tut5103 quiz 2

[YUTONG ZHANG]

(sin⁡y/y-2e^(-x)  sin⁡x )+((cos⁡y+2e^(-x)  cos⁡x)/y)y'=0, u(x,y)=ye^x

Answer:
(sin⁡y/y-2e^(-x)  sin⁡x )+((cos⁡y+2e^(-x)  cos⁡x)/y)y'=0
M=sin⁡y/y-2e^(-x)  sin⁡x
dM/dY=〖-y〗^(-2)  sin⁡y+1/y  cos⁡y
N=(cos⁡y+2e^(-x)  cos⁡x)/y
dN/dX=1/y (-2e^(-x)  cos⁡x+2e^(-x) (-sin⁡x ))

Since dM/dY does not equal dN/dX
The equation is not exact.
Multiply μ=ye^x to both side of the equation
Get:

sin⁡〖(y) e^x 〗-2y sin⁡x+(e^x  cos⁡y+2 cos⁡(x))y^'=0
M=sin⁡〖(y) e^x 〗-2y sin⁡x
dM/dY=e^x  cos⁡y-2 sin⁡x
N=e^x  cos⁡y+2 cos⁡x
dN/dX=e^x  cos⁡y-2 sin⁡x

Now dM/dY=dN/dX
The equation is exact.

There exists ϕ(x,y) s.t. dϕ/dx=M, dϕ/dy=N

ϕ=∫M dx=∫〖sin⁡〖(y)〗 e^x 〗-2y sin⁡(x) dx=sin⁡(y) e^x +2y cos⁡(x)+h(y)
dϕ/dy=e^x  cos⁡y+2 cos⁡x+h'(y)
Also, dϕ/dy=N=e^x  cos⁡y+2 cos⁡x
So h’(y)=0

ϕ(x,y)=sin⁡(y) e^x +2y cos⁡(x)=C