Question

solve differential equation below....
y'+2y+1/s*L(f)=sin(t) where y(0)=0
i manage to get till y=(1/s^2+1)(s/s^2+1)
can someone confirm is correct ?? thanks a lot :)

Answer 1

can I get some brackets

Answer 2

what is L(f) ?

Answer 3

laplace?

Answer 4

ok yes, it's laplace, but your problem confuses me...

Answer 5

yup turingtest.....that clearer...

Answer 6

cuddlepony i attached a file on my post that will be easier to read the question...ty

Answer 7

oh wow, frankly I've never seen a DE that had a Laplace already in it
I'll be curious to see the solution too, sorry I doubt I can help
I'll try though :D

Answer 8

i see....but i check my notes that is equivalent to 1/s L(f) .....maybe we assume that correct...haha

Answer 9

@ash2326 can you try to solve this question for me ty....??

Answer 10

Let me try @angalc557

Answer 11

@angalc557 Is this your question?
\[(y'+2y+1) L(t)=\sin (t)\]
I'm sure there can't be 1/s or L(s) before taking Laplace Transform!!!

Answer 12

hmm the you can refer to the pic i attached to this post for clearer view....ty

Answer 13

Ok
we have
\[ y'+y +\int _{0} ^{t} y(\tau) d\tau= \sin t \]
Let's take Laplace tranform of this
\[ sy(s)-y(0)+y(s) +\frac{1}{s} y(s)= \frac{1}{s^2+1} \]
y(0)=0
so
\[ y(s)(s+1+\frac{1}{s}) =\frac{1}{s^2+1}\]
we get
\[y(s)( \frac{s^2+s+1}{s})=\frac{1}{s^2+1}\]
\[y(s)= \frac{s}{(s^2+s+1)(s^2+1)}\]
s can be written as
\[s=(s^2+s+1)-(s^2+1)\]
so
we have
\[y(s)= \frac{(s^2+s+1)-(s^2+1)}{(s^2+s+1)(s^2+1)}\]
now we get
\[y(s)=\frac{1}{(s^2+1)}-\frac{1}{(s^2+s+1)}\]
now
\[s^2+s+1=s^2+s+1/4-1/4+1=(s+1/2)^2-(\frac{\sqrt{3}}{2})^2\]
so we have now
\[y(s)=\frac{1}{(s^2+1)}-\frac{1}{(s+1/2)^2-(\frac{\sqrt{3}}{2})^2}\]
Now take inverse Laplace Transform
we get
\[\large{y(t)= \sin t- e^{-\frac{1}{2}t} \times \sin(\frac{\sqrt 3}{2} t)}\]

Answer 14

Sorry it 'd be
\[y(s)= \frac{1}{s^2+1}-\frac{1}{(s+1/2)^2+(\frac{\sqrt{3}}{2})^2}\]
so we'll get
\[\large{y(t)=\sin t -e^{-\frac{1}{2} t} sin(\frac{\sqrt3}{2} t)}\]

Answer 15

Sorry I mistakenly put a minus sign instead of a plus.
Did you get it though?

Answer 16

i thought hmm 2y is the question not y ?

Answer 17

Oh no:( Can you do it following the same method or should I do it again??

Answer 18

y(s)(s+2+1s)=1s2+1 the middle one become two right? just confirm for me?

Answer 19

Yeah , keep going

Answer 20

is it y=s/(s^2+1)(s/s^2+1)

Answer 21

first one is 1/(s^2+1)(s/s^2+1) right for y(s)=?

Answer 22

We have
\[y(s)[s+2+ \frac{1}{s}]= \frac{1}{s^2+1}\]
we get
\[ y(s)[\frac{s^2+2s+1}{s}]=\frac{1}{s^2+1}\]
so we get
\[ y(s)= \frac{s}{(s^2+2s+1)(s^2+1)}\]

Answer 23

i see alright agreed...then the part you said s can be written as how u get that ? the next step?

Answer 24

Now
\[s= \frac{1}{2}((s^2+2s+1)-(s^2+1))\]
We have
\[ y(s)= \frac{(s^2+2s+1)-(s^2+1)}{2(s^2+2s+1)(s^2+1)}\]
we get now
\[ y(s)= \frac{1}{2(s^2+1)}-\frac{1}{2(s^2+2s+1)}\]
we know
\[(s^2+2s+1)=(s+1)^2\]
so we get
\[ y(s)= \frac{1}{2(s^2+1)}-\frac{1}{2(s+1)^2}\]
Now taking inverse Laplace Transform
we get
\[ y(t)= \frac{1}{2}\sin t-\frac{1}{2}e^{-t} t\]
Using the following
\[ LT[t]=\frac{1}{s^2}\]
and
\[LT[e^{-t}t]=\frac{1}{(s+1)^2}\]

Answer 25

@angalc557 Did you understand it?

Answer 26

hmm maybe i not so good i cant figure out how u get the half there where it came from? could u explain sorry....

Answer 27

No need to be sorry @angalc557
we have s in the numerator
I want to write it as a difference of the terms in denominator
If I write \((s^2+2s+1)-(s^2+1)\) I'll get 2s
to get s, I have divided it by 2
\[\huge{\frac{(s^2+2s+1)-(s^2+1)}{2}=s}\]

Answer 28

y(t)=1/2 sin t - 1/2 te^-t right? i understand adi till the end.....ty a lot :)