Question

Hi everyone! Could someone please read the question I am posting and tell me why they say that P(t)=0?

Shouldn't P(t)=1 instead?

And also, can you please try and explain why either they are correct or why I am correct?

By the way, I don't need anyone to help me solve the problem but if you would care to discuss the theory I would like that. Thanks! :o)

Answer 1

I also think that p(t)=1

Answer 2

Hi Michele! Well that's good to hear. But can you explain why we both think that?
My best reason is the if p(t) is a generic place holder, and nothing is showing there, there still is an implied "1" as the function of t sitting in front of y''....
Can you say it in any other way?

Answer 3

Thanks @SinginDaCalc2Blues I think that p(t) is the coefficient of the second order derivative of y(t)

Answer 4

can you post your Theorem 3.2.1?

Answer 5

Yes, I agree. I think I will leave the question open for awhile in case someone disagrees and would like to explain why.

Answer 6

ok!

Answer 7

please wait, can p(t) be the coefficient of the first order derivative y(t)?

Answer 8

I don't have the theorem...this is from another school

Answer 9

https://math.berkeley.edu/~ogus/old/Math_54-05/HW%20solutions/homework408.pdf

Answer 10

I think that can be this:
\[y'' + p\left( t \right)y' + q\left( t \right)y = g\left( t \right)\]

Answer 11

actually, I think they are referring to:
p(x)y''+q(x)y'+r(x)y=g(x)
at least that the general assumption in my own class

Answer 12

I mean I suppose you might be able to make the argument that since there is no "function of x" in front of y'', then maybe you can say since it isn't there they said p(t)=0
But I really don't know how to honestly justify that statement...doesn't make much sense since to me, there is an "implied one" there

Answer 13

yes! Since we are studying the second order differential equations, we can consider your p(x) a non null polynomial, so we can divide both sides of your equation by p(x) and we get:
\[y'' + \frac{{q\left( x \right)}}{{p\left( x \right)}}y' + \frac{{r\left( x \right)}}{{p\left( x \right)}}y = \frac{{g\left( x \right)}}{{p\left( x \right)}}\]

Answer 14

and we can write this:
\[y'' + P\left( x \right)y' + Q\left( x \right)y = G\left( x \right)\]

Answer 15

interesting...but since p(t) divided by p(t) = one
p(t) should NOT equal ZERO correct?

Answer 16

yes! correct!

Answer 17

i see what you did...you just renamed the functions of x
you just have to be careful with that unless you note it because we have already used p(x) for example and it gets confusing :o)

Answer 18

you are right!

Answer 19

so then I think we may be able to conclude that the people at Berkeley who wrote that are wrong and dumb then? :o)

Answer 20

if we start from this equation:
\[p\left( x \right)y'' + q\left( x \right)y' + r\left( x \right)y = g\left( x \right)\]
then divdidng both sides by p(x), we get:
\[y'' + P\left( x \right)y' + Q\left( x \right)y = G\left( x \right)\]
where:
\[\begin{gathered}
P\left( x \right) = \frac{{q\left( x \right)}}{{p\left( x \right)}} \hfill \\
Q\left( x \right) = \frac{{r\left( x \right)}}{{p\left( x \right)}} \hfill \\
G\left( x \right) = \frac{{g\left( x \right)}}{{p\left( x \right)}} \hfill \\
\end{gathered} \]

Answer 21

yes that is clear!

Answer 22

capital letters!

Answer 23

yes!

Answer 24

So when they say "the largest interval containing 1 on which p,q
and g are defined and continuous"

By "1" they mean g(t) right?

Answer 25

I don't think so

Answer 26

so you don't think they were referring to the entire 2nd order differential equation which also equals "1" ?

Answer 27

I think that 1 refers to the point x=1, since the function and its first derivative are evaluated at x=1 as initial conditions

Answer 28

okay, let me think about that for a second...

Answer 29

for example, think about the theorem of existence and uniqueness of the solution of a differential equation

Answer 30

I was actually trying to find the theorem of existence and uniqueness of the solution of a differential equation in my book to better understand it and my book is crazy and I haven't found it yet :o(

Answer 31

but yes keep going

Answer 32

please wait I'm going to find my mathematics textbook

Answer 33

remind me to ask you my "super-duper-insightful-question" after you are done explaining...I think it might be important, I just don't wanna forget to ask! :o)