Toronto Math Forum

Problem(3pt). Find all points of continuity of the given function;
$$f(z)=
\begin{cases}\frac{z^4-1}{z-i},& z\neq i\\4i, & z=i
\end{cases}$$

Answer:

f(z) is continuous when $z\neq i$.

When z = i, then

$z^4-1=i^4-1=1-1=0$

$z - i = i-i = 0$

Now use the L'Hospital's Rule we have:
\begin{align*}
    \lim_{z \to i} \frac{z^4-1}{z-i} &= \lim_{z \to i} \frac{4z^3}{1}\\
    &= 4i^3\\
    &= -4i\\
    & \neq 4i
\end{align*}
Therefore f(z) is not continuous at z = i.

Question: Find an integrating factor and solve the given equation.
$$(3x^2y+2xy+y^3)+(x^2+y^2)y'=0$$

\begin{align}
    M(x,y)=3x^2y+2xy+y^3 &\implies M_y=3x^2+2x+3y^2\notag\\
    N(x,y)=x^2+y^2 &\implies N_x=2x\notag
\end{align}

Since $M_y \neq N_x$, this implies the given differential equation is not exact, so we need to find $\mu(x,y)$ such that the equation $\mu(3x^2+2xy+y^3)+\mu(x^2+y^2)y'=0$ is exact.

$$R(x)=\frac{M_y-N_x}{N}=\frac{3x^2+3y^2}{x^2+y^2}=3$$

then we can write $\mu$:
$$\mu(x,y)=e^{\int R(x)dx}=e^{\int 3 dx}=e^{3x}$$

multiply the given differential equation by $\mu$:
\begin{align}
    \mu(3x^2y+2xy+y^3)+\mu(x^2+y^2)y' &= 0\notag\\
    e^{3x}(3x^2y+2xy+y^3)+e^{3x}(x^2+y^2)y' &= 0\notag
\end{align}

Which is now an exact differential equation, this implies there exist $\varphi(x,y)$ such that $\varphi_x=M$ and $\varphi_y=N$.
\begin{align}
    \varphi_y(x,y)=e^{3x}(x^2+y^2) &\implies \varphi(x,y)
    =\int e^{3x}(x^2+y^2)dy\notag\\
    &\implies \varphi(x,y)
    = e^{3x}x^2y+\frac{1}{3}e^{3x}y^3+f(x)\notag
\end{align}
$$\varphi_x(x,y)=2e^{3x}xy+3e^{3x}x^2y+e^{3x}y^3+f'(x) \implies f'(x)=0 \implies f(x)=C\notag$$

Therefore:
$$\varphi(x,y)=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3=C$$
$$e^{3x}x^2y+\frac{1}{3}e^{3x}y^3=C$$