Messages

1

MAT244--Lectures & Home Assignments / Real Repeated Eigenvalue

« on: December 11, 2018, 03:37:44 PM »

Has anyone encountered an example in which a matrix $A$ has two independent eigenvectors with eigenvalue $\lambda$, and the phase portrait would therefore be an unstable or stable proper node?

If so, please share! If it's in the textbook, a page number is fine!

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MAT244--Lectures & Home Assignments / Re: Term Test 2 Question 1 Solve Integral

« on: December 05, 2018, 09:30:19 PM »

So I used variation of parameter and concluded that the particular solution follows:
$$y_p = 2\cos (2t) \int{\tan (t) \sin (2t) dt} + 2 \sin (2t) \int{\tan(t) \cos(2t) dt}$$
I am not sure how to solve the two integrals $\int{\tan (t) \sin (2t) dt}$ and $\int{\tan(t) \cos(2t) dt}$ since it involves the $2t$ term inside the sine and cosine function. Can anyone help me solve this?

First, recall the double-angle trig identity, $\sin(2t)=2\sin(t)\cos(t)$. Thus, the integrand $\tan(t)\sin(2t)=\tan(t)(2\sin(t)\cos(t))=2\sin^2(t)$.


The half-angle trig identity allows us to rewrite $2\sin^2(t)$ as $1-\cos(2t)$.

Now, we have
$$\int{1}dt-\int{\cos(2t)}dt$$

Let $u=2t \Rightarrow du=2dt$
$$\int{1}dt-\int{\cos(2t)}dt=t-\frac{1}{2}\int{\cos(u)}du=t-\frac{1}{2}\sin(u)=t-\frac{1}{2}\sin(2t)=t-\sin(t)\cos(t)$$

The process is similar for $\int{\tan(t)\cos(2t)}dt$. Hope that helps!

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MAT244--Lectures & Home Assignments / Re: integral

« on: November 06, 2018, 11:04:01 PM »

$$-\frac{e^t}{2}\int{e^{-t}\tanh{(t)}} dt$$

For the integrand ${e^{-t}\tanh{t}}$, write $\tanh{t}$ as $\frac{e^t-e^{-t}}{e^{-t}+e^t}$.

(Note that if this doesn't look familiar, you should review hyperbolic sine and cosine)

$$-\frac{e^t}{2}\int{e^{-t}\tanh{(t)}}dt=-\frac{e^t}{2}\int{\frac{e^{-t}(e^t-e^{-t})}{e^{-t}+e^t}}dt$$

Expanding the integrand gives

$$=-\frac{e^t}{2}\int{\left(\frac{e^t}{e^{2t}+1}-\frac{e^{-t}}{e^{2t}+1}\right)}dt=-\frac{e^t}{2}\int{\left(\frac{e^t}{e^{2t}+1}-\frac{e^{-t}}{e^{2t}+1}\right)}dt$$

For the integrand $\left(\frac{e^t}{e^{2t}+1}\right)$, substitute $u=e^x$ and $du=e^x dx$, and for the integrand $\left(\frac{e^{-t}}{e^{2t}+1}\right)$, substitute $v=e^x$ and $dv=e^x dx$.

Try proceeding this way. If you need help beyond this point, just lmk

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Quiz-5 / Re: Q5 TUT 0601

« on: November 02, 2018, 04:23:29 PM »

I think c1 should be -2 and c2 should be 2.

ah yes, this is a typo. thanks

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MAT244--Lectures & Home Assignments / Re: Section 7.1 Q 11

« on: November 02, 2018, 04:18:05 PM »

In this problem, $x_1=3\cos{2t}+4\sin{2t}$ and $x_2=-\sin{2t}+4\cos{2t}$.

$${x_1}^2+{x_2}^2=(3\cos{2t}+4\sin{2t})^2+(-\sin{2t}+4\cos{2t})^2=9(\cos^2{2t}+\sin^2{2t})+16(\cos^2{2t}+\sin^2{2t})=25$$

By definition, this is the equation of a circle centered at the origin with radius 5.

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Quiz-5 / Re: Q5 TUT 0601

« on: November 02, 2018, 04:04:30 PM »

Transform the given system into a single equation of second order and find the solution $(x_1(t), x_2(t))$, satisfying initial conditions

$$\left\{\begin{aligned} & x'_1= -0.5x_1 + 2x_2, &&x_1(0) = -2,\\ &x'_2= -2x_1 - 0.5x_2, &&x_2(0) = 2 \end{aligned}\right.$$


a. Transform the given system into a single equation of second order.

Solving the first equation for $x_2$, we have
$$x_2=\frac{1}{2}x’_1+\frac{1}{4}x_1\tag{1}$$

Plugging $x_2$ into the second equation gives

$$\left(\frac{1}{2}x’_1+\frac{1}{4}x_1\right)’=-2x_1-\frac{1}{2}\left(\frac{1}{2}x’_1+\frac{1}{4}x_1\tag{1}\right)$$
$$x''_1+x'_1+\frac{17}{4}x_1=0$$

b. Find $x_1$ and $x_2$ that also satisfy the given initial conditions.

The characteristic equation is
$$r^2+r+\frac{17}{4}=0$$
$$r_1,2=\frac{1}{2}\pm 2i.$$

The general solution is
$$x_1(t)=e^{-\frac{1}{2}t}\left(c_1\cos{2t}+c_2\sin{2t}\right).$$

It follows that the solution for $x_2$ is

$$x_2(t)=-\frac{1}{2}e^{-\frac{1}{2}t}(c_1\left(-\frac{1}{4}\cos{2t}-2\sin{2t}\right)+c_2\left(2\cos{2t}-\frac{1}{4}\sin{2t}\right))$$

Satisfying the given initial conditions  $x_1(0)=-2$ and $x_2(0)=2$, we have

$$\left\{\begin{aligned} &c_1=-2\\ &c_2=2 \end{aligned}\right.$$

The solutions that satisfy the given initial conditions are

$$x_1(t)=e^{-\frac{1}{2}t}(2\sin{2t}-2\cos{2t})$$
$$x_2(t)=e^{-\frac{1}{2}t}(2\cos{2t}+2\sin{2t}).$$

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Quiz-5 / Re: Q5 TUT 0601

« on: November 02, 2018, 03:40:40 PM »

In the attachment.

Fyi, your posts are all missing attachments!

8

Quiz-5 / Re: Q5 TUT 5101

« on: November 02, 2018, 03:36:29 PM »

Transform the given system into a single equation of second order and find the solution (x_1(t), x_2(t)), satisfying initial conditions
$$\left\{\begin{aligned} &x'_1= 2x_2, &&x_1(0) = 3,\\ &x'_2= -2x_1, &&x_2(0) = 4. \end{aligned}\right.$$


a. Transform the given system into a single equation of second order.

Solving the first equation for $x_2$, we have
$$x_2=\frac{1}{2}x'_1$$

Plugging $x_2$ into the second equation gives

$$\left(\frac{1}{2}x’_1\right)’=-2x_1$$
$$\frac{1}{2}x’’_1+2x’_1=0$$
$$x’’_1+4x’_1=0$$

b. Find $x_1$ and $x_2$ that also satisfy the given initial conditions.

The characteristic equation is
$$\frac{1}{2}r^2+2=0$$
$$r^2+4=0$$
$$ r_1,2=\pm 2i$$

The general solution is
$$x_1(t)=c_1\cos{2t}+c_2\sin{2t}.$$

It follows that the solution for $x_2$ is

$$x_2(t)=\frac{1}{2}(-2c_1\sin{2t}+2c_2\cos{2t})=-c_1\sin{2t}+c_2 \cos{2t}$$

Satisfying the given initial conditions  $x_1(0)=3$ and $x_2(0)=4$, we have

$$\left\{\begin{aligned} &c_1=3\\ &c_2=4\end{aligned}\right.$$


The solutions that satisfy the given initial conditions are

$$x_1(t)=3\cos{2t}+4\sin{2t}$$
$$x_2(t)=-3\sin{2t}+4\cos{2t}.$$

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Quiz-5 / Re: Q5 TUT 0101

« on: November 02, 2018, 03:18:05 PM »

a. Transform the given system into a single equation of second order; and
b. Find $x_1$ and $x_2$ satisfying the initial conditions.

$$\left\{\begin{aligned} &x'_1 = 3x_1 - 2x_2, &&x_1(0) = 3,\\ &x'_2= 2x_1 - 2x_2, &&x_2(0) = \frac{1}{2}. \end{aligned}\right.$$

a. Transform the given system into a single equation of second order.

Solving the first equation for $x_2$, we have
$$x_2=-\frac{1}{2}x’_1+\frac{3}{2}x_1\tag{1}$$

Plugging $x_2$ into the second equation gives

$$\left(-\frac{1}{2}x’_1+\frac{3}{2}x_1\right)’=2x_1-2\left(-\frac{1}{2}x’_1+\frac{3}{2}x_1\right)$$
$$-\frac{1}{2}x’’_1+\frac{3}{2}x’_1=2x_1+x’_1-3x_1$$
$$x’’_1-x’_1-2x_1=0.$$

b. Find $x_1$ and $x_2$ satisfying initial conditions.

The characteristic equation is
$$r^2-r-2=0$$
$$(r+1)(r-2)=0 \Rightarrow r_1=-1, r_2=2.$$

The general solution is
$$x_1(t)=c_1e^{-t}+c_2e^{2t}.$$

It follows that the solution for $x_2$ is

$$x_2(t)=-\frac{1}{2}(-c_1e^{-t}+2c_2e^{2t})+\frac{3}{2}(c_1e^{-t}+c_2e^{2t})=2c_1e^{-t}+\frac{1}{2}c_2e^{2t}$$

Satisfying the given initial conditions  $x_1(0)=3$ and $x_2(0)=\frac{1}{2}$, we obtain the following system

$$\left\{\begin{aligned} &c_1+c_2=3\\ &2c_1+\frac{1}{2}c_2=\frac{1}{2} \end{aligned}\right.$$

Solving this system, we get

$$\cases{c_1=-\frac{2}{3}\\c_2=\frac{11}{3}}$$

The solutions that satisfy the given initial conditions are

$$x_1(t)=-\frac{2}{3}e^{-t}+\frac{11}{3}e^{2t}$$
$$x_2(t)=-\frac{4}{3}e^{-t}+\frac{11}{6}e^{2t}.$$

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MAT244--Lectures & Home Assignments / Re: sketch a graph

« on: November 02, 2018, 10:17:47 AM »

For assigned exercises 8-12 in 7.1, the solutions are equations in parametric form, $x_1(t)$ and $x_2(t)$.

Each value of $t$ gives a point $(x_1(t), x_2(t))$ on the curve in the $(x_1,x_2)$-plane.

One way to sketch the graph of the solution in the $(x_1,x_2)$-plane for $t \ge0$ is to evaluate $x_1(t)$ and $x_2(t)$ at different values of $t \ge 0$, where $x_1$ is the horizontal axis and $x_2$ is the vertical axis.

Not sure if there is a more efficient way of doing this.

11

MAT244--Lectures & Home Assignments / Re: section 4.4

« on: November 02, 2018, 08:46:07 AM »

The method of undetermined coefficients will only work for a fairly small class of functions. More specifically, the inhomogeneous term (the right side) must be an exponential, polynomial, sine or cosine, or a combination of these. Also, it is generally only useful for constant coefficient DEs.

The method of variation of parameters is a more general method that can be used for many more cases, although it is a messier process. (I don't want to mislead you by saying it works in all cases, just in case there is some exception outside the scope of this course)

12

Quiz-4 / Re: Q4 TUT 0801

« on: October 26, 2018, 07:34:19 PM »

Find the general solution of the given differential equation.

$$4y'' + y = 2 \sec\left(\frac{t}{2}\right)\tag{1},\qquad -\pi  < t < \pi.$$

Dividing $(1)$ by $4$, we have

$$y''+\frac{1}{4}y=\frac{1}{2}\sec{\frac{t}{2}},$$


where $p(t)=0$, $q(t)=\frac{1}{4}$, and $g(t)=\frac{1}{2}\sec{\frac{t}{2}}$ are continuous on $(-\pi, \pi)$.

The corresponding homogeneous equation is
$$4y''+y=0,$$

with characteristic equation
$$4r^2+1=0 \Rightarrow r_{1,2}=\pm\frac{i}{2}.$$

The homogeneous solution is $y_h(t)=c_1\cos\frac{t}{2}+c_2\sin\frac{t}{2}$


Computing the Wronskian,


$$W\left(\cos{\frac{t}{2}}, \sin{\frac{t}{2}}\right)(t)=\begin{array}{|c c|}\cos{\frac{t}{2}}& \sin{\frac{t}{2}} \\ -\frac{1}{2}\sin{\frac{t}{2}} &\frac{1}{2}\cos{\frac{t}{2}}\end{array}=\frac{1}{2}\ne0,$$

we verify that $y_1(t)=\cos{\frac{t}{2}}$ and $y_2(t)=\sin{\frac{t}{2}}$ form a fundamental set of solutions.


Calculating the parameters $u_1$ and $u_2$, we have

$$u_1=-\int{\frac{y_2(t)g(t)}{W(y_1,y_2)(t)}}dt=-\int{\tan{\frac{t}{2}}}dt=2\ln\left(\cos{\frac{t}{2}}\right)+c_1$$

$$u_2=\int{\frac{y_1(t)g(t)}{W(y_1,y_2)(t)}}dt=\int{1}dt=t+c_2.$$

It follows that the general solution is

$$y(t)=\cos{\frac{t}{2}}\left(2\ln{\left(\cos{\frac{t}{2}}\right)}+c_1\right)+\sin{\frac{t}{2}}(t+c_2)=2\cos{\frac{t}{2}}\ln{\left(\cos{\frac{t}{2}}\right)}+t\sin{\frac{t}{2}}+c_1\cos{\frac{t}{2}}+c_2\sin{\frac{t}{2}}.$$

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Quiz-4 / Re: Q4 TUt 0401

« on: October 26, 2018, 06:22:16 PM »

Verify that $y_1(t)$ and $y_2(t)$ satisfy the corresponding homogeneous equation, then find a particular solution of the given nonhomogeneous equation,

$$\begin{gather*} t^2y'' - t(t + 2)y' + (t + 2)y = 2t^3,\qquad t > 0;\\ y_1(t) = t,\quad y_2(t) = te^t. \end{gather*}$$

1. Verify that $y_1(t)=t$ and $y_2(t)=te^{t}$ satisfy the corresponding homogeneous equation,

$$t^2y'' - t(t + 2)y' + (t + 2)y =0.\tag{2}$$

$$\cases{y_1(t)=t\\y'_1(t)=1\\y''_1(t)=0}$$

Plugging $y_1$, $y'_1$, and $y''_1$ into $(2)$,

$$t^2(0)- t(t + 2)(1) + (t + 2)(t) =0 \Rightarrow -t^{2}-2t+t^{2}+2t=0$$

$y_1$ satisfies $(2)$ $\Rightarrow$ $y_1$ is a solution to the corresponding homogeneous equation.


Similarily,

$$\cases{y_2(t)=te^{2}\\y'_2(t)=te^{t}+e^{t}\\y''_2(t)=te^{t}+2e^{t}}$$

Plugging $y_2$, $y'_2$, and $y''_2$ into $(2)$,

$$t^2(te^t+2e^t) - t(t + 2)(te^t+e^t) + (t + 2)(te^t) =0 \Rightarrow t^3e^t-t^3e^t+3t^2e^t-3t^2e^t+2te^t-2te^t=0$$

$y_2$ satisfies $(2)$ $\Rightarrow$ $y_2$ is a solution to the corresponding homogeneous equation.


2. Find a particular solution of the given nonhomogeneous equation,

$$t^2y'' - t(t + 2)y' + (t + 2)y = 2t^3, t>0.\tag{1}$$

Dividing $(1)$ by $t^{2}$, we have

$$y''-\frac{t+2}{t}y'+\frac{t+2}{t^{2}}=2t$$

Note that $p(t)=-1-\frac{2}{t}$, $q(t)=\frac{1}{t}+\frac{2}{t^2}$, and $g(t)=2t$ are continuous on $(0, +\infty)$

Computing the Wronskian,

$$W(t, te^{t})(t)=\begin{array}{|c c|}t& te^{t} \\ 1 & te^t+e^t\end{array}=t^2e^t\ne0,$$

we verify that $y_1(t)=t$ and $y_2(t)=te^{t}$ form a fundamental set of solutions $\forall t>0$

Calculating the parameters $u_1$ and $u_2$, we get

$$u_1=-\int{\frac{y_2(t)g(t)}{W(y_1,y_2)(t)}}dt=-\int{\frac{te^{t}2t}{t^2e^t}}dt=-\int{2}dt=-2t+c_1$$

$$u_2=\int{\frac{y_1(t)g(t)}{W(y_1,y_2)(t)}}dt=\int{\frac{t2t}{t^2e^t}}dt=2\int{e^{-t}}dt=-2e^{-t}+c_2$$

The particular solution is

$$y_p(t)=t(-2t+c_1)+te^{t}(-\frac{2}{e^t}+c_2)=-2t^{2}-2t+tc_1+te^tc_2$$

$$y_p(t)=-2t^{2}.$$

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Quiz-4 / Re: Q4 TUT0101

« on: October 26, 2018, 05:43:58 PM »

Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.

$$\begin{gather*} t^2y''-2y=3t^2-1,\qquad t>0\tag{1};\\  y_1(t)=t^2,\quad  y_2(t)=t^{-1}. \end{gather*}$$

1. Verify that $y_1(t)=t^{2}$ and $y_2(t)=t^{-1}$ satisfy the corresponding homogeneous equation,

$$t^{2}y''-2y=0.\tag{2}$$

$$\cases{y_1(t)=t^{2}\\y'_1(t)=2t\\y''_1(t)=2}$$

Plugging $y_1$, $y'_1$, and $y''_1$ into $(2)$,

$$t^{2}(2)-2(t^{2})=0 \Rightarrow 2t^{2}-2t^{2}=0$$

$y_1$ satisfies $(2)$ $\Rightarrow$ $y_1$ is a solution to the corresponding homogeneous equation.


Similarily,

$$\cases{y_2(t)=t^{-1}\\y'_2(t)=-t^{-2}\\y''_2(t)=2t^{-3}}$$

Plugging $y_2$, $y'_2$, and $y''_2$ into $(2)$,

$$t^{2}\left(\frac{2}{t^{3}}\right)-2\left(\frac{1}{t}\right)=0 \Rightarrow \frac{2}{t}-\frac{2}{t}=0$$

$y_2$ satisfies $(2)$ $\Rightarrow$ $y_2$ is a solution to the corresponding homogeneous equation.


2. Find a particular solution of the given nonhomogeneous equation,

$$t^{2}y''-2y=3t^{2}-1, t>0.\tag{1}$$

Dividing $(1)$ by $t^{2}$, we have

$$y''-\frac{2}{t^{2}}y=3-\frac{1}{t^{2}}$$

Note that $p(t)=0$, $q(t)=-\frac{2}{t^{2}}$, and $g(t)=3-\frac{1}{t^{2}}$ are continuous on $(0, +\infty)$

Computing the Wronskian,

$$W(t^{2}, t^{-1})(t)=\begin{array}{|c c|}t^{2}& t^{-1} \\ 2t & -t^{-2}\end{array}=-1-2=-3\ne0,$$

we verify that $y_1(t)=t^{2}$ and $y_2(t)=t^{-1}$ form a fundamental set of solutions $\forall t>0$

Calculating the parameters $u_1$ and $u_2$, we get

$$u_1=-\int{\frac{y_2(t)g(t)}{W(y_1,y_2)(t)}}dt=-\int{\frac{t^{-1}(3-t^{-2})}{-3}}dt=\frac{1}{3}\int{\frac{3}{t}-\frac{1}{t^{3}}}dt=\ln{t}+\frac{1}{6t^{2}}+c_1$$

$$u_2=\int{\frac{y_1(t)g(t)}{W(y_1,y_2)(t)}}dt=\int{\frac{t^{2}(3-t^{-2})}{-3}}dt=-\frac{1}{3}\int{3t^{2}-1}dt=-\frac{t^{3}}{3}+\frac{t}{3}+c_2$$

The particular solution is

$$y_p(t)=t^{2}\left(\ln{t}+\frac{1}{6t^{2}}+c_1\right)+\frac{1}{t}\left(-\frac{t^{3}}{3}+\frac{t}{3}+c_2\right)=t^{2}\ln{t}+\frac{1}{6}+c_1t^{2}-\frac{t^{2}}{3}+\frac{1}{3}+\frac{c_2}{t}=t^{2}\ln{t}+\frac{1}{2}+c_1t^{2}+\frac{c_2}{t}$$

$$y_p(t)=t^{2}\ln{t}+\frac{1}{2}.$$

15

Term Test 1 / Re: TT1 Problem 2 (morning)

« on: October 16, 2018, 11:44:27 AM »

Hi, there is my solution for Problem 2

Hey Doris, you wrote the definition of the Wronskian incorrectly, $W(y_1,y_2)(x)\ne y_1(x)y_2’(x)+y_2(x)y_1'(x)$.

Instead,
$$W(y_1,y_2)(x)=\begin{array}{|c c|}y_1(x)& y_2(x) \\ y_1’(x) & y_2’(x)\end{array}=y_1(x)y_2’(x)-y_2(x)y_1'(x).$$

Hope that helps!