Where did I go wrong with my solution to this first order ODE?

[jellicorse]

Homework Statement

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}

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=1i)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&#039;-2ty=1<br /> <br /> 2te^{t^2}+\frac{1}{2}t^{-2} -2t(2te^t{2}+\frac{1}{2}t^{-2})<br /> <br /> 2te^{t^2}+\frac{1}{2}t^{-2}-4t^2e^{t^2}-t^{-1}Which is not what is required.

 

[Saitama]

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}

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:
$$\int_0^t e^{-s^2}\,ds$$
?

 

[jellicorse]

Thanks a lot Pranav!

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

 

[Saitama]

Thanks a lot Pranav!

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

Oh, I forgot. Welcome to PF!

 

[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!

 

[Saitama]

Thanks, Pranav!

Your welcome. :)

I see that comes to e^{-t^2} according to the Fundamental Theorem of Calculus.

Yes.

 

[hilbert2]

There exists no antiderivative for $e^{-x^2}$.

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

 

[Saitama]

...but you can't represent it in terms of elementary functions.

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