Question

y" + 3y' + 2y = 6 y(0)=0 y'(0) = 2
with laplace

Answer 1

y'' = sy^2 - sy(0) -y'(0)

Answer 2

y' = sy - y(0)

Answer 3

(sy^2 -2) +3(sy) +2y = 6

Answer 4

solve the quadratic

Answer 5

yes, im stuck in integral (6+2s) / s(s+2)(s+1)

Answer 6

First write everything in terms of the laplace transform.
THen solve the equation by converting back.

Answer 7

no, so you got y to be that

Answer 8

Use particla fractions.

Answer 9

then you must use partial fractions

Answer 10

6+2s = As + B(s+2)+C(s+1)
then?

Answer 11

= 3/s -4/(s+1) +1/(s+2)

Answer 12

I created and went straight to wolframa for the partial fractions

Answer 13

cheated

Answer 14

partial fractions are very standard

Answer 15

then you convert them back

Answer 16

the 3/s goes to 3 from memory

Answer 17

and the other two are time shifted exponentials

Answer 18

Laplace is all about matching and partial fractions, at least in solving simple ODE systems.

Answer 19

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?

Answer 20

1/(s-a) = e^(at) ( I googled this lol )

Answer 21

s=0 lol

Answer 22

remember you can pick any value for s, just that some values will make the simultaneous eqns alot easierto solve

Answer 23

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?

Answer 24

y= 3-4e^(-t) + e^(-2t)

Answer 25

you make no sense at all lol

Answer 26

this is why people need to pay attention in high school and first year uni maths course , so they absolutely hammer in the basics

Answer 27

could u help me?

Answer 28

its just simultaneous eqns , takes for ever, you need to set up a matrix etc.

Answer 29

it will take me like 10mins to type it up , I aint doing it lol

Answer 30

matrix? really?

Answer 31

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 \(y(t)=3-4e^{-t}+e^{-2t}\).

Answer 32

Hello suzi!! Does the answer make sense to you?

Answer 33

thank you anwara

Answer 34

You're welcome!