### Author Topic: TUT0301 Quiz2  (Read 776 times)

#### Siyan Chen

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##### TUT0301 Quiz2
« on: October 04, 2019, 02:01:10 PM »
$$1 + (\frac{x}{y}-\sin(y))y’ = 0$$
Let $M=1, N=\frac{x}{y}-\sin(y)$
Then, we can get: $M_{y}=0, N_{x}=\frac{1}{y}$
Define $R=\frac{M_{y}-N_{x}}{M}=\frac{-\frac{1}{y}}{1}=-\frac{1}{y}$,
=> The integrating factor should be: $\mu (x,y) = e^{-\int R dy} = e^{ln|y|} = y$

Multiple both sides by y, we can define the new $M = y, N = x-\sin(y)y$,

Since it’s exact now, there exists a function $\phi (x,y)$, such that: $\phi_{x}(x,y) = M(x,y)\ and\ \phi_{y}(x,y) = N(x,y)$
Integrating with respect to x:
$\phi(x,y) = \int M(x,y) \ dx \\ i.e. \phi(x,y) = \int y\ dx \\ \phi(x,y) = xy + h(y)$
Then, we have: $\phi_{y} = x + h’(y) = N = x-y\sin(y)$, i.e. $h’(y)=-y\sin(y)$
=> $h(y) = -\int y \sin(y) \ dy$

Using integrating by parts,  $u = y, \ v = -\cos(y),\\du = 1, \ dv = \sin(y)$

=> $h(y) = -(-y \cos(y)- \int(-\cos(y))\ dy) = y \cos(y) - \sin(y)$

So, the solution of the given equation is: $\phi(x,y) = xy + y \cos(y) - \sin(y) = C$, where C is the constant