## Some Differential Equation help needed

Of the Partial Kind
Using d'Alemberts soltuion for the vibrating string in one dimension

Find u(1/2,3/2), when l-=1, c=1, f(x) = 0, g(x) = x(1-x)
Now i tried simply substituting this into the solution that is (since f(x)=0)
$$u(x,t) = \frac{1}{2} \int_{x-t}^{x+t} g(x) dx$$
but it yields the wrong answer.
Does the length of the string have anything to do with the answer?

 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target

Recognitions:
Homework Help
 Quote by stunner5000pt Of the Partial Kind Using d'Alemberts soltuion for the vibrating string in one dimension Find u(1/2,3/2), when l-=1, c=1, f(x) = 0, g(x) = x(1-x) Now i tried simply substituting this into the solution that is (since f(x)=0) $$u(x,t) = \frac{1}{2} \int_{x-t}^{x+t} g(x) dx$$ but it yields the wrong answer. Does the length of the string have anything to do with the answer? Thank you in advance for your help!
Stunner, to use D'Alembert's forumua, you need to remember to use the "odd extensions" of both f(x) and g(x). Now, I know that's not pretty but that's just how it is. Remember when I said that $Sin[\pi x]$ was already an odd-extension and so we didn't have to do anything about it? That's not the case with g(x)=x(1-x) over the interval you're integrating from. Look at the first plot. That's g(x) un-extended. We wish to make an odd periodic function of g(x) over the interval of integration. In your case thats 1/2-3/2 to 1/2+3/2 or the interval [-1,2]. So, first thing is to "odd-extend" what the function looks like in [0,1] to the interval [-1,0]. Well, that's the second plot and the equation for it is:

$$g_1[x]=x(1+x)$$

The equation for the interval [0,1] is just g(x):

$$g_2(x)=x(1-x)$$

Now I wish to do that again for the interval [1,2], that is an odd extension of g(x) which would just be flipping it over into the interval [1,2]. The equation for that one would be:

$$g_3(x)=-(x-1)+(x-1)^2$$

The third plot is all three. So:

\begin{align*} u(1/2,3/2) &=\frac{1}{2} \int_{-1}^{2}\tilde{g_0}(\tau)d\tau \\ &= \frac{1}{2}\left(\int_{-1}^0 g_1(\tau)d\tau+\int_0^1 g_2(\tau)d\tau+\int_1^2 g_3(\tau)\tau \right) \end{align}

I get -1/12. Is that what you get?

Edit: Stunner, I initially made a typo on g3 but corrected it above.

Edit2: Forgot the 1/2 in front of the integral sign. Suppose that's -1/12 now. Sorry.
Attached Thumbnails

 it is quite clear that we are not being taught (or the material's presentation) correctly. I did not know how to extend the functions. I understand now... thank you very much!