Switching to polar coordinates, (d r)/(d t)=(1)/(r)(x (d x)/(d t)+y (d y)/(d t))=(1)/(r)[-x

Switching to polar coordinates, (d r)/(d t)=(1)/(r)(x...

Switching to polar coordinates, \[ \begin{array}{l} \frac{d r}{d t}=\frac{1}{r}\left(x \frac{d x}{d t}+y \frac{d y}{d t}\right)=\frac{1}{r}\left[-x y+x^{2}\left(1-r^{2}\right)+x y+y^{2}\left(1-r^{2}\right)\right]=r\left(1-r^{2}\right) \\ \frac{d \theta}{d t}=\frac{1}{r^{2}}\left(-y \frac{d x}{d t}+x \frac{d y}{d t}\right)=\frac{1}{r^{2}}\left[y^{2}-x y\left(1-r^{2}\right)+x^{2}+x y\left(1-r^{2}\right)\right]=1 \end{array} \] Now $d r / d t=r-r^{3}$ or $(d r / d t)-r=-r^{3}$ is a Bernoulli differential equation. Following the procedure in Section 2.5 of the text, we let $w=r^{-2}$ so that $w^{\prime}=-2 r^{-3}(d r / d t) .$ Therefore $w^{\prime}+2 w=2,$ a linear first order differential equation. It follows that $w=1+c_{1} e^{-2 t}$ and so $r^{2}=1 /\left(1+c_{1} e^{-2 t}\right) .$ The general solution can be written as \[ r=\frac{1}{\sqrt{1+c_{1} e^{-2 t}}}, \quad \theta=t+c_{2} \] If $\mathbf{X}(0)=(1,0), r=1$ and $\theta=0$ when $t=0 .$ Therefore $c_{1}=0=c_{2}$ and so $x=r \cos t=\cos t$ and $y=r \sin t=\sin t .$ This solution generates the circle $r=1 .$ If $\mathbf{X}(0)=(2,0), r=2$ and $\theta=0$ when $t=0$ Therefore $c_{1}=-3 / 4, c_{2}=0$ and so \[ r=\frac{1}{\sqrt{1-\frac{3}{4} e^{-2 t}}}, \quad \theta=t \] This solution spirals toward the circle $r=1$ as $t$ increases.

Switching to polar coordinates,
\[
\begin{array}{l}
\frac{d r}{d t}=\frac{1}{r}\left(x \frac{d x}{d t}+y \frac{d y}{d t}\right)=\frac{1}{r}\left[-x y+x^{2}\left(1-r^{2}\right)+x y+y^{2}\left(1-r^{2}\right)\right]=r\left(1-r^{2}\right) \\
\frac{d \theta}{d t}=\frac{1}{r^{2}}\left(-y \frac{d x}{d t}+x \frac{d y}{d t}\right)=\frac{1}{r^{2}}\left[y^{2}-x y\left(1-r^{2}\right)+x^{2}+x y\left(1-r^{2}\right)\right]=1
\end{array}
\]
Now $d r / d t=r-r^{3}$ or $(d r / d t)-r=-r^{3}$ is a Bernoulli differential equation. Following the procedure in Section 2.5 of the text, we let $w=r^{-2}$ so that $w^{\prime}=-2 r^{-3}(d r / d t) .$ Therefore $w^{\prime}+2 w=2,$ a linear first order differential equation. It follows that $w=1+c_{1} e^{-2 t}$ and so $r^{2}=1 /\left(1+c_{1} e^{-2 t}\right) .$ The general solution can be written as
\[
r=\frac{1}{\sqrt{1+c_{1} e^{-2 t}}}, \quad \theta=t+c_{2}
\]
If $\mathbf{X}(0)=(1,0), r=1$ and $\theta=0$ when $t=0 .$ Therefore $c_{1}=0=c_{2}$ and so $x=r \cos t=\cos t$ and $y=r \sin t=\sin t .$ This solution generates the circle $r=1 .$ If $\mathbf{X}(0)=(2,0), r=2$ and $\theta=0$ when $t=0$ Therefore $c_{1}=-3 / 4, c_{2}=0$ and so
\[
r=\frac{1}{\sqrt{1-\frac{3}{4} e^{-2 t}}}, \quad \theta=t
\]
This solution spirals toward the circle $r=1$ as $t$ increases.

Key Concepts

Polar Coordinate Transformation

This concept involves rewriting equations originally expressed in Cartesian coordinates into polar coordinates, where a point's position is defined by a radius (r) and an angle (?). This transformation simplifies many problems, especially those with circular symmetry, by decoupling the radial and angular dynamics.

Bernoulli Differential Equation

A Bernoulli differential equation is a type of nonlinear ordinary differential equation of the form dy/dt + P(t)y = Q(t)y^n, where n is a real number. It can be transformed into a linear differential equation through an appropriate substitution, making it easier to solve.

Substitution Method for Differential Equations

This method involves making an appropriate change of variables that simplifies the given differential equation. In the context of the Bernoulli equation, a common substitution transforms the nonlinear equation into a linear first order differential equation, allowing standard solution techniques to be applied.

Linear First-Order Differential Equation

A linear first-order differential equation is of the form dy/dt + P(t)y = Q(t), where integration factors are typically used to find the solution. This technique involves multiplying the entire equation by an integrating factor that simplifies the left-hand side into the derivative of a product of functions, which can then be integrated directly.

Initial Value Problem and Constant Determination

In solving differential equations, particularly when a general solution is obtained, initial conditions are used to determine the unknown constants. This process ensures that the solution fits the specific scenario or initial state of the system being modeled.

Transcript

00:01 Ladies and gentlemen, today we're looking at section 4 .5 number one.

00:10 And now this is the case where we're starting to look at non -homogeneous equations and how to solve them.

00:20 So in this case, we have the following.

00:26 We have two, we have a solution to this equation.

00:31 And this equation.

00:33 So we're given one solution to this equation and one solution to this equation.

00:40 And we're asked to apply what is called the superposition principle to solve the following questions.

00:48 First off, the question is, what is the superposition principle? while the superposition principle says that if i have a solution, to this to an equation of this form.

01:06 So if i have one solution to an equation of this form and i have a and i have a one solution to an equation of this form, then for any real number c1 and c2, this solution here, or this function, solves this linear or ode, so, i mean, it's pretty straightforward.

01:36 It looks pretty straightforward that, you know, you have a solution here.

01:41 You have a solution there.

01:43 Then a linear combination solves any ode that's equal to a linear combination of these two guys.

01:55 So, i mean, it looks pretty straightforward.

01:58 And, you know, in general it is.

02:00 It's just good to know that, you know, this.

02:03 Holds because you'd like it's a very simple thing that you'd like to hold.

02:09 So with that in mind now, we're really each of these three problems, a, b, and c is really just about applying it and more or less the application is direct.

02:19 So if we look at the first case, we have five sign t.

02:26 And of course, so we have this differential equation equals to five sign t.

02:32 And when we go, we go back to the original one, we see that cosine t is a solution.

02:40 So by the homogen, or by the superposition principle, we would have five cosine t solves a.

02:50 And that's, that's all we need here.

02:53 We're not looking for general solutions or anything like that.

02:57 We're just, the question is just what does the superposition tell us about solution to here? and the next question, of course, it's the same type of thing, but in this case we have this sign t plus, i guess, negative 3e to the 2t.

03:21 And of course, again, that's of this form here.

03:24 Oops.

03:25 So what do we do? well, we apply the same thing and we get this for our answer.

03:32 Now, you'll notice that in my answer, i leave it in this form.

03:41 I actually don't encourage students to cancel or to simplify in general.

03:49 So, i mean, obviously what i'm saying is that you could simplify this further and say, well, it's just equal to cosine t minus e to the 2 t and of course it is um you know that this is true um you know and that's fine but i'll tell you um a lot of times when students do something like this they make mistakes so obviously in this case it's pretty obvious...

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