Author Topic: Q2 TUT 0401 and TUT 0601  (Read 2069 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2563
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Q2 TUT 0401 and TUT 0601
« on: October 05, 2018, 05:15:06 PM »
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$
x^2y^3 + x(1 + y^2)y' = 0,\qquad  \mu(x, y) = 1/xy^3.
$$

Pengyun Li

  • Full Member
  • ***
  • Posts: 20
  • Karma: 14
    • View Profile
Re: Q2 TUT 0401 and TUT 0601
« Reply #1 on: October 05, 2018, 06:14:52 PM »
(x2y3)+x(1+y2)y'= 0

Let M = x2y3, N = x(1+y2)

My = $\frac{d(M)}{dy}$ = 3x2y2

Nx = $\frac{d(N)}{dx}$ = 1+y2

Since My ≠ Nx, hence not exact, and thus we need to use the integrating factor μ = 1/(xy3)

Multiply μ on both sides of the original equation,

1/(xy3)x2y3 + 1/(xy3)x(1+y2)y' = 0

Now let M' = 1/(xy3)x2y3 = x,

N' = 1/(xy3)x(1+y2) = y-3 + y-1,

M'y = $\frac{d(M')}{dy}$ = 0,

N'x = $\frac{d(N')}{dx}$ = 0,

M'y = N'x, hence exact now.

There exist φ(x,y) s.t. φx = M', φy = N'

φ(x,y) = ∫M'dx =$\frac{1}{2}$x2 + h(y)

φy = h'(y) = N' = y-3 + y-1

Thus h'(y) = y-3 + y-1, h(y) = -$\frac{1}{2}$y-2 + ln|y| + C

Therefore, φ(x,y) =$\frac{1}{2}$x2 -$\frac{1}{2}$y-2 + ln|y|= C



Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2563
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Q2 TUT 0401 and TUT 0601
« Reply #2 on: October 05, 2018, 07:27:34 PM »
You need to learn $\LaTeX$ to avoid a shappy html math

And no double-dipping next time (just 1 quiz)