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Messages - Meng Wu

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76
Quiz-3 / Re: Q3-T0101
« on: February 11, 2018, 09:53:26 AM »
First, we divide both sides of the equation by $t^2$:
$$y''+{t+2\over t}y'+{t+2\over t^2}y=0$$
Now the given second-order differential equation has the form:
$$L[y]=y''+p(t)y'+q(t)y=0$$
Noting if we let $p(t)={t+2\over t}$ and $q(t)={t+2\over t^2}$, then $p(t)$ is continuous everywhere except at $t=0$, 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
$$\begin{align}W[y_1,y_2](t)&= cexp(-\int{p(t)dt})\\&=cexp(-\int{{t+2\over t}dt})\\&=ce^{t+2ln|t|}\\&=ct^2e^t\end{align} $$

77
Quiz-3 / Re: Q3-T0201
« on: February 11, 2018, 09:34:16 AM »
$$y’’+3y’+2y=0$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$r^2+3r+2=(r+1)(r+2)=0$$
$$\cases{r_1=-1\\r_2=-2}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Therefore, the general solution of the given differential equation is
$$y=c_1e^{-t}+c_2e^{-2t}$$

78
Quiz-3 / Re: Q3-T0701
« on: February 11, 2018, 09:33:51 AM »
$$y''-2y'-2y=0$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$r^2-2r-2=0$$
We use the quadratic formula which is
$$r={-b\pm \sqrt{b^2-4ac}\over 2a}$$
Hence,
$$\cases{r_1={1+\sqrt{3}}\\r_2=1-\sqrt{3}}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Therefore, the general solution of the given differential equation is
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$

79
Quiz-3 / Re: Q3-T5101
« on: February 11, 2018, 09:32:26 AM »
$$y''-2y'-2y=0$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$r^2-2r-2=0$$
We use the quadratic formula which is
$$r={-b\pm \sqrt{b^2-4ac}\over 2a}$$
Hence,
$$\cases{r_1={1+\sqrt{3}}\\r_2=1-\sqrt{3}}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Therefore, the general solution of the given differential equation is
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$

80
Quiz-3 / Re: Q3-T0401
« on: February 11, 2018, 09:30:55 AM »
$$y''+4y'+3y=0,y(0)=2,y'(0)=-1$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$r^2+4r+3=(r+1)(r+3)=0$$
Hence,
$$\cases{r_1=-1\\r_2=-3}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Then the general solution of the given differential equation is
$$y=c_1e^{-t}+c_2e^{-3t}$$
We also need $y'$ for the IVP,
$$y'=-c_1e^{-t}-3c_2e^{-3t}$$
To satisfy the first initial condition, we set $t=0$ and $y=2$, thus
$$c_1+c_2=2$$
To satisfy the second initial condition, set $t=0$ and $y'=-1$, thus
$$-c_1-3c_2=-1$$
Hence,
$$\cases{c_1+c_2=2\\-c_1-3c_2=-1} \implies \cases{c_1={5\over 2}\\c_2=-{1\over 2}}$$
Therefore, the solution of the initial value problem is
$$y={5\over 2}e^{-t}-{1\over 2}e^{-3t}$$
Note: $y \rightarrow 0$ as $t \rightarrow \infty$.

81
Quiz-3 / Re: Q3-T0801
« on: February 11, 2018, 09:30:21 AM »
$$2y''+y'-4y=0,y(0)=0,y'(0)=1$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$2r^2+r-4=0$$
We use the quadratic formula which is
$$r={-b\pm \sqrt{b^2-4ac}\over 2a}$$
Hence,
$$\cases{r_1={-1+\sqrt{33}\over 4}\\r_2={-1-\sqrt{33}\over 4}}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Then the general solution of the given differential equation is
$$y=c_1e^{{-1+\sqrt{33}\over 4}t}+c_2e^{{-1-\sqrt{33}\over 4}t}$$
We also need $y'$ for the IVP,
$$y'={-1+\sqrt{33}\over 4}c_1e^{{-1+\sqrt{33}\over 4}t}+{-1-\sqrt{33}\over 4}c_2e^{{-1-\sqrt{33}\over 4}t}$$
To satisfy the first initial condition, we set $t=0$ and $y=0$, thus
$$c_1+c_2=0$$
To satisfy the second initial condition, set $t=0$ and $y'=1$, thus
$${-1+\sqrt{33}\over 4}c_1+{-1-\sqrt{33}\over 4}c_2=1$$
Hence,
$$\cases{c_1+c_2=0\\{-1+\sqrt{33}\over 4}c_1+{-1-\sqrt{33}\over 4}c_2=1} \implies \cases{c_1={2\over \sqrt{33}}\\c_2=-{2\over \sqrt{33}}}$$
Therefore, the solution of the initial value problem is
$$y={2\over \sqrt{33}}e^{{-1+\sqrt{33}\over 4}t}-{2\over \sqrt{33}}e^{{-1-\sqrt{33}\over 4}t}$$
Note: $y \rightarrow \infty$ as $t \rightarrow \infty$.

82
Quiz-3 / Re: Q3-T0501
« on: February 11, 2018, 09:29:22 AM »
$$2y’’-3y’+y=0$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$2r^2-3r+1=(2r-1)(r-1)=0$$
Hence,
$$\cases{r_1={1\over2}\\r_2=1}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Therefore, the general solution of the given differential equation is
$$y=c_1e^{{1\over2}t}+c_2e^{t}$$

83
Quiz-1 / Re: TUT 0601
« on: January 26, 2018, 03:07:47 PM »
$${dy\over dx}=-{4x+3y\over 2x+y}$$
If we divide right side of the given equation by $x$, we get:
$${dy\over dx}=-{4+3{y\over x} \over 2+{y\over x}}$$
Hence, given equation is a Homogeneous.$\\$
Let $v={y\over x}$, $v$ is a function of $x$.$\\$
Thus, $${dy\over dx}=-{4+3v\over 2+v}$$
Since $v={y\over x} \rightarrow y=vx$   $\\$
By differentiating both sides with respect to $x$, we get:
$${dy\over dx}=v+x{dv\over dx}=-{4+3v\over 2+v}$$
$$x{dv\over dx}={-(4+3v)\over 2+v}-v$$
$$x{dv\over dx}={-(4+3v)\over 2+v}-{v(2+v)\over 2+v}$$
$$x{dv\over dx}={-(v^2+5v+4)\over 2+v}$$
$${dv\over dx}={-(v+1)(v+4)\over 2+v}\cdot {1\over x}$$
$${2+v\over -(v+1)(v+4)}\cdot {dv\over dx}={1\over x}$$
$$-{1\over x}+{2+v\over -(v+1)(v+4)}\cdot {dv\over dx}=0$$
$${1\over x}+{2+v\over (v+1)(v+4)}\cdot {dv\over dx}=0$$
Notice that above equation has the form of $M(x)+N(y){dy\over dx}=0$ $\\$
Hence the equation is Seperable. $\\$
Rewrite the equation, we get:
$${2+v\over (v+1)(v+4)}dv=-{1\over x}dx$$
Integrating both sides, we get:
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{-{1\over x}dx}$$
For $\int{{2+v\over (v+1)(v+4)}dv}$, we use Partial Fraction:
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{({A\over v+1}+{B\over v+4})dv}$$
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{A(v+4)+B(v+1)\over (v+1)(v+4)}$$
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{(A+B)v+(4A+B)\over (v+1)(v+4)}$$
Hence, $$\begin{cases} A+B=1 \\4A+B=2\\ \end{cases} \implies
\begin{cases} A={1\over3}\\ B={2\over3} \end{cases}$$
Thus, rewrite gets us:
$$\int{{2+v\over (v+1)(v+4)}dv}=\int{{1\over 3(v+1)}+{2\over 3(v+4)}dv}=-\int{1\over x}dx$$
$${1\over 3}ln|v+1|+{2\over3}ln|v+4|=-ln|x|+c$$
where c is arbitrary constant. $\\$
Multiply both sides by $3$, we get:
$$ln|v+1|+ln|v+4|^2=-3ln|x|+3c$$
Now substitute $v={y\over x}$ back in:
$$ln|{y\over x}+1|+ln|{y\over x}+4|^2=-3ln|x|+3c$$
$$ln|{y+x\over x}|+ln|{y+4x\over x}|^2=-3ln|x|+3c$$
$$ln|y+x|-ln|x|+2ln|y+4x|-2ln|x|=-3ln|x|+3c$$
$$ln|y+x|+ln|y+4x|^2=3c$$
$$ln|y+x||y+4x|^2=3c$$
$$exp^{ln|y+x||y+4x|^2}=exp^{3c}$$
$$|y+x||y+4x|^2=c$$
where c is another arbitrary constant.
Therefore, we get the solution to the given differential equation.

84
Quiz-1 / Re: Q1-T0401
« on: January 26, 2018, 09:19:01 AM »
$$\cases{
t^3y'+4t^2y=e^{-t}, & t<0\cr
y(-1)=0}$$
First, we divide both sides of given equation by $t^3$, we get:
$$y'+{4\over t}y={e^{-t}\over t^3}$$
Since given differential equation has the form
$$y'+p(t)y=g(t)$$
Hence $p(t)={4\over t}$ and $g(t)={e^{-t}\over t^3}$$\\$
First, we find the integrating factor $\mu(t)$ $\\$
As we know. $\mu(t)=\exp^{\int{p(t)dt}}$ $\\$
Thus, $\mu(t)=\exp^{\int{{4\over t}dt}}=e^{4ln|t|}=t^4$$\\$
Then mutiply $\mu(t)$ to both sides of the equation, we get:
$$t^4y'+4t^3y=te^{-t}$$
and $$(t^4y)'=te^{-t}$$
Integrating both sides:
$$\int{(t^4y)'}=\int{te^{-t}}$$
Thus, $$t^4y=\int{te^{-t}}$$
For $\int{te^{-t}}$, we use Integration By Parts:$\\$
Let $u=t, dv=e^{-t}$.$\\$
Then $du=dt, v=-e^{-t}$$\\$
Hence, $$\int{te^{-t}}=uv-\int{vdu}$$
$$\int{te^{-t}}=-te^{-t}-\int{-e^{-t}dt}$$
$$\int{te^{-t}}=-te^{-t}-e^{-t}+c$$
Thus $$t^4y=-te^{-t}-e^{-t}+c$$
where $c$ is arbitrary constant.$\\$
Now we divide both sides by $t^4$, we get the general solution:
$$y=(-te^{-t}-e^{-t}+c)/t^4$$
To satisfy the initial condition, we set $t=-1$ and $y=0$$\\$
Hence, $$0={-(-1)e^{-(-1)}-e^{-(-1)}+c\over (-1)^4}$$
$$0=e-e+c$$
so $$c=0$$
Therefore, the solution of the initial problem is
$$y=(-te^{-t}-e^{-t})/t^4, \space t<0$$

85
Quiz-1 / Re: TUT 0501
« on: January 25, 2018, 06:22:29 PM »
$$ty'+2y=sint$$
First, we divide both sides of the given equation by $t$, we get: $\\$
$$y'+{2\over t}y={sint\over t}$$
Now the differential equation has the form
$$y'+p(t)y=g(t)$$
Hence $p(t)={2\over t}$ and $g(t)={sint\over t}$$\\$
First, we find the integrating factor $\mu(t)$ $\\$
As we know. $\mu(t)=\exp^{\int{p(t)dt}}$ $\\$
Thus, $\mu(t)=\exp^{\int{{2\over t}dt}}=e^{2ln|t|}=e^{ln|t|}\cdot e^{ln|t|}=t\cdot t=t^{2}$$\\$
Then mutiply $\mu(t)$ to both sides of the equation, we get:
$$t^2y'+2ty=tsint$$
and $$(t^2y)'=tsint$$
Integrating both sides:
$$\int{(t^2y)'}=\int{tsint}$$
Thus, $$t^2y=\int{tsint}$$
For $\int{tsin(t)}$, we use Integration By Parts:$\\$
Let $u=t, dv=sint$.$\\$
Then $du=dt, v=-cost$$\\$
Hence, $$\int{tsint}=uv-\int{vdu}$$
$$\int{tsint}=-tcost-\int{-costdt}$$
$$\int{tsint}=-tcost+\int{costdt}$$
$$\int{tsint}=-tcost+sint+c$$
Thus $$t^2y=-tcost+sint+c$$
where $c$ is arbitrary constant.$\\$
Now we divide both sides by $t^2$, we get the general solution:
$$y={(sint-tcost+c)/t^2}$$
Since given $t>0$, $y\rightarrow 0$ as $t \rightarrow \infty$.

86
Quiz-1 / Re: Q1-TUT0501
« on: January 25, 2018, 06:22:17 PM »
$$ty'+2y=sint$$
First, we divide both sides of the given equation by $t$, we get: $\\$
$$y'+{2\over t}y={sint\over t}$$
Now the differential equation has the form
$$y'+p(t)y=g(t)$$
Hence $p(t)={2\over t}$ and $g(t)={sint\over t}$$\\$
First, we find the integrating factor $\mu(t)$ $\\$
As we know. $\mu(t)=\exp^{\int{p(t)dt}}$ $\\$
Thus, $\mu(t)=\exp^{\int{{2\over t}dt}}=e^{2ln|t|}=e^{ln|t|}\cdot e^{ln|t|}=t\cdot t=t^{2}$$\\$
Then mutiply $\mu(t)$ to both sides of the equation, we get:
$$t^2y'+2ty=tsint$$
and $$(t^2y)'=tsint$$
Integrating both sides:
$$\int{(t^2y)'}=\int{tsint}$$
Thus, $$t^2y=\int{tsint}$$
For $\int{tsin(t)}$, we use Integration By Parts:$\\$
Let $u=t, dv=sint$.$\\$
Then $du=dt, v=-cost$$\\$
Hence, $$\int{tsint}=uv-\int{vdu}$$
$$\int{tsint}=-tcost-\int{-costdt}$$
$$\int{tsint}=-tcost+\int{costdt}$$
$$\int{tsint}=-tcost+sint+c$$
Thus $$t^2y=-tcost+sint+c$$
where $c$ is arbitrary constant.$\\$
Now we divide both sides by $t^2$, we get the general solution:
$$y={(sint-tcost+c)/t^2}$$
Since given $t>0$, $y\rightarrow 0$ as $t \rightarrow \infty$.

87
Quiz-1 / Re: Q1-T0201
« on: January 25, 2018, 11:19:50 AM »
Since given differential equation has the form
$$y'+p(t)y=g(t)$$
Hence $p(t)=-1$ and $g(t)=2te^{2t}$$\\$
First, we find the integrating factor $\mu(t)$ $\\$
As we know. $\mu(t)=\exp^{\int{p(t)dt}}$ $\\$
Thus, $\mu(t)=\exp^{\int{-1dt}}=e^{-t}$$\\$
Then mutiply $\mu(t)$ to both sides of the equation, we get:
$$e^{-t}y'-e^{-t}y=e^{-t}\cdot 2te^{2t}=2te^{t}$$
and $(e^{-t}y)'=2te^{t}$ $\\$
Integrating both sides:
$$\int{(e^{-t}y)'}=\int{2te^{t}}$$
Thus, $$e^{-t}y=\int{2te^{t}}$$
For $\int{2te^{t}}$, we use Integration By Parts:$\\$
Let $u=2t, dv=e^{t}$.$\\$
Then $du=2dt, v=e^{t}$$\\$
Hence, $$\int{2te^{t}}=uv-\int{vdu}$$
$$\int{2te^{t}}=2te^{t}-\int{2e^{t}dt}$$
$$\int{2te^{t}}=2te^{t}-2e^{t}+c$$
Thus $$e^{-t}y=2te^{t}-2e^{t}+c$$
where $c$ is arbitrary constant.$\\$
Now we divide both sides by $e^{-t}$, we get the general solution:
$$y=2te^{2t}-2e^{2t}+ce^{t}$$
To satisfy the initial condition, we set $t=0$ and $y=1$$\\$
Hence, $$1=2(0)e^{2\cdot 0}-2e^{2\cdot 0}+ce^{0}$$
$$1=0-2+c$$
so $$c=3$$
Therefore, the solution of the initial problem is
$$y=2te^{2t}-2e^{2t}+3e^{t}$$

88
Quiz-1 / Re: Q1-T0201
« on: January 25, 2018, 10:03:38 AM »

This answer can not be improved anymore I think :O

89
Quiz-1 / Re: Q1-T0101
« on: January 25, 2018, 09:01:59 AM »
$$y'+3y=t+e^{-2t}$$
Since given differential equation has the form $$y'+p(t)y=g(t)$$
Hence $p(t)=3$ and $g(t)=t+e^{-2t}$ $\\$
First, we find the integrating factor $\mu(t)$$\\$
As we know, $\mu(t)=\exp^{\int p(t)dt}$ $\\$
$\mu(t)=\exp^{\int 3dt}=e^{3t}$$\\$
Mulitply $\mu(t)$ to both sides of the equation, we get:$\\$
$e^{3t}y'+3e^{3t}y=te^{3t}+e^{-2t}\cdot e^{3t}=te^{3t}+te^{t}$ $\\$
and $(e^{3t}y)'=te^{3t}+te^{t}$ $\\$
Integrating both sides: $\\$
$$e^{3t}y=\int{te^{3t}+te^{t}}$$ $\\$
which is also:
$$e^{3t}y=\int{te^{3t}}+\int{te^{t}}$$ $\\$
Now, $\int{te^{t}}=e^{t}+c$, where $c$ is arbitrary constant $\\$
For $\int{te^{3t}}$ we use Integration By Parts: $\\$ Let $u=t, dv=e%{3t}$. $\\$Then we get: $du=dt$ and $v={1\over3}e^{3t}$. $\\$
Thus, $$\int{te^{3t}}=uv-\int{vdu}$$
$$\int{te^{3t}}={t\over3}e^{3t}-\int{{1\over3}e^{3t}dt}$$
$$\int{te^{3t}}={t\over3}e^{3t}-{1\over9}e^{3t}$$ $\\$
Thus, $$e^{3t}y={t\over3}e^{3t}-{1\over9}e^{3t}+e^{t}+c$$ where $c$ is arbitrary constant. $\\$
Now divide both side by $e^{3t}$, we get the general solution: $$y={t\over3}-{1\over9}+e^{-2t}+ce^{-3t}$$ $\\$
Note: $y$ is asymptotic to ${t\over3}-{1\over9}$ as $t\rightarrow \infty$.

90
MAT244--Misc / Re: Week 4 Quiz
« on: January 23, 2018, 10:30:56 PM »
Quiz 1:
Drawn from assignments for Week 1--2; during Tutorials
http://www.math.toronto.edu/courses/mat244h1/20181/homeassignments.html

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