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First Order Differential Equations, given initial value...

nuggz619

New member
Oct 4, 2013
3
I'm having trouble with this problem... Im almost certain that I have the first part correct which is solving the first order DiffEQ using an integrating factor. I think that I am computing the constant incorrectly. I have followed all steps, including the similar problem given on WileyPlus online homework... However, when I enter my final solution WileyPlus insists that I am wrong.

Can someone please look over my work to see If i did it correctly? I've been stuck on this problem for over 2 hours now and I just cant seem to come up with the solution that WileyPlus online wants me to come up with.

The funny thing is that the equivalent question in my textbook has a similar answer to mine, however it is slightly different because my online homework (WileyPlus) gave me the initial condition y(pi/2) = 9.... where as my text gives initial condition y(pi/2) = 1...

Here is a picture of the first part of the problem that I computed... which according to wolfram Alpha plus, is correct.

https://docs.google.com/file/d/0B9dQB0QE48ufNHpRRWptRDRBQkE/edit?usp=sharing

When I computed C, given initial condition y(pi/2) = 9.... I came up with
C = (4 - 9pi^2)/4

for a final answer of...

y(t) = ( sin(t) - tcos(t) + (4-9pi^2)/4 ) / t^2

However, Wiley Plus insists that my answer is incorrect. I have redone this problem probably 5 or 6 times and i keep coming up with the same answer.

I am literally stuck!

Please help :)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The document to which you have posted a link requires permission to view...I have requested permission, under the username gaussian3142. Once I have permission, I will work the problem to see what I find, and post the problem so that others do not have to follow a link. :D
 

nuggz619

New member
Oct 4, 2013
3

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Okay, we are given the IVP:

\(\displaystyle t\frac{dy}{dt}+2y=\sin(t)\) where \(\displaystyle y\left(\frac{\pi}{2} \right)=9\)

Multiply the ODE by $t$:

\(\displaystyle t^2\frac{dy}{dt}+2ty=t\sin(t)\)

The left side is the differentiation of a product:

\(\displaystyle \frac{d}{dt}\left(t^2y \right)=t\sin(t)\)

Integrate with respect to $t$:

\(\displaystyle \int\,d\left(t^2y \right)=\int t\sin(t)\,dt\)

\(\displaystyle t^2y=\sin(t)-t\cos(t)+C\)

Thus, the general solution is:

\(\displaystyle y(t)=\frac{\sin(t)-t\cos(t)+C}{t^2}\)

So far so good, this is equivalent to your working.

Now, using the initial values to determine the value of the parameter $C$, we may write:

\(\displaystyle y\left(\frac{\pi}{2} \right)=\frac{1+C}{\left(\frac{\pi}{2} \right)^2}=9\,\therefore\,C=\frac{9}{4}\pi^2-1\)

This is the negative of the value for the parameter you found.
 

nuggz619

New member
Oct 4, 2013
3
Thank you so much... I cant believe that I made such an idiotic mistake.... BAH!

you are a life saver.