TUT0602 Quiz2

[Yuchen Cong]

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 M_y=3x²y², N_x=1+y²

Since M_y≠N_x, 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
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)x²+h(y)

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

Since φ_y=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|