sec 9.2 question 18

[youjianz]

im not sure how i able to solve this question

can someone give the answer for this

thanks in advance

[Qingyang Wei]

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.

[Qingyang Wei]

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 $H$.

[Qingyang Wei]

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?

[Victor Ivrii]

May be it is exact? Or you can find an integrating factor.