# Homework Help: Verify that this is the solution to an ordinary differential equation

1. Jan 13, 2010

### NCyellow

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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)
Thank you!

2. Jan 13, 2010

### Staff: Mentor

Is this your y(t)?
$$y(t)~=~\int_0^t e^{t^2}(e^{-s^2})ds~+ e^{t^2}$$

3. Jan 13, 2010

### NCyellow

Yes, but the first e^(t^2) is outside and multiplying the integral.

4. Jan 13, 2010

### HallsofIvy

So
$$y(t)= e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2}$$?
No, the derivative is NOT $e^{t^2}$.
Differentiating,
$$y'= 2te^{t^2}\int_t^0 e^{-s^2}ds- e^{t^2}e^{-t^2}+ 2te^{t^2}$$
$$= 2t(e^{t^2}\int_t^0 e^{-s^2}ds+ e^{t^2})- 1= 2ty- 1$$.
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. $y(0)= 0+ e^{0^2}= 1$. 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 $e^{2x}\int e^{-2s}ds$ and the Fundamental theorem of Calculus to differentiate $\int e^{-2s}ds$ itself.)

5. Jan 13, 2010

### NCyellow

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. Jan 13, 2010

### Staff: Mentor

y and y' aren't "numbers" to plug in - they are functions. It's very important to communicate clearly, even moreso 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. Jan 13, 2010

### dacruick

when i do this i have the integral of e^(-t^2). which from my understanding is not a very fun integral. I dont understand the method you guys took to doing this. can someone explain??

8. Jan 13, 2010

### NCyellow

Hi, Thank you so much for helping me. I will try to be more clear in the future.

9. Jan 14, 2010

### HallsofIvy

The derivative of
$$\int_0^x e^{-t^2}dt$$
is just $e^{-x^2}$- that's the "Fundamental Theorem of Calculus". But $\int_0^x e^{x^2}e^{-t^2}dt= e^{x^2}\int_0^t e^{-t^2}dt$ is a product of functions of x and you have to use the chain rule.