Question

Solve the first order equation y_(k+1)=ky_k+k^2 recursively, given the initial condition y_0=1.

Answer 1

\[y _{k+1}=ky _{k}+k ^{2}\]

Answer 2

initial condition \[y _{0}=1\]

Answer 3

hi friend, do you have any idea?

Answer 4

i try to use nonhomogeneous recurrence relation method...but i didnt get the right answer...

Answer 5

\[y _{k}=ky _{k-1}+k ^{2}\]
solve homo part:
\[y _{k}=ky _{k-1}\]
we get
\[r ^{1}=ky _{0}=k\]
\[r=k , r-k=0\]

thus,\[y _{k}^{h}=c _{1}k ^{k}\]

Answer 6

for nonhomo \[F(n)=k ^{2}\]
so F(n) in the form of equation
\[P(k)=ak ^{2}+bk+c\]

Answer 7

so we solve by letting \[P(k)=y _{k}\]

compare both side, then find value a, b and c

Answer 8

the i get the solution for non homo part

\[F(n)=\frac{ 2 }{ 5} k ^{2}+\frac{1}{ 5 }k\]

Answer 9

after combining and find \[c ^{1}\] still dont get correct answer...

Answer 10

hold on. I think the LHS has problem. the characteristic equation you get from the original one may be wrong.

Answer 11

I got L^2 -kL = 0 (homogenous part) ---> L (L -k) = 0 ---> L =0 and L =k.
Ok, sorry, end up at the same result, just different way. ok, go ahead

Answer 12

I assume that yours is right, so the combination of both homogenous and nonhomogenous part is?

Answer 13

y\[y _{k}=c _{1}k ^{k}+\frac{ 2 }{ 5 }k ^{2}+\frac{ 1 }{5 }k\]

find c1 by subtituting \[y _{1}=0\]
we get c1=-3/5

Answer 14

your initial condition is yo not y1?

Answer 15

yup...but you can find y1 by substitute k=0 in the original eq...

Answer 16

sory let k=1 and you get y2=0

Answer 17

y2=1 sorry...

Answer 18

no, not that, you don't have C yet, how can you claim the form and then using your claim to calculate something else?

Answer 19

My prof said that we have to calculate everything based on the given condition, claim the equation and use it to check for every values of k

Answer 20

from the original eq \[y _{k+1}=ky _{k}+k ^{2}\]

so i substitute k=1 then get y2=1

Answer 21

i find the value of y starting with k=0 then k=1 and so on...

Answer 22

don't forget yo means k = -1, should we step back one term? to get the series starts at k =0?

Answer 23

hurm...

Answer 24

Maybe I am wrong, but your k doesn't match with the y at all. if this way wrong, try another way. one more question, did you try generating function for this stuff?

Answer 25

It's never never simple, but what we study, try again friend, don't give up. we still have our profs, right?

Answer 26

when do you need this stuff? I can ask my prof via email. I will send you the answer right after I get it, is it ok?

Answer 27

this is my assingment...so my duedate is tommorow 4 pm....my lecturer just give us 1 day to finish it....

Answer 28

but anyway....thanks for ur support..... =) i will try finding another way....

Answer 29

hey..
this sounds straight forward!!

Answer 30

start with k = 0;
\(y_1=(0)y_0+(0)^2=0\)

k=1
\(y_2=(1)y_1+(1)^2=1\)

k=2
\(y_3=(2)y_2+(2)^2=6\)

k=3
\(y_4=(3)y_3+(3)^2=27\)
and so on

Answer 31

the given equation is itself the solution.
it is not given as a differential equation.

Answer 32

yes....but we need to find the solution in the form of y(k) or something...but not in recursive form...

Answer 33

so, "y" is a function of "k"??

Answer 34

i'm not sure wether k is a constant or an unknown...thats make me confuse....

Answer 35

should be \[y _{k+1} - ky _{k}=k ^{2}\]

Answer 36

oh yea

Answer 37

ok....then....

Answer 38

\[
y_{k+1}=ky_k+k^2\\\;\\
y_{k+1}=k[(k-1)y_{k-1}+(k-1)^2]+k^2=k(k-1)y_{k-1}+k^2+k\times (k-1)^2\\
\implies y_{k+1}=k(k-1)y_{k-1}+k[k+(k-1)^2]\\\;\\
y_{k+1}=k(k-1)[(k-2)y_{k-2}+(k-2)^2]+k[k+(k-1)^2]\\
\implies y_{k+1}=k(k-1)(k-2)y_{k-2}+k(k-1)(k-2)^2+k[k+(k-1)^2]\\
\implies y_{k+1}=k(k-1)(k-2)y_{k-2}+k[k+(k-1)[(k-1)+(k-2)^2]\\;\\
y_{k+1}=k!\times y_1+k\sum_{n=1}^kn\\\;\\
y_{k+1}=k!\times y_1+k\frac{k(k+1)}{2}
\]

Answer 39

@Mertsj

Answer 40

This is a question for the smart people.

Answer 41

did I make sense in the last one?

Answer 42

@Mertsj could you bring in the task force? :)

Answer 43

Try Phi or Satellite or Turing Test

Answer 44

@phi @ParthKohli calling all the backup!! heavy artillary

Answer 45

@hartnn @Callisto @Luis_Rivera Calling chopper backup here

Answer 46

@amistre64 @satellite73

Answer 47

@Mertsj they both offline!

Answer 48

But one never knows when they will show up and see their notifications.

Answer 49

@TuringTest

Answer 50

My base willl be destroyed by then!

Answer 51

@narsha what if we take the "Z" transform of this thing? it'd be way easier that way!

Answer 52

"Z" transform...meaning????

Answer 53

transform the system to complex plane.

Answer 54

??? kind of confusing....now im trying to use method of generating function because i stuck with nonhomogeneous recurrence relation.....

Answer 55

is there anyway to check the answer?

Answer 56

@amistre64 could you look at this pls

Answer 57

@electrokid , can i ask something about the answer that u just type above....how u generate from \[y _{k-2}\] to \[y _{1}\] ?????????

Answer 58

my bad. I did not use it. and deleted that post.

Answer 59

my logic starts right after that!

Answer 60

i think when u use back substitution is correct....but i dont get how u simplify from \[y _{k+1}=k(k−1)(k−2)yk−2+k[k+(k−1)[(k−1)+(k−2)2]\]

becomes \[y _{k+1}=k! \times y _{1}+k(\frac{ k(k+1) }{ 2 })\]

Answer 61

left part, as you keep substituting, you reach the k!

Answer 62

the second part, I am not sure. I thought it'd reach the artimatic series sum.

Answer 63

\[
y_{k+1}=k(k-1)(k-2)[(k-3)y_{k-3}+(k-3)^2]+k[k+(k-1)[(k-1)+(k-2)^2]\\
\quad=k(k-1)(k-2)(k-3)y_{k-3}+k[k+(k-1)[(k-1)+(k-2)^2+(k-2)(k-3)^2]\\
\quad=k(k-1)(k-2)(k-3)y_{k-3}+k[k+(k-1)[(k-1)+(k-2)[(k-2)+(k-3)^2]
\]

Answer 64

how can the second part be represented!!!!

Answer 65

@narsha do you see a pattern there?

Answer 66

\[\Large{
y_{k+1}=k(k-1)(k-2)(k-3)y_{k-3}+\\\qquad\quad\color{red}{k[k+(k-1)[(k-1)+(k-2)[(k-2)+(k-3)^2}}]
\]

Answer 67

\[
k^2+k(k-1)^2+k(k-1)(k-2)^2+k(k-1)(k-2)(k-3)^2+\dots
\]

Answer 68

i see that pattern....but at thelast answer, how you get \[y _{1}\] ???

Answer 69

ooooh
\[
=\sum_{n=1}^k (k-n)!(n)^2
\]

Answer 70

\[
y_{k+1}=k!\times y_1+\sum_{n=1}^k (k-n)!(n)^2
\]

Answer 71

noo. has to be rational.

Answer 72

\[
=\sum_{n=1}^k {k!\over n!}(n)^2
\]

Answer 73

yep. this satisfies good.

Answer 74

\[\boxed{\Large
y_{k+1}=k!\times y_1+\sum\limits_{n=1}^k \huge \frac{k!\times n^2}{n!}
}\]

Answer 75

I can do back substitution till y_1.
what happens when we do the next one to get y_0?

Answer 76

ermmm....\[y _{0}\] is already given as initial cond....so i dont think we need to find it....just go on with y_2 , y_3 and so on....

Answer 77

nono.
the general solution \(y_{k+1}\) should be explressed in terms of \(y_0\)

Answer 78

if we want to express \[y _{0}\] then we have to let k=-1.....but here k should start with 0.. cause when we substitute k in original eq

Answer 79

\[y _{k+1}=ky _{k}+k ^{2}\] we cant find it

Answer 80

\[y _{k+1}=ky _{k}+k ^{2}~:~y_0=1\]
\[y _{1}=0y _{0}+0 ^{2}\]
\[y _{2}=1(y _{0})+1^2\]
\[y _{3}=2(y _{0}+1^2)+2^2\]
\[y _{4}=3(2(y _{0}+1^2)+2^2)+3^2\]
\[y _{5}=4(3(2(y _{0}+1^2)+2^2)+3^2)+4^2 \]
\[y _{6}=5(4(3(2(y _{0}+1^2)+2^2)+3^2)+4^2)+5^2\\~~~~=5\cdot4\cdot3\cdot2\cdot y_0+5\cdot4\cdot3\cdot2\cdot 1^2+5\cdot4\cdot3\cdot2^2+5\cdot4\cdot3^2+5\cdot4^2+5^2 \]
\[y _{n}=(n-1)!\cdot y_0+(\frac{1(n-1)!}{0!} +\frac{2(n-1)!}{1!}+\frac{3(n-1)!}{2!}+...+\frac{(n-1)(n-1)!}{(n-2)!}) \]
\[y _{n}=(n-1)!~y_0+(n-1)!~(\frac{1}{0!} +\frac{2}{1!}+\frac{3}{2!}+...+\frac{(n-1)}{(n-2)!})~:~y_0=1\]
\[y _{n+1}=4~n!+n!~(\frac{3}{2!}+\frac{4}{3!}+...+\frac{n}{(n-1)!})\]

@electrokid if you havent gotten it by now, then i think your on the right track. Im sure i missed something while trying to keep track of the latex