Author Topic: TUT0601 Quiz2  (Read 389 times)

Joy Zhou

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TUT0601 Quiz2
« on: October 05, 2019, 06:31:35 PM »
Find an integrating factor and solve the given equation.
$$
\left(3 x+\frac{6}{y}\right)+\left(\frac{x^{2}}{y}+3 \frac{y}{x}\right) \frac{d y}{d x}=0
$$

We want to find an integrating factor $\mu$ as a function of $xy$ such that
$(\mu M)_{y}=(\mu N)_{x}$, Let $z=xy$. Thus, $\mu(x y)=\mu(z(x, y))$ Then

$
\mu_{x}(x y)=\frac{d \mu}{d z} \frac{\partial z}{\partial x}=y \frac{d \mu}{d z}
$
$\mu_{y}(x y)=\frac{d \mu}{d z} \frac{\partial z}{\partial y}=x \frac{d \mu}{d z}$

Therefore,
$$
(\mu M)_{y}=(\mu N)_{x}
$$
$$
\mu M_{y}+x M \frac{d \mu}{d z}=\mu N_{x}+y N \frac{d \mu}{d z}
$$
$$
\mu\left(M_{y}-N_{x}\right)=\frac{d \mu}{d z}(y N-x M)
$$
$$
\frac{\mathrm{d} \mu}{\mathrm{d} z}=\mu\left(\frac{N_{x}-M_{y}}{x M-y N}\right)
$$

Therefore,

$
\mu(z)=\exp \left(\int R(z) \mathrm{d} z\right)
$
\quad where $R(z)=R(x y)=\frac{N_{x}-M_{y}}{x M-y N}$

Returning to our original differential equation, let

$M(x, y)=3 x+\frac{6}{y}$ \quad and \quad $N(x, y)=\frac{x^{2}}{y}+3 \frac{y}{x}=0$

Then

$\frac{\partial}{\partial y} M(x, y)=\frac{-6}{y^{2}}$ \quad and \quad $\frac{\partial}{\partial x} N(x, y)=\frac{2 x}{y}-\frac{3 y}{x^{2}}$

We can see that this equation is not exact, however, note that
$$\frac{N_{x}-M_{y}}{x M-y N}=\frac{\frac{2 x}{y}-\frac{3 y}{x^{2}}+\frac{6}{y^{2}}}{x\left(3 x+\frac{6}{y}\right)-y\left(\frac{x^{2}}{y}+3 \frac{y}{x}\right)}=\frac{\frac{2 x}{y}-\frac{3 y}{x^{2}}+\frac{6}{y^{2}}}{2 x^{2}+\frac{6 x}{y}-\frac{3 y^{2}}{x}}=\frac{\frac{2 x}{y}-\frac{3 y}{x^{2}}+\frac{6}{y^{2}}}{x y\left(\frac{2 x}{y}-\frac{3 y}{x^{2}}+\frac{6}{y^{2}}\right)}=\frac{1}{x y} $$
Let $xy=z$

Thus, we have an integrating factor
$$
\mu(x y)=\exp \left(\int \frac{1}{z} \mathrm{d} z\right)=e^{\log |z|}=z=x y
$$

Multiplying the original differential equation through by our integrating factor, we have
 $$\left(3 x^{2} y+6 x\right)+\left(x^{3}+3 y^{2}\right) \frac{d y}{d x}=0$$
We can see that this differential equation is exact because
$$
\frac{\partial}{\partial y}\left(3 x^{2} y+6 x\right)=3 x^{2}=\frac{\partial}{\partial x}\left(x^{3}+3 y^{2}\right)
$$
Thus, there exists a function $\psi(x, y)$ such that
\begin{equation}
\psi_{x}(x, y)=3 x^{2} y+6x
\tag{1}
\end{equation}
\begin{equation}
\psi_{y}(x, y)=x^{3}+3 y^{2}
\tag{2}
\end{equation}
Integrating (1) with respect to $x$, we get
$$
\psi(x, y)=x^{3} y+3 x^{2}+h(y)
$$
for some function $h$ of $y$. Next, differentiating with respect to $y,$ and equating
with ( 2)$,$ we get
$$
\psi_{y}(x, y)=x^{3}+h^{\prime}(y)
$$
Therefore,
$$
h^{\prime}(y)=3 y^{2}
$$
$$h(y)=y^{3}$$
and we have
$$
\psi(x, y)=x^{3} y+3 x^{2}+y^{3}
$$
Thus, the solutions of the differential equation are given implicitly by
$$
x^{3} y+3 x^{2}+y^{3}=C
$$