## melmel 2 years ago linear D.E of nth order with constant coefficient...help guys .

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1. melmel

|dw:1381226444353:dw|

2. terenzreignz

This WILL be messy... are you up for it? :D

3. melmel

yes

4. terenzreignz

For these kinds of differential equations, the general form of the solution is given by $\Large f^{(h)}(x) + f^{(p)}(x)$ where the f^h is the general solution to the homogeneous equation while f^p is any particular solution of this equation. Can you find the general solution to the homogeneous equation?

5. melmel

$m ^{^{2}}=11$

6. terenzreignz

What does that mean? D:

7. melmel

$y=e ^{a}$

8. terenzreignz

Oh wait... there's no y' term?

9. melmel

get the higher order first D.E

10. terenzreignz

So basically, it's $\Large y'' - 11y = xe^x$?

11. melmel

$m ^{2} +1 -12 =0$

12. melmel

then m= square root of 11

13. melmel

then get the vlue of y

14. melmel

$y=C{1}^{+\sqrt{11}x}+C{2}^{-\sqrt{11}x}$

15. melmel

Im right? what is the next get the other y which from Xe^x

16. melmel

whic is y= Ae^u

17. melmel

take derivative until second derivative

18. melmel

ang get the value of A I stocked here

19. melmel

help guys if Im wrong :(

20. terenzreignz

Well, if it doesn't work with only one undetermined coefficient, did you try 2? Specifically this: $\Large y = \color{blue}Axe^x + \color{red}Be^x$

21. melmel

subst the value of Y" ,Y' and Y to the y′′+ y -12y

22. melmel

where did you get this y=Axex+Bex

23. melmel

@UnkleRhaukus

24. UnkleRhaukus

are you sure the second term in the question isn't a derivative?

25. terenzreignz

Odd, isn't it :D

26. terenzreignz

Oh by the way, I didn't actually "get" that Axe^x + Be^x, for all intents and purposes, it IS just a guess... see if it gets you the coefficients you want, since Axe^x alone doesn't seem to do it.

27. UnkleRhaukus

if not (yes strange), then i think you (@melmel ) got the complementary homogenous solution $\large f^h(x)=y_c(x)=C_{1}e^{+\sqrt{11}x}+C_{2}e^{-\sqrt{11}x}$ {forgot to type the e's in the bases }

28. UnkleRhaukus

Which method to calculate the particular solution $f^h(x)=y_p(x)$ are you going to use; undetermined coefficients, operator D , or variation of parameters, @melmel ?

29. terenzreignz

apparently offline :( Oh well, onwards and upwards XD

30. terenzreignz

^_^ thanks