## suzi20 Group Title y" + 3y' + 2y = 6 y(0)=0 y'(0) = 2 with laplace 3 years ago 3 years ago

1. elecengineer Group Title

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

2. elecengineer Group Title

y' = sy - y(0)

3. elecengineer Group Title

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

4. elecengineer Group Title

5. suzi20 Group Title

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

6. math1234 Group Title

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

7. elecengineer Group Title

no, so you got y to be that

8. math1234 Group Title

Use particla fractions.

9. elecengineer Group Title

then you must use partial fractions

10. suzi20 Group Title

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

11. elecengineer Group Title

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

12. elecengineer Group Title

I created and went straight to wolframa for the partial fractions

13. elecengineer Group Title

cheated

14. elecengineer Group Title

partial fractions are very standard

15. elecengineer Group Title

then you convert them back

16. elecengineer Group Title

the 3/s goes to 3 from memory

17. elecengineer Group Title

and the other two are time shifted exponentials

18. math1234 Group Title

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

19. suzi20 Group Title

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?

20. elecengineer Group Title

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

21. elecengineer Group Title

s=0 lol

22. elecengineer Group Title

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

23. suzi20 Group Title

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?

24. elecengineer Group Title

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

25. elecengineer Group Title

you make no sense at all lol

26. elecengineer Group Title

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

27. suzi20 Group Title

could u help me?

28. elecengineer Group Title

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

29. elecengineer Group Title

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

30. suzi20 Group Title

matrix? really?

31. AnwarA Group Title

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

32. AnwarA Group Title

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

33. suzi20 Group Title

thank you anwara

34. AnwarA Group Title

You're welcome!