Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz2 => Topic started by: Lan Cheng on October 04, 2019, 02:00:01 PM

show the given equation is not exact but becomes exact when multiplied by the given integrating factor, then solve the equation.
x^(2) * y^(3) + x * (1+y^2) * y' = 0, u(x, y) = 1/(xy^3).
First, let's show the given DE x^(2) * y^(3) + x * (1+y^2)y' = 0 is not exact.
Define M = x^(2) * y^(3), N = x * (1 + y^2).
M_y = d/(dy) [x^(2) * y^(3)] = 3x^(2) * y^(2)
N_x = d/(dx) [x * (1 + y^2)] = 1 + y^2
Since 3x^(2)y^(2) != 1 + y^2, the given DE is not exact.
multiply each side of given DE by integrating factor u, we get
1/(xy^3) * x^(2) * y^(3) + 1/(xy^3) * x * (1 + y^2)y' = x + (y^(3) + y^(1)) * y' = 0
Let the new M = 1/(xy^3) x^(2) * y^(3), new N = 1/(xy^3) * x * (1 + y^2)
M_y = 0, N_x = 0, M_y = N_x and the new DE is exact.
there exist \phy (x, y) such that
\phy x = M, \phy y = N.
\phy x = M = 1/(x * y^3) * x^(2) * y^(3)
Integrating both side by x, we have
\phy = 1/2 * x^2 + h(y)
\phy y = h'(y) = N = 1/(xy^3) * x * (1 + y^2)
h'(y) = 1/(xy^3) * x * (1 + y^2)
Integrating both side by y, we have
h(y) = 0.5 * y^(2) + lny + C
Altogether, we have
\phy (x,y) = 0.5 * x^2  0.5 * y^(2) + lny + C
So our general solution is
0.5 * x^2  0.5 * y^(2) + lny = C.