# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-2 => Topic started by: Kun Zheng on October 04, 2019, 02:31:08 PM

Title: TUT0602 QUIZ2
Post by: Kun Zheng on October 04, 2019, 02:31:08 PM
Hi everyone, the problem is to prove (x^2)(y^3) + x(1+y^2)y'=0 is not exact, then multiply μ(x,y)=1/(xy^3) to solve the equation

First of all, we observe it follows M(x,y)+N(x,y)∂y/∂x = 0 form, we can get M(x,y)=(x^)(2y^3), N(x,y)= x(1+y^2)=x+xy^2
Then we get My=∂/∂y((x^2)(y^3))=3(x^2)(y^2) and Nx=∂/∂x(x(1+y^2))=1+y^2
As we see My is not equal to Nx, this is not an exact equation.

Second, we try multiplying the integrating factor μ(x,y)=1/(xy^3) on both sides: x+((1/y^3) + (1/y))∂y/dx=0
We do the first step again: M(x,y)=x and N(x,y)=1/y^3 + (1/y)
Derivative them: My=∂/∂y(x) =0 ; Nx=∂/∂x((1/y^3) + (1/y))=0
As we see My=Nx, now this is an exact equation.

Now we need to get the solution: ∫ M∂x+N∂x=(1/2)x^2 - (1/2)y^(-2)+ln|y|=2ln|y|+(x^2)-(1/y^2)=c