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Topics - Junhong Zhou

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Quiz 2 / Quiz2-6101 6D
« on: October 02, 2020, 02:22:25 PM »
Problem(3pt). Find all points of continuity of the given function;
\begin{cases}\frac{z^4-1}{z-i},& z\neq i\\4i, & z=i


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

When z = i, then


$z - i = i-i = 0$

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

Quiz-3 / tut0402 quiz3
« on: October 11, 2019, 02:03:35 PM »
Question: Find the Wronskian of the given pair of functions.
$$cos^2(x), 1+cos(2x)$$

$$W = \det
cos^2(x) & 1+cos2x \\
-2sin(x)cos(x) & -2sin(2x)
= \det
cos^2(x) & 2cos^2(x) \\
-sin(2x) & -2sin(2x)

    \implies W &= -2cos^2(x)sin2(x)-(-sin(2x))(2cos^2(x))\notag\\
    &= 2sin(2x)(-cos^2(x)+cos^2(x))\notag\\
    &= 2sin(2x)\cdot0\notag\\
    &= 0\notag

Quiz-2 / TUT0402 quiz2
« on: October 04, 2019, 02:00:07 PM »
Question: Find an integrating factor and solve the given equation.

    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

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.


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 $\mu$:
    \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

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


Quiz-1 / TUT0402 quiz1
« on: September 27, 2019, 02:00:07 PM »

write this into standard equation:

which here we have
$$\mu(x)=e^{\int p(t)dt}=e^{\int\frac{2}{t}dt}=e^{2\ln{t}}=t^2$$

multiply $\mu(x)$ to both sides of the standard equation we have:
    (t^2)y'+(t^2)\frac{2}{t}y &= (t^2)\frac{sin(t)}{t}\notag\\
    t^2y'+2ty &= tsin(t)\notag\\
    \int t^2y'+2tydt &=\int tsin(t)dt\notag\\
    t^2y &= \int tsin(t) \notag

integrate $\int tsin(t)dt$ by parts, let $u=t$ and $dv=sin(t)dt$:
    \int tsin(t)dt &= t(-cos(t))-\int -cos(t)dt\notag\\
    &= -tcos(t)+\int cos(t)dt\notag\\
    &= -tcos(t)+sin(t)+C\notag

so we have:
    t^2y &= -tcos(t)+sin(t)+C\notag\\
    y &= \frac{-tcos(t)+sin(t)+C}{t^2}\notag


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