Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz2 => Topic started by: Yuchen Cong on October 07, 2019, 04:32:59 PM

Q: x^{2}y^{3}+x(1+y^{2})y'=0, μ(x,y)=1/xy^{3}
Solution:
Define that M(x,y)=x^{2}y^{3}, N(x,y)=x(1+y^{2})
Then M_{y}=3x^{2}y^{2}, N_{x}=1+y^{2}
Since M_{y}≠N_{x}, so the given DE is not exact
Multiply μ(x,y)=1/xy^{3} to both sides:
(1/xy^{3})(x^{2}y^{3})+(1/xy^{3})x(1+y^{2})y'=0
x+(1/y^{3}+1/y)y'=0
Then we have M'(x,y)=x, N'(x,y)=1/y^{3}+1/y
M'_{y}=0, N'_{x}=0
Since M'_{y}=N'_{x}, the DE is exact
Thus, there exists a function φ(x,y) such that
φ_{x}=M'(x,y) and φ_{y}=N'(x,y)
Since φ_{x}=M'(x,y)=x
Integrating both sides with respect to x, we get
φ(x,y)=(1/2)x^{2}+h(y)
Differentiating both sides with respect to y:
φ_{y}=h'(y)
Since φ_{y}=N'(x,y)=1/y^{3}+1/y
Then h'(y)=1/y^{3}+1/y
h(y)=(1/2)y^{2}+lny+C
Thus, φ(x,y)=(1/2)x^{2}(1/2)y^{2}+lny+C
Therefore, C=(1/2)x^{2}(1/2)y^{2}+lny