Author Topic: TUT0401 QUIZ2  (Read 363 times)

Juntian Ye

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TUT0401 QUIZ2
« on: October 04, 2019, 06:17:22 PM »
Question: Find an integrating factor and solve the given equation.
      (3x3y + 2xy +y3) + (x2 +y2)y’ = 0

Solution:
let M=3x3y + 2xy +y3, N= x2 +y2
My=3x3+2x+3y2, Nx=2x
since My≠Nx —> not exact equation
we need to find the integrationg factor µ(x,y)
multiply µ on both side of the equation
µ(3x3y + 2xy +y3) + µ(x2 +y2)y’ = 0
let M1= µ(3x3y + 2xy +y3) = µM, N1= µ(x2 +y2) = µN
we want M1y=N1x —>(µM)y=(µN)x
µyM+µMyxN+µNx
suppose µ is only depend on x
then µ = µ (x), µy=0
—>0+µMyxN+µNx
   µx/µ= (My-Nx)/N=3
—>µ=e∫3dx
   =e3x
we get e3x(3x3y + 2xy +y3) + e3x(x2 +y2)y’ = 0
M1= e3x(3x3y + 2xy +y3), N1= e3x(x2 +y2)
M1y=3x3e3x+2xe3x+3y2e3x,N1x=3x3e3x+2xe3x+3y2e3x
M1y=N1x —> it is exact equation now
By thm, there exist ϕ(x,y) s.t. ϕx = M1, ϕy = N1
ϕ(x,y) = ∫e3x(3x3y + 2xy +y3)dx
   =x2ye3x+(1/3)e3xy3+h(y)
ϕy =x2e3x+e3xy2+h’(y)
—>h’(y)=0 —>h(y) is constant
thus, ϕ(x,y)=x2ye3x+(1/3)e3xy3=C