### Author Topic: Q2 TUT 0201  (Read 1797 times)

#### Victor Ivrii

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##### Q2 TUT 0201
« 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.
$$(x + 2) \sin(y) + x \cos(y)y' = 0,\qquad \mu (x, y) = xe^x.$$

#### Pengyun Li

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##### Re: Q2 TUT 0201
« Reply #1 on: October 05, 2018, 05:54:21 PM »
(x+2)sin(y) + xcos(y)y' = 0

Let M = (x+2)sin(y), N = xcos(y)

My = $\frac{d((x+2)sin(y))}{dy}$ = (x+2)cos(y)

Nx = $\frac{d(xcos(y))}{dx}$ = cos(y)

Since My ≠ Nx, 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 = xex, which is the integrating factor.

Multiply μ on both sides of the original equation,

xex(x+2)sin(y) + x2cos(y)exy' = 0

Now let M' = xex(x+2)sin(y), N' = x2cos(y)ex,

M'y = $\frac{d(M')}{dy}$ = (x+2)xexcos(y),

N'x = $\frac{d(N')}{dx}$ = (x+2)xexcos(y),

M'y = N'x, hence exact now.

There exist φ(x,y) s.t. φx = M', φy = N'

φ(x,y) = ∫M'dx =x2exsin(y) + h(y)

φy = φ'(x,y) = x2excos(y)+h'(y) = N'

x2excos(y) + h'(y) = x2excos(y)

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

Therefore, φ(x,y) = x2exsin(y) = C

#### Victor Ivrii

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