Toronto Math Forum

MAT244--2018F => MAT244--Lectures & Home Assignments => Topic started by: youjianz on December 05, 2018, 08:14:47 PM

Title: sec 9.2 question 18
Post by: youjianz on December 05, 2018, 08:14:47 PM
im not sure how i able to solve this question

can someone give the answer for this

thanks in advance
Title: Re: sec 9.2 question 18
Post by: Qingyang Wei 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.
Title: Re: sec 9.2 question 18
Post by: Qingyang Wei 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 $H$.
Title: Re: sec 9.2 question 18
Post by: Qingyang Wei 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?
Title: Re: sec 9.2 question 18
Post by: Victor Ivrii on December 06, 2018, 02:49:21 PM
May be it is exact? Or you can find an integrating factor.