TUT0102 Quiz2

[Changhao Jiang]

Question: e^x+（e^x  cot⁡(y)+2y csc⁡(y) ） y^'=0, find an integrating factor and solve the given equation.
Solution:
My=0.
Nx=∂/∂x [e^x  cot⁡(y)+2y csc⁡(y) ]=cot⁡(y) e^x.
We know the given equation is not exact, so we need to find μ(t).
R=(My-Nx)/M=(0-cot⁡(y) e^x)/e^x =-cot⁡(y).
μ(t)=e^(-∫Rdy)=e^(-∫cot⁡y dy )=e^ln⁡|sin⁡y | =sin⁡y.
Multiply sin⁡y on both sides.
We get sin⁡y e^x+(e^x  cos⁡y+2y) y^'=0.
There exists ψ(x,y) such that ψ_x (x,y)=sin⁡y e^x.
Integrating on both sides with x, ψ(x,y)=sin⁡y e^x+g(y).
Differentiate on both sides with y, ψ_y (x,y)=cos⁡y e^x+g^' (y).
Since ψ_y (x,y)=N=e^x  cos⁡y+2y, then g^' (y)=2y and so g(y)=y^2+c.
Combine all results above, we get cos⁡y e^x+y^2=c.