suzi20
y" + 3y' + 2y = 6 y(0)=0 y'(0) = 2
with laplace
Delete
Share
This Question is Closed
elecengineer
Best Response
You've already chosen the best response.
0
y'' = sy^2 - sy(0) -y'(0)
elecengineer
Best Response
You've already chosen the best response.
0
y' = sy - y(0)
elecengineer
Best Response
You've already chosen the best response.
0
(sy^2 -2) +3(sy) +2y = 6
elecengineer
Best Response
You've already chosen the best response.
0
solve the quadratic
suzi20
Best Response
You've already chosen the best response.
0
yes, im stuck in integral (6+2s) / s(s+2)(s+1)
math1234
Best Response
You've already chosen the best response.
0
First write everything in terms of the laplace transform.
THen solve the equation by converting back.
elecengineer
Best Response
You've already chosen the best response.
0
no, so you got y to be that
math1234
Best Response
You've already chosen the best response.
0
Use particla fractions.
elecengineer
Best Response
You've already chosen the best response.
0
then you must use partial fractions
suzi20
Best Response
You've already chosen the best response.
0
6+2s = As + B(s+2)+C(s+1)
then?
elecengineer
Best Response
You've already chosen the best response.
0
= 3/s -4/(s+1) +1/(s+2)
elecengineer
Best Response
You've already chosen the best response.
0
I created and went straight to wolframa for the partial fractions
elecengineer
Best Response
You've already chosen the best response.
0
cheated
elecengineer
Best Response
You've already chosen the best response.
0
partial fractions are very standard
elecengineer
Best Response
You've already chosen the best response.
0
then you convert them back
elecengineer
Best Response
You've already chosen the best response.
0
the 3/s goes to 3 from memory
elecengineer
Best Response
You've already chosen the best response.
0
and the other two are time shifted exponentials
math1234
Best Response
You've already chosen the best response.
0
Laplace is all about matching and partial fractions, at least in solving simple ODE systems.
suzi20
Best Response
You've already chosen the best response.
0
6+2s = As + B(s+2)+C(s+1)
s=-1 --> 4 = -A+B
s=-2 --> 2 = -2A-C
for the last s what number should i choose?
elecengineer
Best Response
You've already chosen the best response.
0
1/(s-a) = e^(at) ( I googled this lol )
elecengineer
Best Response
You've already chosen the best response.
0
s=0 lol
elecengineer
Best Response
You've already chosen the best response.
0
remember you can pick any value for s, just that some values will make the simultaneous eqns alot easierto solve
suzi20
Best Response
You've already chosen the best response.
0
6+2s = As + B(s+2)+C(s+1)
s=-1 --> 4 = -A+B
s=-2 --> 2 = -2A-C
s=0 --> 6 = 2B+C
----------------------
4+A = B
6 = 2(4+A)+C
6 = 8 +2A+C
elimination
-2 = 2A+C
2 = -2A -C
infinity?
elecengineer
Best Response
You've already chosen the best response.
0
y= 3-4e^(-t) + e^(-2t)
elecengineer
Best Response
You've already chosen the best response.
0
you make no sense at all lol
elecengineer
Best Response
You've already chosen the best response.
0
this is why people need to pay attention in high school and first year uni maths course , so they absolutely hammer in the basics
suzi20
Best Response
You've already chosen the best response.
0
could u help me?
elecengineer
Best Response
You've already chosen the best response.
0
its just simultaneous eqns , takes for ever, you need to set up a matrix etc.
elecengineer
Best Response
You've already chosen the best response.
0
it will take me like 10mins to type it up , I aint doing it lol
suzi20
Best Response
You've already chosen the best response.
0
matrix? really?
AnwarA
Best Response
You've already chosen the best response.
1
I can see that you reach the point \(Y(s)=\frac{2s+6}{s(s+1)(s+2)}\). Now we should use partial fractions to write the expression in a form that can be easily to find its inverse laplace transform. That's
\[{2s+6 \over s(s+1)(s+2)}={a \over s}+{b \over s+1}+{c \over s+2}\]. Multiplying both sides by \(s(s+1)(s+2)\) gives:
\[2s+6=a(s+1)(s+2)+bs(s+2)+cs(s+1)\]
Plugging \(s=0\) gives \(2a=6 \implies a=3\); \(s=-1\) gives \(-b=4 \implies b=-4\), and \(s=-2\) gives \(2c=2 \implies c=1\). So, \(Y(s)=3\frac{1}{s}-\frac{4}{s+1}+\frac{1}{s+2}\). Hence \(y(t)=3-4e^{-t}+e^{-2t}\).
AnwarA
Best Response
You've already chosen the best response.
1
Hello suzi!! Does the answer make sense to you?
suzi20
Best Response
You've already chosen the best response.
0
thank you anwara
AnwarA
Best Response
You've already chosen the best response.
1
You're welcome!