Toronto Math Forum
MAT244--2018F => MAT244--Tests => Quiz-2 => Topic started by: Victor Ivrii on October 05, 2018, 05:15:06 PM
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Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$
x^2y^3 + x(1 + y^2)y' = 0,\qquad \mu(x, y) = 1/xy^3.
$$
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(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
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You need to learn $\LaTeX$ to avoid a shappy html math
And no double-dipping next time (just 1 quiz)