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Question Number 125054 by micelle last updated on 08/Dec/20

x(dy/dx)−6(y/x)=7x^3 help me

$${x}\frac{{dy}}{{dx}}−\mathrm{6}\frac{{y}}{{x}}=\mathrm{7}{x}^{\mathrm{3}} \\ $$$${help}\:{me} \\ $$

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

(dy/dx)−((6y)/x^2 )=7x^4 I.F=e^(∫−(6/x^2 )) =e^(6/x) ye^(6/x) =7∫x^2 e^(1/x) dx y=7e^(−(6/x)) Σ_(n=0) ^∞ (x^(1−n) /(n!(1−n)))+7Ce^(−(6/x))

$$\frac{{dy}}{{dx}}−\frac{\mathrm{6}{y}}{{x}^{\mathrm{2}} }=\mathrm{7}{x}^{\mathrm{4}} \\ $$$${I}.{F}={e}^{\int−\frac{\mathrm{6}}{{x}^{\mathrm{2}} }} ={e}^{\frac{\mathrm{6}}{{x}}} \:\:\:\:\:\:\:\: \\ $$$${ye}^{\frac{\mathrm{6}}{{x}}} =\mathrm{7}\int{x}^{\mathrm{2}} {e}^{\frac{\mathrm{1}}{{x}}} {dx} \\ $$$${y}=\mathrm{7}{e}^{−\frac{\mathrm{6}}{{x}}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{1}−{n}} }{{n}!\left(\mathrm{1}−{n}\right)}+\mathrm{7}{Ce}^{−\frac{\mathrm{6}}{{x}}} \\ $$

Answered by TITA last updated on 08/Dec/20

(dy/dx)−6(y/x^(2 ) )=7x^2 using intergrating factor ∅=e^(∫((−6)/x^(2 ) )dx) =e^(6/x) hence e^(6/x) (dy/dx)−6e^(6/x) (y/x^2 )=7e^(6/x) x^2 ((d(ye^(6/x) ))/dx)=7x^2 ⇒ye^(6/x) =7∫e^(6/x) x^2 dx

$$\frac{{dy}}{{dx}}−\mathrm{6}\frac{{y}}{{x}^{\mathrm{2}\:} }=\mathrm{7}{x}^{\mathrm{2}} \:{using}\:{intergrating}\:{factor} \\ $$$$\emptyset={e}^{\int\frac{−\mathrm{6}}{{x}^{\mathrm{2}\:\:} }{dx}} ={e}^{\frac{\mathrm{6}}{{x}}} \:\:\:{hence}\:{e}^{\frac{\mathrm{6}}{{x}}} \frac{{dy}}{{dx}}−\mathrm{6}{e}^{\frac{\mathrm{6}}{{x}}} \frac{{y}}{{x}^{\mathrm{2}} }=\mathrm{7}{e}^{\frac{\mathrm{6}}{{x}}} {x}^{\mathrm{2}} \\ $$$$\frac{{d}\left({ye}^{\frac{\mathrm{6}}{{x}}} \right)}{{dx}}=\mathrm{7}{x}^{\mathrm{2}} \:\Rightarrow{ye}^{\frac{\mathrm{6}}{{x}}} =\mathrm{7}\int{e}^{\frac{\mathrm{6}}{{x}}} {x}^{\mathrm{2}} {dx} \\ $$

Commented by TITA last updated on 08/Dec/20

please continue from there

$${please}\:{continue}\:{from}\:{there} \\ $$

Commented by benjo_mathlover last updated on 08/Dec/20

it should be ye^(6/x) = ∫ 7x^2 e^(6/x) dx. your solution is wrong

$${it}\:{should}\:{be}\:{ye}^{\frac{\mathrm{6}}{{x}}} \:=\:\int\:\mathrm{7}{x}^{\mathrm{2}} {e}^{\frac{\mathrm{6}}{{x}}} \:{dx}. \\ $$$${your}\:{solution}\:{is}\:{wrong} \\ $$

Commented by TITA last updated on 08/Dec/20

ok thanks

$${ok}\:{thanks} \\ $$

Commented by mohammad17 last updated on 08/Dec/20

false solution

$${false}\:{solution} \\ $$

Answered by Bird last updated on 08/Dec/20

e→y^′ −(6/x^2 )y=7x^2 h→(y^′ /y)=(6/x^2 ) ⇒ln∣y∣=−(6/x)+c ⇒ y=k e^(−(6/x)) lavrange method→ y^′ =k^(′ ) e^(−(6/x)) +k((6/x^2 ))e^(−(6/x)) e⇒k^(′ ) e^(−(6/x)) +((6k)/x^2 )e^(−(6/x)) −(6/x^2 )ke^(−(6/x)) =7x^2 ⇒k^′ =7x^2 e^(−(6/x)) ⇒ k=∫ 7x^(2 ) e^(−(6/x)) dx+λ ⇒ y(x)=(∫^x 7t^2 e^(−(6/t)) dt+λ)e^(−(6/x))

$${e}\rightarrow{y}^{'} −\frac{\mathrm{6}}{{x}^{\mathrm{2}} }{y}=\mathrm{7}{x}^{\mathrm{2}} \\ $$$${h}\rightarrow\frac{{y}^{'} }{{y}}=\frac{\mathrm{6}}{{x}^{\mathrm{2}} }\:\Rightarrow{ln}\mid{y}\mid=−\frac{\mathrm{6}}{{x}}+{c}\:\Rightarrow \\ $$$${y}={k}\:{e}^{−\frac{\mathrm{6}}{{x}}} \:\:\:{lavrange}\:{method}\rightarrow \\ $$$${y}^{'} ={k}^{'\:} {e}^{−\frac{\mathrm{6}}{{x}}} \:+{k}\left(\frac{\mathrm{6}}{{x}^{\mathrm{2}} }\right){e}^{−\frac{\mathrm{6}}{{x}}} \\ $$$${e}\Rightarrow{k}^{'\:} {e}^{−\frac{\mathrm{6}}{{x}}} +\frac{\mathrm{6}{k}}{{x}^{\mathrm{2}} }{e}^{−\frac{\mathrm{6}}{{x}}} −\frac{\mathrm{6}}{{x}^{\mathrm{2}} }{ke}^{−\frac{\mathrm{6}}{{x}}} =\mathrm{7}{x}^{\mathrm{2}} \\ $$$$\Rightarrow{k}^{'} =\mathrm{7}{x}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{6}}{{x}}} \:\Rightarrow \\ $$$${k}=\int\:\mathrm{7}{x}^{\mathrm{2}\:} {e}^{−\frac{\mathrm{6}}{{x}}} {dx}+\lambda\:\Rightarrow \\ $$$${y}\left({x}\right)=\left(\int^{{x}} \mathrm{7}{t}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{6}}{{t}}} {dt}+\lambda\right){e}^{−\frac{\mathrm{6}}{{x}}} \\ $$

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