Messages

1

MidTerm / Re: MT Problem 3

« on: March 16, 2013, 12:51:20 PM »

Just got my midterm back on Friday and looked carefully through..


In the official 2013Midterm answers (both versions on Forum and on CourseSite), why we, when using variation-method, have
   v₁ = - ∫ (t^2 + 1) g(t) / Wronskian  dt   ??  what is (t^2 +1) ?! Should that not be y₂ = t^2 ?!

And how do we, from this step, get the next step, where (t^2 +1) changes to t with no reason ?

I see the results of v1 and v2 are correct, but the steps are totally incomprehensible and WRONG.

And why Wronskian = -t^2 ? should it not be t^2 ?

2

MidTerm / Re: MT Problem 2b

« on: March 06, 2013, 10:22:36 PM »

$$
y(t) =(c_1 \sqrt{t^2-1} (1-t)^{3/2})/\sqrt{t+1}+(c_2 t \sqrt{t^2-1} \sqrt{1-t})/((t-1) \sqrt{t+1})
$$

[Don't know how to get the codes working..]

3

MidTerm / Re: MT Problem 2b

« on: March 06, 2013, 10:19:28 PM »

The homogeneous sol'ns were y₁=t, y₂=t² right?

But why Wolfram alpha gives y(t) = (c_1 sqrt(t^2-1) (1-t)^(3/2))/sqrt(t+1)+(c_2 t sqrt(t^2-1) sqrt(1-t))/((t-1) sqrt(t+1))
...

the 2nd term of Wolfram is equivalent {if regardless of the signs (consider all things in roots being positive)} ; but I have no idea what the first term is ... hold on, it can be written as    constant *(t-1)^2    , so it's still correct