Question

ODE, find the solution of the given ivp

Answer 1

\[y''-2y+y=te^t+4,~~~~y(0)=1,~~~y'(0)=1\] so this is the same as last time, but when I try \[y_p(t)=t^1e^t(A_0+A_1t) \] and replace it in the equation I don't quite get the right answer, maybe I'm just doing it wrong

Answer 2

\[y_p(t)=A_0te^t+A_1t^2e^t\]
\[y'_p(t)=A_0e^t+A_0te^t+2A_1te^t+A_1t^2e^t\] then finding the second derivative etc.. I seem to get \[A_1=\frac{ t }{ 2 }+2e^{-t}\] I don't really know what do here, maybe I'm thinking about this wrong

Answer 3

what is the homogeneous solutoin ?

Answer 4

should be \[y=c_1e^t+c_2te^t\]

Answer 5

so you're trying to solve particular solution for the right hand side, \(g(t) = te^t\) first.

since \(te^t\) is part of the homogeneous solution, you need to multiply your initial guess \(y_p=e^t(A_0+A_1t)\) by \(t\) and you're doint that...

Answer 6

Oh wait, so I don't do \[y_p=\color{red}t e^t(A_0+A_1t)\]

Answer 7

you have to do that

Answer 8

find the second derivative also and plug it in the equation ?

Answer 9

Yeah my works a messy, let me just redo it here, \[y_p(t) = A_0te^t+A_1t^2 e^t\]
\[y'_p(t)=A_0e^t+A_0te^t+2A_1te^t+A_1t^2e^t\]
\[y''_p(t) = 2A_0e^t+A_0te^t+2A_1e^t+4A_1te^t+A_1t^2e^t\] oh I should've simplified the first derivative a bit, but yes, I did plug in all the derivatives, so I have something nasty like this \[2A_0e^t+A_0te^t+2A_1e^t+4A_1+A_1t^2e^t-2(A_0e^t+A_0te^t+2A_1te^t+A_1t^2e^t)+...\]\[+A_0te^t+A_1t^2e^t=te^t+4\]

Answer 10

careful, we're working y_p for the right hand side \(te^t\) for now.

we will work the \(4\) later and put the solutions together

Answer 11

If you want to work the particular solution for entire right hand side, \(g(t) =te^t\color{red}{+4}\), at once,
the guess must be \(y_p =te^t(A_0+A_1t) \color{Red}{+ A_2} \)

Answer 12

get 3 equations by comparing the coefficients both sides and solve..
its pretty lengty, but definetely not hard right ?

Answer 13

The math isn't too bad, but I don't think I understand the guessing part, so what exactly did I do wrong, all of that should just equal te^t?

Answer 14

Ok wait, I think I get it, as what I was doing is to solve te^t, so all that should just equal te^t right, and then we can do a similar way for the 4, that will probably get the other coefficient

Answer 15

teamviewer ?

Answer 16

Haha I deleted that

Answer 17

\(A_0,A_1,A_2\) are constants here, we're expecting real numbers as solutions here

Answer 18

We can use twiddla though

Answer 19

can we talk on twiddla ?

Answer 20

Not sure, kai and I used to just write bunch of stuff on it

Answer 21

like paint

Answer 22

and yeah I know we want real numbers, i feel I have to use partial fraction decomposition kind of thing

Answer 23

okay lets just do it here and you don't need partial fractions or anything
you should be solving 3 linear equations

Answer 24

I'll take a pic and attach, wait

Answer 25

I'll download teamviewer again later to, kk

Answer 26

i tried solving...i got 0 for all co-eff.
I think we are supposed to multiply our initial guess by t^2,
if we multiply by t, we still have te^t term which is repeated in the homog. solution...

Answer 27

we get zero if the terms are repeated in the complementory solution

Answer 28

Ah, wow, that seems much easier than I thought, it makes a lot of sense, thanks so much @ganeshie8 !!

Answer 29

I thought of making it 2-3 pages but decided to save you from the distracting algebra ;)

Answer 30

So it seems with the A2 term, it's much easier

Answer 31

is it clear why we had to multiply \(t\) to our initial guess ?

Answer 32

Yeah, I see, and I know where I was making my mistakes to

Answer 33

why ?

Answer 34

i mean why we had to multiply the initial guess for \(te^t\) by \(t\) ?

just want to see your version of it

Answer 35

Well I just used the method of undetermined coeff for it, as I learnt using the cases \[\large y_p(x) = x^S e^{\alpha x}(A_0+A_1x+...+A_nx^n)\] but basically because our alpha is a simple solution of the characteristic equation, from what I can tell

Answer 36

It is a member of homogeneous solution, so obviously when you plugin, it makes the left hand side evaluate to 0.

It never gives you \(te^t\)

Answer 37

Yup! That's a much better way of explaining it haha, thanks a ton

Answer 38

So for higher order we do the same?

Answer 39

if we have to use this method

Answer 40

Never had to use this method for beyong 2nd order, but I don't see how it could fail...

Answer 41

As you might know, it only works for few simple right hand sides whose solutions form we can guess simply by looking at them..

Answer 42

Haha, well...I have to use it for y^(6) which will be interesting, I think it requires the wronskian

Answer 43

6 roots

Answer 44

whats the differential eqn ?

Answer 45

\[y^{(6)}+y'''=t\]

Answer 46

whats your guess for \(y_p \) ?

Answer 47

Oh it should be actually easier than this one looks like \[A_0+A_1t\]

Answer 48

It is easy to see that the DE has a minimum of 3rd deriviatve term, it kills the that y_p...

Answer 49

so we need to change our guess

Answer 50

since the minimum order of terms in the DE is 3, a better guess would be :
\(y_p = t^3(A_0+A_1t)\)

Answer 51

give it a try

Answer 52

plug it in the differential equation,
compare coefficients both sides and solve the constants

Answer 53

Ah, ok I think I'm starting to get this, will have to take derivatives up to 6

Answer 54

Yes, all easy derivatives

Answer 55

actually we just need to take the derivative 3 times
because it vanishes after that

Answer 56

\[y_p=A_0t^3+A_1t^4\]
\[y'_p=3A_0t^2+4A_1t^3\]\[y''_p=6A_0t+12A_1t^2\]\[y'''_p=6A_0+24A_1t\]\[y^{4}_p=24A_1\] Ha, well at least 5 and 6 are gone

Answer 57

\[0+(6A_0+24A_1t) =t \implies A_0 = 0, A_1 = \frac{ 1 }{ 24 }\]

Answer 58

Yep, essentially you have solved \(y^{'''}=t\)

Answer 59

Yes I see, now I just need to know how the homogenous equation looks like haha, probably something gross

Answer 60

\[R^3(R^3+1) = 0 \implies R^3(R+1)(R^2-R+1)=0\] ewww

Answer 61

It doesn't matter how gross it looks when we know that there is a definite method that works always for constant coefficients...

Answer 62

Agreed

Answer 63

just a polynomial equation, let wolfram solve it

Answer 64

Haha, ok but 3 roots are 0 so that's nice, but yeah, so in its entirety, I have to also pay attention to the order of the DE

Answer 65

This will take some time, I'll go over the undetermined coeff section sometime later and try to just figure out yp

Answer 66

for all the questions

Answer 67

and not necessarily solve them, but figure out the yp

Answer 68

I should start paying you/ sending you wine bottles thanks as always haha

Answer 69

LOL
@ganeshie8 nice hand writing :)

Answer 70

A0=4 A1=0 A2=1/6
\[y_p= 4te^t + 1/6\]
\[ -2y'= -8te^t -8e^t\]
\[ y''= 4te^t + 8e^t\]
add them: y''-2y'+y= 1/6

Answer 71

i'll just quote my text: we multiply our first guess by the least positive power of 'x' that suffices to eliminate duplication between terms of the trial solution and the complementary function.
\[y_c=c_1te^t + c_2e^t\]
\[y_p= A_0te^t +A_1t^2e^t + A_2\]
the first terms of both are a duplication

Answer 72

bump up the A0 and A1 terms by one more power of 't'
\[y_p=A_0t^2e^t +A_1t^3e^t+A2\]
substitute,
I got
\[y_p=\frac{ t^3e^t }{ 6 } + 4\]
which fits the differential equation

Answer 73

@Astrophysics @ganeshie8

Answer 74

we can quickly tell from inspection that A2 has to be 4 because y' and y'' will turn constants to zero

Answer 75

and because of e^t, A0 and A1 will never be constants no matter how much you differentiate

Answer 76

@baru looks i made a typo, it should be

A2=4 A1=0 A0=1/6

Answer 77

wait let me go thru again

Answer 78

my work has a mistake in the guess itself yeah, it should be multiplied by \(t^2\) as you're doing !
thanks for catching :)

Answer 79

:)

Answer 80

Ah ok, thanks @baru

Answer 81

\(\color{blue}{\text{Originally Posted by}}\) @baru
i'll just quote my text: we multiply our first guess by the least positive power of 'x' that suffices to eliminate duplication between terms of the trial solution and the complementary function.
\[y_c=c_1te^t + c_2e^t\]
\[y_p= A_0te^t +A_1t^2e^t + A_2\]
the first terms of both are a duplication
\(\color{blue}{\text{End of Quote}}\)

So is this a rule of thumb sort of? The method I used to get the guess didn't seem to work I guess, because I'm still thinking te^t(...)

Answer 82

yep,
lets have a look at our first guess...
\[y_p=A_0e^t+A_1te^t+A_2\]
\[y_c=c_1e^t + c_2te^t\]
notice that the A0,C1 and A1,C2 terms are duplications, so we need to multiply by t until that duplication is eliminated,


if we multiply by t, \[y_p=A_0te^t+A_1t^2e^t+A_2\]
we still have a duplication which is A0 term and C2 term
so we multiply again by t..

Answer 83

Ah ok, but the first one being a duplicate doesn't matter?

Answer 84

for your second example,
your first guess was A0+A1t,
but in your homogeneous soln, you have c1+c2t+c2t^2, so you have to multiply by t^3

Answer 85

A2 should be left alone,
its actually the superposition principle at work,
A0 and A1 terms are the guess for the 'te^t' term on RHS, and A2 is our guess for the '4' that occurs on RHS. any modifications we make, we consider them separately

Answer 86

which one are you talking about?

Answer 87

\[A_1t^2e^t \] and \[c_2te^t\]

Answer 88

they are not a duplication: A1 has t^2 and C2 has 't'

Answer 89

its a duplication if you can do this (C2+A1)( )

Answer 90

Oops never mind, neither of them are duplicates

Answer 91

Sorry about that

Answer 92

I was looking at this \[y_p=A_0e^t+A_1te^t+A_2 \] haha

Answer 93

:p
i guess we dont want duplicates, because then those terms are solutions to the homogeneous equation, thus will have to give zero when we substitute

Answer 94

That's very helpful to know, thanks a lot baru!!

Answer 95

i actually didn't know this...so I looked it up in the text and luckily noticed that tidbit about duplication xD

Answer 96

Haha, yeah I didn't read this section in the text, I probably should...as this is probably one of the confusing parts about DE