## anonymous 4 years ago (D^2 + 9)y =5e^x - 162x i need to check if my answer is right.

1. anonymous

@AnimalAin this.

2. anonymous

Awesome! u did it write

3. anonymous

this is my answer: 2Ae^x + B + Cx = 5e^x -162x

4. anonymous

lets see...... AWESOME AWESOME :) thats right good job

5. anonymous

It's been a minute since I've done these, but I think the way is to consider the homogeneous solution by using an auxilliary (sp?) equation, then fit a particular solution that will match the RHS.

6. anonymous

lilmissmindset is right thats what i got for my answer

7. anonymous

That's the problem we did before.....

8. anonymous

yes yes, this is the first question i asked. it's just that the book says the general solution for this is $=C _{1} \cos 3x+ C _{2} \sin3x + \frac{ 1 }{ 2 } e ^{x} -18xy$

9. anonymous

there's no y after -18x SORRY

10. anonymous

Lets look at this in more detail.

11. anonymous

and our answer is 5/2 e^x -162x

12. anonymous

i thought, maybe there's something wrong with me, getting the solution after all

13. anonymous

$(D^2 + 9)y =5e^x - 162x$Consider$(D^2 + 9)y =0$Aux equation$r^2+9=0 \implies r= \pm 3i \implies y _{h}=C _{1} \cos 3x +C_2 \sin3x$OK so far?

14. anonymous

yeps. ok.

15. anonymous

ok with that part,

16. anonymous

Then we find the particular solution. Consider the first part of the RHS by itself, so$(D^2 + 9)y_p =5e^x \implies y_p=\frac{e^x}{2}$

17. anonymous

This is true since the derivative of the e^x term is always itself. That means the second derivative is the same as the original function. Added to the nine copies of the original function, we get 10ysubp=5e^x.

18. anonymous

owwww. i understand,

19. anonymous

Now consider the polynomial part. The second derivative of a linear function is zero, so we get$9y_{p2} =- 162x \implies y_{p2} =-19x$

20. anonymous

So we get something like$y=y_h+y_p=C_1\cos 3x+C_2\sin3x+\frac{e^x}{2}-19x$

21. anonymous

i forgot to multiply $y _{p}$ with 9 when i substitute, right? that's why i did ot get the answer,

22. anonymous

not*

23. anonymous

Actually, my division is horrible today.... should be 18 in the last term; exactly as your text had said.

24. anonymous

I hope you've got this; you're wearing me out....LOL

25. anonymous

actually, it's the same for the two of us. LOL hey, thanks a lot. you've been helping me large. Sorry for the stress,

26. anonymous

SORRY!

27. anonymous

i get it, i'm just confused on some parts.

28. anonymous

No stress. It's fun working with students that work hard and get it. Do math every day.

29. anonymous

i'll practice harder, :)

30. anonymous

I'm sure you'll get it.

31. anonymous

i hope i get it the soonest possible, my exam is tomorrow. O.O

32. anonymous

may i ask what book the problem referred?