## suzi20 4 years ago y" + 3y' + 2y = 6 y(0)=0 y'(0) = 2 with laplace

1. elecengineer

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

2. elecengineer

y' = sy - y(0)

3. elecengineer

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

4. elecengineer

5. suzi20

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

6. math1234

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

7. elecengineer

no, so you got y to be that

8. math1234

Use particla fractions.

9. elecengineer

then you must use partial fractions

10. suzi20

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

11. elecengineer

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

12. elecengineer

I created and went straight to wolframa for the partial fractions

13. elecengineer

cheated

14. elecengineer

partial fractions are very standard

15. elecengineer

then you convert them back

16. elecengineer

the 3/s goes to 3 from memory

17. elecengineer

and the other two are time shifted exponentials

18. math1234

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

19. suzi20

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

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

21. elecengineer

s=0 lol

22. elecengineer

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

23. suzi20

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

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

25. elecengineer

you make no sense at all lol

26. elecengineer

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

could u help me?

28. elecengineer

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

29. elecengineer

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

30. suzi20

matrix? really?

31. AnwarA

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

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

33. suzi20

thank you anwara

34. AnwarA

You're welcome!