Author Topic: TUT0602 Quiz2  (Read 520 times)

Yuchen Cong

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TUT0602 Quiz2
« on: October 07, 2019, 04:32:59 PM »
Q: x2y3+x(1+y2)y'=0, μ(x,y)=1/xy3

Solution:
Define that M(x,y)=x2y3, N(x,y)=x(1+y2)

Then My=3x2y2, Nx=1+y2

Since My≠Nx, so the given DE is not exact

Multiply μ(x,y)=1/xy3 to both sides:
(1/xy3)(x2y3)+(1/xy3)x(1+y2)y'=0
x+(1/y3+1/y)y'=0

Then we have M'(x,y)=x, N'(x,y)=1/y3+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
φx=M'(x,y) and φy=N'(x,y)

Since φx=M'(x,y)=x
Integrating both sides with respect to x, we get
φ(x,y)=(1/2)x2+h(y)

Differentiating both sides with respect to y:
φy=h'(y)

Since φy=N'(x,y)=1/y3+1/y
Then h'(y)=1/y3+1/y
h(y)=(-1/2)y-2+ln|y|+C

Thus, φ(x,y)=(1/2)x2-(1/2)y-2+ln|y|+C

Therefore, C=(1/2)x2-(1/2)y-2+ln|y|