Messages

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MAT244--Lectures & Home Assignments / How to draw phase portraits by hand effectively?

« on: December 06, 2018, 01:44:44 PM »

The only method of drawing phase portrait that I am aware of is to evaluate each particular solutions of the system at a large amount of points, and "connect the dots" to produce a trajectory for the solution. But that seems to be very time consuming. Is there any more efficient way to draw this by hand?

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MAT244--Lectures & Home Assignments / Re: sec 9.2 question 18

« on: December 05, 2018, 09:34:57 PM »

For the question 18 on the 11th edition, we can write down $$\frac{dy}{dx} = \frac{-2xy^2 + 6xy}{2x^2y - 3x^2 - 4y}$$
And we can rearrange the equation as:
$$ (2x^2y - 3x^2 - 4y)\frac{dy}{dx} + (2xy^2 - 6xy) = 0$$

Now does this equation looks like a type of equations we encountered before? Can you solve this with the methods we learned previously?

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MAT244--Lectures & Home Assignments / Re: sec 9.2 question 18

« on: December 05, 2018, 09:12:04 PM »

Sorry, that previous post was for question 18 of 9.2 on the 10th edition of the book. If you are looking at the 11th edition then that's not the answer for that. Sorry if it causes any confusion.

Just to clarify, is the question you are asking this one?
$$\frac{dx}{dt}=2x^2y - 3x^2 - 4y, \frac{dy}{dt} = -2xy^2 + 6xy$$
a) Find an equation of the form $H(x,y)=c$ and b) plot several level curves of the function $H$.

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MAT244--Lectures & Home Assignments / Re: sec 9.2 question 18

« on: December 05, 2018, 09:04:44 PM »

For part a): We can write $$\frac{dy/dt}{dx/dt} = \frac{dy}{dx} = \frac{-8x}{2y}$$
This is a separable equation, so we can write it as
$$2ydy=-8xdx$$
Integrate both sides and get
$$y^2 +c_1 = -4x^2 +c_2$$
rearrange, and let $C = c_2 - c_1$, we  get
$$y^2 + 4x^2 = C$$. This is the expression $H(x,y)=C$ that all trajectories of the system satisfies.

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

« on: December 05, 2018, 08:49:38 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?

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

« on: December 05, 2018, 04:39:52 PM »

For question 1 a) of Term Test 2, i.e. Solve equation $y'' + 4y = 2\tan (t)$, how do you solve the integral to arrive at the particular solution of the equation? I see the solution to the question being posted in the Term test section of the forum but I am not sure how the particular solution $$y_p(t)=-\cos 2t(t-\sin (t)\cos (t))+\sin 2t(\log (\cos (t)) - 1/2 \cos 2t)$$ is calculated, specifically, how the integral is solved to arrive at this answer.