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

Mathematics
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y'' = sy^2 - sy(0) -y'(0)
y' = sy - y(0)
(sy^2 -2) +3(sy) +2y = 6

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Other answers:

solve the quadratic
yes, im stuck in integral (6+2s) / s(s+2)(s+1)
First write everything in terms of the laplace transform. THen solve the equation by converting back.
no, so you got y to be that
Use particla fractions.
then you must use partial fractions
6+2s = As + B(s+2)+C(s+1) then?
= 3/s -4/(s+1) +1/(s+2)
I created and went straight to wolframa for the partial fractions
cheated
partial fractions are very standard
then you convert them back
the 3/s goes to 3 from memory
and the other two are time shifted exponentials
Laplace is all about matching and partial fractions, at least in solving simple ODE systems.
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?
1/(s-a) = e^(at) ( I googled this lol )
s=0 lol
remember you can pick any value for s, just that some values will make the simultaneous eqns alot easierto solve
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?
y= 3-4e^(-t) + e^(-2t)
you make no sense at all lol
this is why people need to pay attention in high school and first year uni maths course , so they absolutely hammer in the basics
could u help me?
its just simultaneous eqns , takes for ever, you need to set up a matrix etc.
it will take me like 10mins to type it up , I aint doing it lol
matrix? really?
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 \(y(t)=3-4e^{-t}+e^{-2t}\).
Hello suzi!! Does the answer make sense to you?
thank you anwara
You're welcome!

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