### Author Topic: TUT0402 quiz2  (Read 1007 times)

#### yuhan cheng

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• Karma: 0 ##### TUT0402 quiz2
« on: October 04, 2019, 02:04:45 PM »
$$1+(\frac{x}{y}-\sin(y))y^{\prime}=0.$$

\noindent Let
$$M(x,y)=1\quad \text{and}\quad N(x,y) =(\frac{x}{y}-\sin(y))$$
Then
$$\frac{\partial}{\partial y}M(x,y)=0\quad\text{and}\quad \frac{\partial}{\partial x}N(x,y)=\frac{1}{y}$$
We can see that this equation is not exact, however, note that
$$\frac{d\mu}{d y}=(\frac{N_x-M_y}{M})\mu=\frac{\mu}{y}\quad\rightarrow\quad \mu=y$$
Multiplying our original euqation by$\mu(y)$, we have
$$y+(x-y\sin(y))y^{\prime}=0$$
We can see that this equaion is exact, since
$$\frac{\partial}{\partial y}(y)=1=\frac{\partial}{\partial x}(x-y\sin(y))$$
Thus, there exists a function $\psi(x,y)$ such that
\begin{align}
\psi_x (x,y)&=y\\
\psi_y (x,y)&=x-y\sin(y)
\end{align}
Integating (1) with respect to x, we get
$$\psi(x,y)=xy+h(y)$$
for some arbitary function $h$ of $y$. Next, differentiating with respect to $y$, and equating with (2), we get
$$\psi_y(x,y)=x+h^{\prime}(y)$$
Therefore,
$$h^{\prime}(y)=-y\sin(y)\quad\rightarrow\quad h(y)=s\int y\sin(y)dy=y\cos(y)-\sin(y)$$
and we have
$$\psi(x,y)=xy+y\cos(y)-\sin(y)$$
Thus, the solutions of the differential equation are given implicity by
$$xy+y\cos(y)-\sin(y)=C$$