### Author Topic: TUT0801 Quiz2  (Read 419 times)

#### XiaolongZhao

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• Karma: 0 ##### TUT0801 Quiz2
« on: October 04, 2019, 02:00:05 PM »
Q：Prove the equation is not exact, and becomes exact after multiplying the integrating factor, and then solve the equation:
x^2 y^3+x(1+y^2 ) y'=0   μ(x,y)=1/xy^3
A:
Step1: set M(x,y) and N(x,y), then get M_y and N_x to prove it is not exact
x^2 y^3 dx+x(1+y^2 )dy=0
Set M=x^2 y^3,N=x+xy^2
M_y=3x^2 y^2,N_x=1+y^2
Since M_y≠N_x, I’ve proven it is not exact

Step2: Multiply the integrating factor to both sides of the equation, and then get new M, N, M_y, and N_x, to prove it becomes exact
xdx+(1+y^2)/y^3  dy=0
Set M'=x, N'=(1+y^2)/y^3
M'_y=0, N'_x=0
Since M'_y=N'_x , I’ve proven it becomes exact

Step3: Solve the equation
Since the equation is exact, ∃φ(x,y)  s.t.φ_x=M',φ_y=N'
φ=∫M'dx=∫xdx=x^2/2+h(y)
φ_y=0+h' (y)=N'=(1+y^2)/y^3
h' (y)=(1+y^2)/y^3
h(y)=∫〖[(1+y^2)/y^3]dy〗=∫〖(1/y^3 )dy〗+∫〖(1/y)dy〗= -y^(-2)/2+ln|y|+C
φ=x^2/2 - y^(-2)/2 + ln|y|+C
Thus, x^2/2 - y^(-2)/2 + ln|y|=C is the solution of the differential equation

The clearer answer is shown in the picture below.