Toronto Math Forum
MAT2442018F => MAT244Tests => Quiz2 => Topic started by: Victor Ivrii on October 05, 2018, 05:09:19 PM

Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$
(x + 2) \sin(y) + x \cos(y)y' = 0,\qquad \mu (x, y) = xe^x.
$$

(x+2)sin(y) + xcos(y)y' = 0
Let M = (x+2)sin(y), N = xcos(y)
M_{y} = $\frac{d((x+2)sin(y))}{dy}$ = (x+2)cos(y)
N_{x} = $\frac{d(xcos(y))}{dx}$ = cos(y)
Since M_{y} ≠ N_{x}, hence not exact, and thus we need to find the integrating factor.
Let $\frac{My  Nx}{N}$. we can derive $\frac{(x+2)cos(y)cos(y)}{cos(y)}$ = $\frac{x+1}{x}$, which is a function of x only.
μ = e^{∫$\frac{x+1}{x}$dx} = xe^{x}, which is the integrating factor.
Multiply μ on both sides of the original equation,
xe^{x}(x+2)sin(y) + x^{2}cos(y)e^{x}y' = 0
Now let M' = xe^{x}(x+2)sin(y), N' = x^{2}cos(y)e^{x},
M'_{y} = $\frac{d(M')}{dy}$ = (x+2)xe^{x}cos(y),
N'_{x} = $\frac{d(N')}{dx}$ = (x+2)xe^{x}cos(y),
M'_{y} = N'_{x}, hence exact now.
There exist φ(x,y) s.t. φ_{x} = M', φ_{y} = N'
φ(x,y) = ∫M'dx =x^{2}e^{x}sin(y) + h(y)
φ_{y} = φ'(x,y) = x^{2}e^{x}cos(y)+h'(y) = N'
x^{2}e^{x}cos(y) + h'(y) = x^{2}e^{x}cos(y)
Thus h'(y)=0, h(y) is a constant.
Therefore, φ(x,y) = x^{2}e^{x}sin(y) = C

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