Simple ODE, but how to approach?

In summary: Just substitute in the dy for y' and you're good to go. In summary, Daniel is explaining that the general solution is based upon the substitution of y'dt for y' and that there is no need to worry about the dt.
  • #1
BlkDaemon
15
0
I'm creeping my way through DiffEq, and recently started reading Paul Dawkins' PDF, which is actually pretty helpful. He does, however, tend to assume that his readers know how to approach what he calls simple problems.

Well, one of 'em has me stumped.

2t*y' + 4y = 3

I need to find the general solution. Here's what I tried:

1) I tried isolating y. I get y=(3 - 2t*y')/4. That doesn't work because I've got y' as part of the solution. Eeek.

2) I tried isolating y'. That didn't work either, for the same reason. I got

y' = (3-4y)/2t

But if I take the derivative of that, I get weirdness. Now, it's really possible that I'm just calculating the derivative incorrectly.

I also can't figure out how to ditch the "t", but when I look at the answer, there's t, as well as C, so I know that (because of the arbitrary constant) there's a derivative involved, and that means the "t" isn't going anywhere.

Anyone have any suggestions on how to work this? I don't need the answer. I need to know how to approach the problem.

Thanks in advance.
 
Physics news on Phys.org
  • #2
2ty' + 4y = 3
2ty' = 3 - 4y
y'/(3 - 4y) = 1/(2t)
dy/(3 - 4y) = dt/2t
-0.25ln(3 - 4y) = 0.5ln(t) + C
ln(3 - 4y) = -2ln(t) + D
3 - 4y = E/t²
-4y = E/t² - 3
y = F/t² + 0.75

This is the general solution. It just uses separation of variables. F varies over R, and given an initial value, you can find the particular solution by solving for F.
 
Last edited:
  • #3
Thanks for the response. Now I see where I missed a step!
 
  • #4
Wait, I'm confused again...

Essentially, y' is dy. I get that. But where does dt come from? This has me stumped. If I interpret the change from y' to dy as a simple change in notation, I can't just pull dt out of thin air.

And when I perform that step, I get:

dy/(3-4y) = 1/2t

which yields:

(-ln |4y-3|)/4 = (t^2)/4 + C

Right? Then the 4s cancel out, and I'm left with

- ln |4y-3| = t^2 + C

But this isn't right. Where am I missing it?

I realize this is a pretty simple separable equation, but I'm missing some crucial steps that are interfering with my understanding of the overall concept.

Hopefully that makes sense.

Daemon
 
  • #5
Nope.Rewrite it and take care with that "dt" and that "t".

[tex] \frac{dy}{3-4y}=\frac{dt}{2t} [/tex]

Daniel.
 
  • #6
BikDamon:
The trick is based upon use of the integration technique "substitution", which is, essentially mirrors the chain rule of differentiation.
Suppose you can write your differential equation as:
[tex]f(y(t))\frac{dy}{dt}=g(t)[/tex]
Now, integrate this between arbitrary instants "0" and "t":
[tex]\int_{0}^{t}f(y(\tau))\frac{dy}{d\tau}d\tau=\int_{0}^{t}g(\tau)d\tau[/tex]
where I've introduced the dummy variable [tex]\tau[/tex] for clarity (and rigour).

Furthermore assume it exists a function F(x), so that F'(x)=f(x).
We may then rewrite our equation as:
[tex]\int_{0}^{t}\frac{dF}{dx}\mid_{x=y(\tau)}\frac{dy}{d\tau}d\tau=\int_{0}^{t}g(\tau)d\tau[/tex]
That is, we may write this as:
[tex]\int_{0}^{t}(\frac{d}{d\tau}F(y(\tau)))d\tau=\int_{0}^{t}g(\tau)d\tau[/tex]
by the chain rule of differentiation.
But the left hand side equals now by FOTC:

[tex]\int_{0}^{t}(\frac{d}{d\tau}F(y(\tau)))d\tau=F(y(t))-F(y(0))[/tex]
which equals, with [tex]y(0)=y_{0}, y(t)=y:
[tex]F(y(t))-F(y(0))=\int_{y_{0}}^{y}\frac{dF}{dy}dy=\int_{y_{0}}^{y}f(y)dy[/tex]
Thus, we have gained the equality:
[tex]\int_{y_{0}}^{y}f(y)dy=\int_{0}^{t}g(\tau)d\tau[/tex]
 
  • #7
I get the theory...

I understand the theory in the last response, but I got lost on the actual application.

Essentially, once I replace y' with dy, I arbitrarily assign dt to the right side of the equation, yes?
 
  • #8
[tex] y'\equiv \frac{dy}{dt} [/tex].You could (even though it's mathematically doubtful) move around those differentials (divide by them,multiply by them).

Daniel.
 
  • #9
You do NOT replace y' with dy; you replace y'dt with dy !
 
  • #10
And one more thing:nothing you do here is arbitrary.Only integration constants have that privilage.

Daniel.
 
Last edited:
  • #11
And, read again my prior response.
Make sure you undestand it properly.
 
  • #12
I think part of my challenge is that the originally stated problem does not include "dy" or "dt" and therefore working towards the solution has become a bit convoluted for me.

The original problem only says: 2ty' + 4y = 3

Do you guys even see where I'm getting lost? If not, I'll move on to the next problem!
 
  • #13
so the problem doesn't specify that y is a function of t? well that sure can be misleading.
 
  • #14
BlkDaemon said:
I think part of my challenge is that the originally stated problem does not include "dy" or "dt" and therefore working towards the solution has become a bit convoluted for me.

The original problem only says: 2ty' + 4y = 3

Do you guys even see where I'm getting lost? If not, I'll move on to the next problem!


You're getting lost at this step [tex] y'\equiv \frac{dy}{dx} [/tex],which is really sad.

Daniel.
 
  • #15
BlkDaemon said:
Thanks for clearing that up. End of discussion, since "that's sad" loosely translates as "too stupid to ask questions here". Thanks for your help, and I'll research my questions before posting in the future.

I believe that what Daniel refers as "that's sad" is the fact that you should know Leibniz notarion by now, way before taking a course on differential equations.
 
  • #16
Gotcha. Actually, I *have* worked with the notation before. What I didn't realize in the problem was that the notation was implied. The problem came from a summary of term definitions in a larger document, and I'm working my way through it.

I'm not trying to belabour the issue; I just wanted to make sure that I understood exactly what was going on.

Thanks.
 
  • #17
It's perfectly common in the beginning to be confused by different notations.
Hopefully, things have cleared out now.
 

Related to Simple ODE, but how to approach?

1. What is a simple ODE?

A simple ODE, or ordinary differential equation, is a mathematical equation that relates a function to its derivatives. It is commonly used to model physical systems in science and engineering, and can be solved using analytical or numerical methods.

2. How do I approach solving a simple ODE?

The approach to solving a simple ODE depends on the specific equation and initial conditions. Generally, it involves identifying the order of the equation, finding a suitable method for solving it (e.g. separation of variables, substitution, etc.), and applying the initial conditions to find the final solution.

3. What is the difference between an initial value problem and a boundary value problem for simple ODEs?

An initial value problem for a simple ODE involves finding the solution for a given function and its initial conditions at a single point. A boundary value problem involves finding the solution for a given function and its boundary conditions over a range of values.

4. Can a simple ODE have multiple solutions?

Yes, a simple ODE can have multiple solutions. This is known as the existence of "general solutions," where a single equation can have multiple solutions that satisfy the same initial conditions. In some cases, the equation may also have multiple solutions for different initial conditions.

5. What are the applications of simple ODEs in science?

Simple ODEs are widely used in various fields of science, including physics, chemistry, biology, and engineering. They are used to model and understand the behavior of physical systems, such as population growth, chemical reactions, and motion of objects. They also play a crucial role in developing scientific theories and making predictions about real-world phenomena.

Similar threads

  • Differential Equations
Replies
5
Views
1K
  • Differential Equations
Replies
4
Views
2K
  • Differential Equations
Replies
4
Views
2K
Replies
2
Views
2K
Replies
3
Views
7K
  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
2
Views
1K
  • MATLAB, Maple, Mathematica, LaTeX
Replies
1
Views
1K
  • Differential Equations
Replies
2
Views
1K
  • Differential Equations
Replies
8
Views
1K
Back
Top