Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

Messages - Qihui Huang

Pages: [1]
Quiz-5 / LEC0101 quiz5
« on: November 01, 2019, 02:03:28 PM »
Question: find a particular solution of the given homogeneous equation:
$$t^2y''+7ty'+5y=t, t>0$$
Rewrite as $$y''+\frac{7}{t}y'+\frac{5}{t^2}y=\frac{1}{t}$$ Let $$y=v(t)y_1(t)=vt^{-1}$$ then $$y'=v't^{-1}-vt^{-2}$$ $$y''=v''t^{-1}-2v't^{-2}+2vt^{-3}$$
$$v''t^{-1}-2v't^{-2}+2vt^{-3}+7v't^{-2}-7vt^{-3}+5vt^{-3}=t^{-1}$$ $$v''t^{-1}+5t^{-2}v'=t^{-1}$$ $$v''+5t^{-1}v'=1$$
Let $$v'=r(t), v''=r'(t)$$ so we get $$r'+\frac{5}{t}r=1$$ Find the integrating factor $$u(t)=e^{\int \frac{5}{t}dt}$$ $$u(t)=t^5$$ $$t^5r'+t^5\frac{5}{t}r=t^5$$ $$\int (t^5r)'=\int t^5 dt $$ $$rt^5=\frac{1}{6}t^6+c_1$$ $$r=\frac{1}{6}t+c_1t^{-5}$$ $$v=\frac{1}{12}t^2-\frac{1}{4}c_1t^{-4}+c_2$$ $$y=\frac{1}{12}t-\frac{1}{4}c_1t^{-4}t^{-1}+c_2t^{-1}$$ $$y=\frac{1}{12}t-\frac{1}{4}c_1t^{-5}+c_2t^{-1}$$ $$y=a_1t^{-5}+a_2t^{-1}+\frac{1}{12}t$$where $a_1$ and $a_2$ are arbitrary constants

Quiz-3 / TUT0702 Quiz3
« on: October 11, 2019, 02:26:03 PM »
Verify that the functions y1 and y2 are solutions of the given differential equation. Do they constitute a fundamental set of solutions?

$$y''-2y+y=0,y_1(t)=e^t, y_2(t)=te^t$$

Differentiate $y_1(t)=e^t, y_2(t)=te^t$ respect to t,
$$y_1'=e^t, y_2'=e^t+te^t$$
Substitute back to the differential equation,
So both $y_1(t)$ and $y_2(t)$ are two valid solutions to the equation.

To check whether it is a fundamental set of solution:
We want  $W(y_1,y_2)(t) \neq 0$
The determinant is not zero, so it is a fundamental set of solutions.

Quiz-2 / TUT0702 Quiz2
« on: October 04, 2019, 02:12:37 PM »
Determine whether the equation is exact or not

Let $$M(x,y)=e^xsin(y)-2ysin(x)$$ and let $$N(x,y)=-3x+e^xsin(y)$$
Then, $$M_y(x,y)=e^xcos(y)-2sin(x)$$ $$N_x(x,y)=-3+e^xsin(y)$$
Since $$M_y \neq N_x$$
so the given differential equation is not exact.

Quiz-1 / TUT 0702 question
« on: September 27, 2019, 02:26:19 PM »
Quiz tut0702

QUESTION: $$y'+y=te^{-t}+1$$

follow the form $$y'+p(t)y=g(t)$$

find the integrating factor: $$u(t)=e^{\int1dt}=e^t$$

mutiply u(t) on both sides $$e^ty'+e^ty=te^te^{-t}+e^t$$

integrating both sides $$\int{(e^ty')}=\int{te^0+e^t dt}$$
$$e^ty=\int{t dt}+\int{e^t dt}$$

So, as $t\rightarrow \infty$, $y \rightarrow 1$, since the term with $e^{-t}$ goes to zero.

Pages: [1]