Question

Need help with the attached 2nd order differential equation.

I have attached my work thus far (it would be a pain to retype, especially if its wrong)

Answer 1

Can you just write it out using LaTex?

Answer 2

is latex the equation editor? I'll type up what I've gotten thus far.

Answer 3

basically, I used the theorem and lemma in the book that say "the difference of any two solutions to the non homogeneous diff eq are solutions to the homogoenous" so I used that to find y1 and y2 by subtraction.

Answer 4

Now, I THINK (not sure, my book is old and not clear at all) that I can plug the y1 and y2 I found into the general solution for the homogeneous equation. Which is of the form \[y(t) = c_1y_1(t)+c_2y_2(t)+\psi(t)\] where psi(t) is any of the solutions they gave.

Answer 5

So you're finding \(\sf \color{blue}{y_g = y_c+y_p}\)

Answer 6

giving me \[y(t) = c_1(-4e^t)+c_2(2e^t+e ^{-t ^{3}}) + (5e^t+e ^{-t ^{3}}+e ^{t ^{2}})\]

Answer 7

I think so abbot, because the form of the nonhomog ends with =g(t) and not 0 right?

Answer 8

And what are you being aksed to do? Lol.

Answer 9

well, first i was wondering if i did that first part right. Then, Im gven 3 initial values which I assume I need to use to find c1 and c2... but Im not getting even close to the answer in the book, so I was hoping someone could tell me where I have gone wrong :D

Answer 10

Well, you need to start by finding y', first :)

Answer 11

ok, thats what I thought, that at this point it should parallel easier IVPs in the book. gimme a sec to type it up.

Answer 12

Yeah, I think you should get a system of equations where you have
(1) c\(_1\)+c\(_2\) = n
(2) c\(_1\) + c\(_2\) = n

not like that, but SOMETHING like that

Answer 13

I got \[y'(t)= c_1(-4e^t)+c_2(2e^t-3t^3e ^{-t ^{3}})+(5e^t-3e ^{-t ^{3}}+2te ^{t ^{2}})\]

Answer 14

well, Im glad I was on the right track. So I plug in my initial values and simplify y(t) and y'(t) ... then, I'll have a system of quations that I can use to find c1 and c2.

Answer 15

Yep! :)
I am sure you can solve a system by now, yes?

Answer 16

yep, I just wasn't quite sure about all this, first time encountering non-homogenous and going between then was making me unsure. I can take it from here I think, but I really appreciate the help!

Answer 17

Yeah. Of course! Best of Luck! Cheers.