# Solution to first order ODE

1. Mar 4, 2014

### jellicorse

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

Can anyone point out where I have gone wrong with this?

Verify that the given function is a solution of the differential equation.

$$y' -2ty =1$$ $$y= e^{t^2}\int^t_0 e^{-s^2}ds+e^{t^2}$$

3. The attempt at a solution

The steps I have taken are the following:

i) Evaluate integral in expression for y
ii) Differentiate the expression for y
iii)Substitute these values into $$\frac{dy}{dt}-2ty=1$$

i)Evaluate the integral

$$\int^t_0e^{-s^2}ds$$

by substitution:

$$u=s^2$$; $$ds=\frac{1}{-2s}du$$

$$\frac{1}{-2s}\int^t_0e^{u}du = \left[\frac{e^{-s^2}}{-2s}\right]^t_0 =\frac{e^{-t^2}}{-2t}$$

ii)Differentiate expression for y

$$y=e^{t^2}\cdot \frac{e^{-t^2}}{-2t}+e^{t^2}$$

$$y=e^{t^2}-\frac{1}{2}t^{-1}$$

$$y'=2te^{t^2}+\frac{1}{2}t^{-2}$$

iii) Substitute these into $$y'-2ty=1 [tex]2te^{t^2}+\frac{1}{2}t^{-2} -2t(2te^t{2}+\frac{1}{2}t^{-2})$$

$$2te^{t^2}+\frac{1}{2}t^{-2}-4t^2e^{t^2}-t^{-1}$$

Which is not what is required.

2. Mar 4, 2014

### Saitama

Hi jellicorse!

The quoted step is incorrect. There exists no antiderivative for $e^{-x^2}$. While evaluating the integral, you took $s$ outside the integral but $s$ is a function of $u$ so what you did is not valid.

For the given problem, you have to differentiate the given function using Fundamental Theorem of Calculus. Can you find the derivative wrt t for:
$$\int_0^t e^{-s^2}\,ds$$
?

3. Mar 4, 2014

### jellicorse

Thanks a lot Pranav!

I need to go and revise the Fundamental Theorem of Calculus again to check...

4. Mar 4, 2014

### Saitama

Oh, I forgot. Welcome to PF!

5. Mar 4, 2014

### jellicorse

Thanks, Pranav!

I see that comes to $$e^{-t^2}$$ according to the Fundamental Theorem of Calculus... I am getting a bit rusty on that so need to revise it a bit!

6. Mar 4, 2014

Yes.

7. Mar 4, 2014

### hilbert2

It sure does exist, and it's called the error function, but you can't represent it in terms of elementary functions.

http://en.wikipedia.org/wiki/Error_function

8. Mar 4, 2014

### Saitama

Yes, that's what I meant, sorry for the confusion.