Verify that this is the solution to an ordinary differential equation

In summary: Let u= x^2 so that the product is e^u\int e^{-u}du. By the chain rule, the derivative of this is e^u\int e^{-u}du+ e^u(-1)e^{-u}= e^{-u}\int e^{-u}du= e^{-u}(e^{-u}+ C)= e^{-x^2}(e^{-x^2}+ C). But that is 0 when x= 0 so the general solution is y= e^{-x^2}(e^{-x^2}+ C). And then, yes, you can substitute that into the original equation to find C. But you should be able to see
  • #1
NCyellow
22
0

Homework Statement


I am given dy/dt -2yt = 1
and y(t) = (e^(t^2))[e^(-s^2)ds] + e^(t^2)
integrate from t to 0 within the brackets.

Homework Equations





The Attempt at a Solution


I know that the derivative of y(t) would equal e^(t^2)
However I do not know how I am supposed to solve after I plug all the numbers in. It becomes a huge mess with 2t * the whole mess that is y(t)
Please advise
Thank you!
 
Physics news on Phys.org
  • #2
Is this your y(t)?
[tex]y(t)~=~\int_0^t e^{t^2}(e^{-s^2})ds~+ e^{t^2}[/tex]
 
  • #3
Mark44 said:
Is this your y(t)?
[tex]y(t)~=~\int_0^t e^{t^2}(e^{-s^2})ds~+ e^{t^2}[/tex]

Yes, but the first e^(t^2) is outside and multiplying the integral.
 
  • #4
NCyellow said:

Homework Statement


I am given dy/dt -2yt = 1
and y(t) = (e^(t^2))[e^(-s^2)ds] + e^(t^2)
integrate from t to 0 within the brackets

Homework Equations





The Attempt at a Solution


I know that the derivative of y(t) would equal e^(t^2).
So
[tex]y(t)= e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2}[/tex]?
No, the derivative is NOT [itex]e^{t^2}[/itex].
Differentiating,
[tex]y'= 2te^{t^2}\int_t^0 e^{-s^2}ds- e^{t^2}e^{-t^2}+ 2te^{t^2}[/tex]
[tex]= 2t(e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2})- 1= 2ty- 1[/tex].
That's exactly why it satisfies dy/dt- 2ty= 1.
But that is NOT the "general solution". The way you have set it up, when t= 0, the integral is from 0 to 0 and so is 0. [itex]y(0)= 0+ e^{0^2}= 1[/itex]. This is the solution of dy/dx- 2ty= 1 with intial condition y(0)= 1. To get the general solution, add "+ C". Then choose C to give whatever specific value it is supposed to have.

However I do not know how I am supposed to solve after I plug all the numbers in. It becomes a huge mess with 2t * the whole mess that is y(t)
Please advise
Thank you!
Plug what numbers in? I have no idea what you are talking about! Do you mean differentiate it? (As I just did? I used the product rule on [itex]e^{2x}\int e^{-2s}ds[/itex] and the Fundamental theorem of Calculus to differentiate [itex]\int e^{-2s}ds[/itex] itself.)
 
  • #5
HallsofIvy said:
So
[tex]y(t)= e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2}[/tex]?
No, the derivative is NOT [itex]e^{t^2}[/itex].
Differentiating,
[tex]y'= 2te^{t^2}\int_t^0 e^{-s^2}ds- e^{t^2}e^{-t^2}+ 2te^{t^2}[/tex]
[tex]= 2t(e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2})- 1= 2ty- 1[/tex].
That's exactly why it satisfies dy/dt- 2ty= 1.
But that is NOT the "general solution". The way you have set it up, when t= 0, the integral is from 0 to 0 and so is 0. [itex]y(0)= 0+ e^{0^2}= 1[/itex]. This is the solution of dy/dx- 2ty= 1 with intial condition y(0)= 1. To get the general solution, add "+ C". Then choose C to give whatever specific value it is supposed to have.


Plug what numbers in? I have no idea what you are talking about! Do you mean differentiate it? (As I just did? I used the product rule on [itex]e^{2x}\int e^{-2s}ds[/itex] and the Fundamental theorem of Calculus to differentiate [itex]\int e^{-2s}ds[/itex] itself.)

Hi, thank you very much. I guess i misread a rule from the book when I thought that differentiating it would be simple. When i said plugging numbers in I meant plugging the y and y' back into the original equation to make sure it works.
 
  • #6
y and y' aren't "numbers" to plug in - they are functions. It's very important to communicate clearly, even more so in the context of a forum where you have to use text only, and there is a lag time between a question and an answer.
 
  • #7
when i do this i have the integral of e^(-t^2). which from my understanding is not a very fun integral. I don't understand the method you guys took to doing this. can someone explain??
 
  • #8
Mark44 said:
y and y' aren't "numbers" to plug in - they are functions. It's very important to communicate clearly, even more so in the context of a forum where you have to use text only, and there is a lag time between a question and an answer.

Hi, Thank you so much for helping me. I will try to be more clear in the future.
 
  • #9
The derivative of
[tex]\int_0^x e^{-t^2}dt[/tex]
is just [itex]e^{-x^2}[/itex]- that's the "Fundamental Theorem of Calculus". But [itex]\int_0^x e^{x^2}e^{-t^2}dt= e^{x^2}\int_0^t e^{-t^2}dt[/itex] is a product of functions of x and you have to use the chain rule.
 

1. What is an ordinary differential equation?

An ordinary differential equation (ODE) is a mathematical equation that describes how a variable changes over time, based on its current value and the rate at which it is changing. It is commonly used in physics, engineering, and other scientific fields to model physical systems.

2. How do you verify that a given solution is the solution to an ODE?

To verify that a solution is correct, you must substitute it into the original ODE and see if it satisfies the equation. This means that when you plug in the solution for the variable, both sides of the equation should be equal.

3. Can a solution to an ODE be incorrect?

Yes, a solution can be incorrect if it doesn't satisfy the original ODE. It is important to carefully check your work and ensure that all steps are correct when solving ODEs.

4. Are there different methods for solving ODEs?

Yes, there are several methods for solving ODEs, including separation of variables, substitution, and integrating factors. The choice of method depends on the specific form of the ODE and the initial conditions given.

5. Is there a way to check the accuracy of a solution to an ODE?

Yes, there are methods for checking the accuracy of a solution to an ODE. These include comparing the solution to known solutions or using numerical methods to approximate the solution. Additionally, graphing the solution can help visualize its behavior and any potential errors.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
515
  • Calculus and Beyond Homework Help
Replies
7
Views
510
  • Calculus and Beyond Homework Help
Replies
7
Views
135
  • Calculus and Beyond Homework Help
Replies
1
Views
663
  • Calculus and Beyond Homework Help
Replies
2
Views
323
  • Calculus and Beyond Homework Help
Replies
4
Views
552
  • Calculus and Beyond Homework Help
Replies
2
Views
463
  • Calculus and Beyond Homework Help
Replies
2
Views
222
  • Calculus and Beyond Homework Help
Replies
3
Views
444
  • Calculus and Beyond Homework Help
Replies
5
Views
846
Back
Top