Q2 TUT 0201

[Victor Ivrii]

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

[Pengyun Li]

(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) + x²cos(y)e^xy' = 0

Now let M' = xe^x(x+2)sin(y), N' = x²cos(y)e^x,

M'_y = $\frac{d(M')}{dy}$ = (x+2)xe^xcos(y),

N'_x = $\frac{d(N')}{dx}$ = (x+2)xe^xcos(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 =x²e^xsin(y) + h(y)

φ_y = φ'(x,y) = x²e^xcos(y)+h'(y) = N'

x²e^xcos(y) + h'(y) = x²e^xcos(y)

Thus h'(y)=0, h(y) is a constant.

Therefore, φ(x,y) = x²e^xsin(y) = C

[Victor Ivrii]

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