Toronto Math Forum
MAT2442018F => MAT244Tests => Quiz2 => Topic started by: Victor Ivrii 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.
$$
x^2y^3 + x(1 + y^2)y' = 0,\qquad \mu(x, y) = 1/xy^3.
$$

(x^{2}y^{3})+x(1+y^{2})y'= 0
Let M = x^{2}y^{3}, N = x(1+y^{2})
M_{y} = $\frac{d(M)}{dy}$ = 3x^{2}y^{2}
N_{x} = $\frac{d(N)}{dx}$ = 1+y^{2}
Since M_{y} ≠ N_{x}, hence not exact, and thus we need to use the integrating factor μ = 1/(xy^{3})
Multiply μ on both sides of the original equation,
1/(xy^{3})x^{2}y^{3} + 1/(xy^{3})x(1+y^{2})y' = 0
Now let M' = 1/(xy^{3})x^{2}y^{3} = x,
N' = 1/(xy^{3})x(1+y^{2}) = 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. φ_{x} = M', φ_{y} = N'
φ(x,y) = ∫M'dx =$\frac{1}{2}$x^{2} + h(y)
φ_{y} = h'(y) = N' = y^{3} + y^{1}
Thus h'(y) = y^{3} + y^{1}, h(y) = $\frac{1}{2}$y^{2} + lny + C
Therefore, φ(x,y) =$\frac{1}{2}$x^{2} $\frac{1}{2}$y^{2} + lny= C

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