}\:{y}^{'} \:={k}^{'} \:{x}^{\mathrm{4}} \:+\mathrm{4}{kx}^{\mathrm{3}} \\ $$$${e}\Rightarrow{k}^{'} \:{x}^{\mathrm{4}} \:+\mathrm{4}{k}\:{x}^{\mathrm{3}} −\mathrm{4}{kx}^{\mathrm{3}} \:=\mathrm{1}+\frac{\mathrm{2}}{{x}}\:\Rightarrow \\ $$$${k}^{'} \:={x}^{−\mathrm{4}} \:+\mathrm{2}{x}^{−\mathrm{5}} \:\Rightarrow \\ $$$${k}\:=−\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\mathrm{3}} \:−\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\mathrm{4}} \:+\lambda\:\Rightarrow \\ $$$${y}\:=\left(−\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\mathrm{4}\:} \:+\lambda\right){x}^{\mathrm{4}} \\ $$$$=−\frac{{x}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}\:+\lambda{x}^{\mathrm{4}} \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 08/Sep/20

I.F=e^(∫((−4)/x)) =(1/x^4 ) (y/x^4 )=∫(1/x^4 )+(2/x^5 ) y=−(x/3)−(1/2)+Cx^4

$${I}.{F}={e}^{\int\frac{−\mathrm{4}}{{x}}} =\frac{\mathrm{1}}{{x}^{\mathrm{4}} } \\ $$$$\frac{{y}}{{x}^{\mathrm{4}} }=\int\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\frac{\mathrm{2}}{{x}^{\mathrm{5}} } \\ $$$${y}=−\frac{{x}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}+{Cx}^{\mathrm{4}} \\ $$

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