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 - Meng Wu

Pages: 1 ... 3 4 [5] 6 7
61
Quiz-4 / Re: Q4-T0501
« on: March 02, 2018, 04:58:00 PM »
$$(1-t)y’’+ty’-y=2(t-1)^2e^{-t}, 0< t<1; y_1(t)=e^t, y_2(t)=t$$
Hence,$$\cases{y_1(t)=e^t\\y_1’(t)=e^t\\y_1’’(t)=e^t} \text{and} \cases{y_2(t)=t\\y_2’(t)=1\\y_2’’(t)=0}$$
Substitute back into the homogeneous equation: $$(1-t)y’’+ty’-y=0$$
Verified that $y_1(t)$ and $y_2(t)$ both satisfy the corresponding homogeneous equation. $\\$
And the complementary solution $y_c(t)=c_1e^t+c_2$ $\\$
Now divide both sides of the original equation by $1-t$:
$$y’’+{t\over 1-t}-{1\over 1-t}=-2(t-1)e^{-t}$$
Then $$p(t)={t\over 1-t}, q(t)=-{1\over 1-t}, g(t)=-2(t-1)e^{-t}$$
$$W[y_1,y_2](t)=\begin{array}{|c c|} y_1(t)&y_2(t)\\y_1’(t)&y_2’(t)\end{array}=(1-t)e^t$$
Since the particular solution has the form: $$Y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$$
and $$\begin{align}u_1(t)&=-\int{{y_2(t)g(t)\over W[y_1,y_2](t)}}dt\\&=-\int{t\cdot (-2(t-1)e^{-t})\over (1-t)e^t}dt\\&=-2\int{te^{-2t}}dt\\&=(t+{1\over2})e^{-2t}\end{align}$$
$$\begin{align}u_2(t)&=\int{y_1(t)g(t)\over W[y_1,y_2](t)}dt\\&=\int{e^t\cdot (-2(t-1)e^{-t})\over (1-t)e^t}dt\\&=2\int{e^{-t}}\\&=-2e^{-t}\end{align}$$
Therefore, $$Y(t)=(t+{1\over2})e^{-2t}\cdot e^t+ (-2e^{-t})\cdot t=({1\over2}-t)e^{-t}$$
Hence, the general solution:
$$\begin{align}y(t)&=y_c(t)+Y(t)\\&=c_1e^t+c_2t+({1\over2}-t)e^{-t}\end{align}$$
Therefore, the particular solution of the given nonhomogeneous equation is $$Y(t)=({1\over2}-t)e^{-t}$$

62
Quiz-4 / Q4-T0501
« on: March 02, 2018, 04:56:49 PM »
Verify the given functions $y_1$ and $y_2$ satisfy the correspending homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
\begin{align*}
&(1-t)y’’+ty’-y=2(t-1)^2e^{-t}, 0< t<1\\
 &y_1(t)=e^t, y_2(t)=t.
\end{align*}

63
MAT244--Misc / Re: Test Paper
« on: February 27, 2018, 03:54:02 PM »
Because too many students failed to write properly their tutorial sections, papers will be distributed at lecture sections.

We didn’t get our test papers during the lecture, does that mean it will be returned during tutorial?  :o

64
MAT244––Home Assignments / Typo from Section 3.6 Problems
« on: February 26, 2018, 11:31:38 AM »
Section 3.6, Question# 16. The original question from 10edition is:
$$(1-t)y’’+ty’-y=2(t-1)^2e^{-t},  0 < t < 1; y_1(t)=e^t, y_2(t)=t$$

65
MAT244--Misc / Term Test papers
« on: February 23, 2018, 07:47:05 PM »
Could we be able to pickup our test papers before our own tutorials? My tutorial is on Thursday. :o

66
Term Test 1 / Re: P3-Morning
« on: February 21, 2018, 10:56:39 AM »
Small Error:
For part(b), $y'(0)$ should be $$y'(0)=3c_1+2c_2+2=0$$
$$\implies\cases{c_1=2\\c_2=-4}$$ and $$y(t)=2e^{3t}-4e^{2t}+2e^t-te^{2t}$$

67
Term Test 1 / Re: P2-Morning
« on: February 21, 2018, 10:28:50 AM »
Cheryl
Almost unreadable.

Meng
(9) is OK, but (10) is not

How about now?

68
Term Test 1 / Re: P1-Day
« on: February 21, 2018, 08:31:10 AM »
(4) is correct but then integrated with an error

Typos. Corrections done.

69
Term Test 1 / Re: P2-Morning
« on: February 16, 2018, 01:15:55 AM »
$(a)$ $\\$
First we divide both sides by $-x^2(ln(x)-1)$:
$$y’’-{x \over x^2(ln(x)-1)}y’+{1 \over x^2(ln(x)-1)}y=0$$
Let $p(t)=-{x \over x^2(ln(x)-1)}$ and $q(t)={1 \over x^2(ln(x)-1)}.$ $\\$
Noting $p(t)$ and $q(t)$ are continuous everywhere except when $x^2(ln(x)-1)\neq0$.
$\\$
By Abel’s Theorem:$$\begin{align}W(y_1,y_2)(x)&=cexp(-\int{p(x)dx})\\&=cexp(-\int-{x \over x^2(ln(x)-1)}dx)\\&=ce^{ln|ln|x|-1|}\\&=c(ln|x|-1)\end{align}$$
Let $c=1$, then $W(y_1,y_2)(x)=ln|x|-1$. $\\$
$(b)$ $\\$
Since $y_1(x)=x,$ then
$$\cases{y_1’(x)=1\\y_1’’(x)=0}$$
Substitute these values back to the ODE:
$$\begin{align}0-{x \over x^2(ln(x)-1)}\cdot 1 + {1 \over x^2(ln(x)-1)}\cdot x=0\end{align}$$
Thus $y_1=(x)$ is indeed a solution. $\\$
Now we need to find $y_2(x)$ $\\$
Since we already know: $$\begin{align}W(y_1,y_2)(x)&=\begin{array}{|c c|}y_1(x)& y_2(x) \\ y_1’(x) & y_2’(x)\end{array}=ln|x|-1\\&=\begin{array}{|c c|}x& y_2(x) \\ 1& y_2’(x)\end{array}=xy_2’(x)-y_2(x)=ln|x|-1\end{align}$$
Hence, we have $$xy_2’-y_2=ln|x|-1$$
Divide both sides by $x$:
$$y_2’-{1\over x}y_2={ln|x|\over x}-{1 \over x}$$
$$\mu(x)=exp(\int-{1\over x}dx)=e^{-ln|x|}=x^{-1}$$
Multiply both sides by $x^{-1}:$ $\\$
$$\begin{align}x^{-1}y_2’-x^{-2}y_2&=x^{-2}ln|x|-x^{-2}\end{align}$$
Hence, $$(x^{-1}y_2)’=x^{-2}ln|x|-x^{-2}$$
Integrating both sides:
$$\int{(x^{-1}y_2)’}dx=\int{x^{-2}\ln|x|-x^{-2}}dx$$
$$\begin{align}x^{-1}y_2&=\int{x^{-2}\ln|x|}dx-\int{x^{-2}}dx\end{align}$$
For $\int{x^{-2}\ln|x|}dx$, we use Integral by parts: $\\$
Let $\cases{u=\ln|x|\\du={1\over x}dx}$   and $\cases{dv=x^{-2}dx\\v=-x^{-1}}$ $\\$
Thus:$$\begin{align}\int{x^{-2}\ln|x|}dx&=-x^{-1}\ln|x|-\int{-x^{-1}\cdot {1\over x}}dx\\&=-x^{-1}\ln|x|+\int{x^{-2}}dx \\&= -x^{-1}\ln|x|-x^{-1}+c\end{align}$$
Therefore,$$\begin{align}x^{-1}y_2&=\int{x^{-2}\ln|x|}dx-\int{x^{-2}}dx\\&=-x^{-1}\ln|x|-x^{-1}+c-\int{x^{-2}}dx\\&=-x^{-1}\ln|x|-x^{-1}+c+x^{-1}\\&=-x^{-1}\ln|x|+c\end{align}$$
Hence,$$y_2=-\ln|x|+cx$$
Let $c=1$, we have $$y_2=-\ln|x|+x$$
Since we already know that $W(y_1,y_2)(x)=\ln|x|-1\neq 0$,
Hence $y_2(x)$ is indeed another linearly independent solution. $\\$
$(c)$ $\\$
For $W(y_1,y_2)(x)=\ln|x|-1\neq 0$, we have general solution $$y=c_1y_1+c_2y_2=c_1x+c_2(-\ln|x|+x)$$ where $c_1$ and $c_2$ are some arbitrary constants.
Now, $$y’=c_1-{c_2\over x}+c_2$$
Let $x=1$ and $y=1$, $x=1$ and $y’=0$, we have
$$\cases{c_1+c_2=1\\c_1=0}\implies \cases{c_1=0\\c_2=1}$$
Therefore, the solution for the IVP is $$y(x)=-\ln|x|+x$$

70
Term Test 1 / Re: P1-Day
« on: February 15, 2018, 11:52:19 PM »
Let $$M(x,y)=2xy, N(x,y)=4x^2+4e^y+ye^y$$
Check exactness:
$${\partial{M} \over \partial{y}}=2x,  {\partial{N} \over \partial{x}}=8x$$
Since, ${\partial{M} \over \partial{y}} \neq {\partial{N} \over \partial{x}}$, we need to find an integrating factor $\mu(x,y)$. $\\$
Now check if $\mu(x,y)$ if depends on only $x$, or depends on only $y$, or depends on only $xy$.
$${N_x-M_y\over M}={8x-2x\over 2xy}={3 \over y}$$
Thus the integrating factor $\mu(x,y)$ is depends on only $y$ which is $\mu(y)={ 3\over y}$. $\\$
Hence $${d\mu\over dy}={3\over y}\mu \implies \mu'-{3\over y}\mu=0$$
Integrating factor $\mu_1(y)=exp^{\int{p(y)}dy}=e^{3ln|y|}=y^3$ $\\$
Now we multiply $\mu_1(y)=y^3$ to both sides of the equation:
$$\begin{align}y^3\cdot 2xy+y^3\cdot(4x^2+4e^y+ye^y)&=0 \\ 2xy^4+(4x^2y^3+4y^3e^y+y^4e^y)y' &=0\end{align}$$
Now the differential equation becomes Exact:$\\$
So there is a function $\psi(x,y)$ such that:
$$\begin{align}\psi_x(x,y)&=2xy^4 \\ \psi_y(x,y)&=4x^2 y^3+4y^3e^y+y^4e^y\end{align}$$
We integrate $\psi_x(x,y)$ with respect to $x$:
$$\psi(x,y)=x^2y^4+h(y)$$
Hence,$$\begin{align}\psi_y(x,y)&=4x^2y^3+h'(y)\\&=4x^2y^3+4y^3e^y+y^4e^y\end{align}$$ and
$$h'(y)=4y^3e^y+y^4e^y \implies h(y)=y^4 e^y$$
Therefore, $$\psi(x,y)=x^2y^4+y^4 e^y$$ and the general solution of this ODE is $$x^2y^4+y^4 e^y=c$$
Now let $x=1$ and $y=1$:
$$(1)^2 (1)^4+(1)^4 e^1=c$$
Therefore, the solution of the IVP is $$x^2y^4+y^4 e^y=e+1$$

71
Term Test 1 / Re: P4-Day
« on: February 15, 2018, 10:35:21 PM »
$(a)$ $\\$

First find the complementary solution for the homogeneous equation:
$$y’’-4y’+5y=0$$
Characteristic equation: $$r^2-4r+5=0 \implies \cases{r_1=2+i\\r_2=2-i}$$
Thus $$y_c(t)=c_1e^{2t}cos(t)+c_2e^{2t}sin(t)$$
Now we need to find the particular solution:
$$y_p(t)=Y_1(t)+Y_2(t)$$
We assume $Y_1(t)=Ae^t$ and there are no duplicates of $y_c(t)$,
$\\$thus $Y_1’(t)=Ae^t$ and $Y_1’’(t)=Ae^t$. $\\$
Substitue theses values back to $y’’-4y’+5y=2e^t$:
$$Ae^t-4Ae^t+5Ae^t=2e^t \implies A=1$$
We assume $Y_2(t)=Bcos(t)+Csin(t)$ and there are no duplicates of $y_c(t)$, $\\$thus $Y_2’(t)=-Bsin(t)+Ccos(t)$ and $Y_2’’(t)=-Bcost(t)-Csin(t)$. $\\$
Substitute these values back to $y’’-4y’+5y=8cos(t)$:
$$-Bcost(t)-Csin(t)+4Bsin(t)-4Ccos(t)+5Bcos(t)+5Csin(t)=8cos(t) \implies \cases{B=1\\C=-1}$$
Thus, $$y_p(t)=Y_1(t)+Y_2(t)=e^t+cos(t)-sin(t)$$
Therefore, the general solution is $$\begin{align}y(t)&=y_c(t)+y_p(t)\\&=c_1e^{2t}cos(t)+c_2e^{2t}sin(t)+e^t+cos(t)-sin(t)\end{align}$$

$(b)$ $\\$
$$y’(t)=2c_1e^{2t}cos(t)-c_1e^{2t}sin(t)+2c_2e^{2t}sin(t)+c_2e^{2t}cos(t)+e^t-sin(t)-cos(t)$$
Set $t=0$ and $y=0$; $t=0$ and $y’(t)=0$:
$$\cases{c_1+0+1+1-0=0\\2c_1-0+0+c_2+1-0-1=0} \implies \cases{c_1=-2\\c_2=4}$$
Therefore, the solution for the IVP is $$y(t)=-2e^{2t}cos(t)+4e^{2t}sin(t)+e^t+cos(t)-sin(t)$$

72
Term Test 1 / Re: P3-Day
« on: February 15, 2018, 09:58:07 PM »
$(a)$ $\\$
First we find the solution to the homogeneous solution for $$y’’(t)-3y’(t)+2y(t)=0$$
Characteristic equation: $$r^2-3r+2$$
Hence, $$\cases{r_1=1\\r_2=2}$$ and the complementary solution for the homogeneous equation is $$y_c(t)=c_1e^t+c_2e^{2t}$$
Now we need to find the particular solution for the non-homogeneous equation
$$y’’(t)-3y’(t)+2y(t)=-6+3e^t$$
Let $y_p(t)$ be the particular solution and $y_p(t)=Y_1(t)+Y_2(t)$.
We assume$Y_1(t)=At+B$, and there are no duplicates of the solution of the homogeneous equation, thus $Y_1’(t)=A$ and $Y_1’’(t)=0$. $\\$
And substitute these back to $y’’(t)-3y’(t)+2y(t)=-6$:
$$0-3A+2(At+B)=-6 \implies \cases{A=0\\B=-3}$$
We assume $Y_2(t)=Ce^t$, and noting there are duplicates of the solution of the homogeneous equation. $\\$ So we need to multiply $t$ for the propose of $Y_2(t)$, thus $Y_2(t)=Cte^t$ which leads to no duplicates. $\\$
Thus $Y_2’(t)=Ce^t+Cte^t$ and $Y_2’’(t)=2Ce^t+Cte^t$.
And substitute these values back to $y’’(t)-3y’(t)+2y(t)=3e^t$:
$$2Ce^t+Cte^t-3Ce^t-3Cte^t+2Cte^t=3e^t \implies C=-3$$
Hence, the particular solution $$y_p(t)=Y_1(t)+Y_2(t)=-3-3e^t$$
Therefore, the general solution is $$y(t)=y_c(t)+y_p(t)=c_1e^t+c_2e^{2t}-3-3te^t$$

$(b)$ $\\$
$$y’(t)=c_1e^t+2c_2e^{2t}-3e^t-3te^t$$
Set $t=0$ and $y=0$; $t=0$ and $y’=0$:
$$\cases{c_1+c_2=3\\c_1+2c_2=3} \implies \cases{c_1=3\\c_2=0}$$
Therefore, the solution for the IVP is $$y(t)=3e^t-3te^t-3$$

73
Term Test 1 / Re: P2-Day
« on: February 15, 2018, 09:16:18 PM »
$(a)$ $\\$
First we divide both sides by $x^2-1$:
$$y’’-{2x\over x^2-1}y’+{2 \over x^2-1}y=0$$
Let $p(t)=-{2x\over x^2-1}$ and $q(t)={2 \over x^2-1}.$ $\\$
Noting $p(t)$ and $q(t)$ are continuous everywhere except when $x=1$ or $x=-1.$
By Abel’s Theorem:$$\begin{align}W(y_1,y_2)(x)&=cexp(-\int{p(x)dx})\\&=cexp(-\int{-{2x\over x^2-1}}dx)\\&=ce^{ln|x^2-1|}\\&=c(x^2-1)\end{align}$$
Let $c=1$, then $W(y_1,y_2)(x)=x^2-1$. $\\$
$(b)$ $\\$
Since $y_1(x)=x,$ then
$$\cases{y_1’(x)=1\\y_1’’(x)=0}$$
Substitute these values back to the ODE:
$$\begin{align}0-{2x\over x^2-1}\cdot 1+{2 \over x^2-1}\cdot x=0\end{align}$$
$$-{2x\over x^2-1}+{2x\over x^2-1}=0$$
Thus $y_1=(x)$ is indeed a solution. $\\$
Now we need to find $y_2(x)$ $\\$
Since we already know: $$\begin{align}W(y_1,y_2)(x)&=\begin{array}{|c c|}y_1(x)& y_2(x) \\ y_1’(x) & y_2’(x)\end{array}=x^2-1\\&=\begin{array}{|c c|}x& y_2(x) \\ 1& y_2’(x)\end{array}=xy_2’(x)-y_2(x)=x^2-1\end{align}$$
Hence, we have $$xy_2’-y_2=x^2-1$$
Divide both sides by $x$:
$$y_2’-{1\over x}y_2=x-{1 \over x}$$
$$\mu(x)=exp(\int-{1\over x}dx)=e^{-ln|x|}=x^{-1}$$
Multiply both sides by $x^{-1}:$ $\\$
$$\begin{align}x^{-1}y_2’-x^{-2}y_2&=1-x^{-2}\end{align}$$
Hence, $$(x^{-1}y_2)’=1-x^{-2}$$
Integrating both sides:
$$\int{(x^{-1}y_2)’}dx=\int{1-x^{-2}}dx$$
$$\begin{align}x^{-1}y_2&=\int{1dx}-\int{x^{-2}}dx\\&=x+x^{-1}+c\end{align}$$
Hence,$$y_2=x^2+1+cx$$
Let $c=1$, we have $$y_2=x^2+x+1$$
Since we already know that $W(y_1,y_2)(x)=x^2-1\neq 0$,
Hence $y_2(x)$ is indeed another linearly independent solution. $\\$
$(c)$ $\\$
For $W(y_1,y_2)(x)=x^2-1\neq 0$, we have general solution $$y=c_1y_1+c_2y_2=c_1x+c_2(x^2+x+1)$$ where $c_1$ and $c_2$ are some arbitrary constants.
Now, $$y’=c_1+2c_2x+c_2$$
Let $x=0$ and $y=1$, $x=0$ and $y’=2$, we have
$$\cases{c_2=1\\c_1+c_2=2}\implies \cases{c_1=1\\c_2=1}$$
Therefore, the solution for the IVP is $$y(x)=x^2+2x+1$$

74
Term Test 1 / Re: P-2
« on: February 14, 2018, 10:22:08 AM »
Solution to Problem 2:

Prof. Victor would prefer you typing out the solutions xD ( that is if you want the bonus mark)

75
Quiz-3 / Re: Q3-T0301
« on: February 11, 2018, 10:06:57 AM »
First, we divide both sides of the equation by $cos(t)$:
$$y''+tan(t)y'-{t\over cos(t)}y=0$$
Now the given second-order differential equation has the form:
$$L[y]=y''+p(t)y'+q(t)y=0$$
Noting if we let $p(t)=tan(t)$ and $q(t)=-{t\over cos(t)}$, then $p(t)$ is continuous everywhere except at ${\pi\over 2}+k\pi$, where $k=0,1,2,\dots$ and $q(t)$ is also continuous everywhere except at $t=0$. $\\$
Therefore, by Abel's Theorem: the Wronskian $W[y_1,y_2](t)$ is given by
$$\begin{align}W[y_1,y_2](t)&= cexp(-\int{p(t)dt})\\&=cexp(-\int{tan(t)dt})\\&=ce^{ln|cos(t)|}\\&=ccos(t)\end{align}$$

Pages: 1 ... 3 4 [5] 6 7