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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

xdydx6yx=7x3helpme

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))

dydx6yx2=7x4I.F=e6x2=e6xye6x=7x2e1xdxy=7e6xn=0x1nn!(1n)+7Ce6x

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

dydx6yx2=7x2usingintergratingfactor=e6x2dx=e6xhencee6xdydx6e6xyx2=7e6xx2d(ye6x)dx=7x2ye6x=7e6xx2dx

Commented by TITA last updated on 08/Dec/20

please continue from there

pleasecontinuefromthere

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

itshouldbeye6x=7x2e6xdx.yoursolutioniswrong

Commented by TITA last updated on 08/Dec/20

ok thanks

okthanks

Commented by mohammad17 last updated on 08/Dec/20

false solution

falsesolution

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))

ey6x2y=7x2hyy=6x2lny∣=6x+cy=ke6xlavrangemethody=ke6x+k(6x2)e6xeke6x+6kx2e6x6x2ke6x=7x2k=7x2e6xk=7x2e6xdx+λy(x)=(x7t2e6tdt+λ)e6x

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