Question

Last one for tonight, Find a particular solution for \[\large \frac{d^{2}y}{dx}-4\frac{dy}{dx}+9y=4\cos(x)\]

Answer 1

Have you considered solving the Characteristic Equation, \(r^{2} - 4r + 9 = 0\)?

Answer 2

yes I did and I got \[\large m=2 \pm \sqrt5 i\] @tkhunny

Answer 3

well in this case \[\large r=2 \pm \sqrt5 i\]

Answer 4

Okay, since they are NOT Real, what's the plan?

Answer 5

as both r's are complex conjugates, I would have to find a corresponding solution, problem is, I don't know which solution :/

Answer 6

\(2 + \sqrt{5}i\implies e^{2t}\cos(\sqrt{5}t)\)

\(2 - \sqrt{5}i\implies e^{2t}\sin(\sqrt{5}t)\)

Answer 7

okay, I see that would be taken from the table. so for a particular solution does that mean we can try, \[\large a*x*cos(kx)+b*x*sin(kx)\] ?

Answer 8

Well, you're not going to get \(cos(x)\) out of any of that. Are you SURE there IS a solution?

Answer 9

to be honest, I'm not 100% sure if there is one

Answer 10

the table always confuses me :S

Answer 11

If the only GENERAL solutions you are allowed are \(\cos(\sqrt{5}x)\;and\;\sin(\sqrt{5}x)\), I'm not seeing how we'll manage \(\cos(x)\).

Answer 12

I'm a little slow on that :( could you please clarify ?

Answer 13

\(\dfrac{d}{dx}\cos(\sqrt{5}x) = -\sqrt{5}\sin(\sqrt{5}x)\)

\(\dfrac{d^{2}}{dx^{2}}\cos(\sqrt{5}x) = -5\cos(\sqrt{5}x)\)

That pesky \(\sqrt{5}\) in the argument won't magically go away.

Answer 14

ohhh I see now, yes it will have to be incorporated through out the calculation :/

Answer 15

why did you second differentiate the \[\cos(\sqrt5 x)\] ?

Answer 16

oh never mind I see why

Answer 17

okay from having \[\large y'\] and \[\large y''\] where do we go from here?

Answer 18

Well, we know the general solution to the homogeneous equation. We have only to add sines and cosines to the general solution to find a solution to the nonhomogeneous equation.

I get the homogeneous solution shown above, with some parameters to generalize and \(\dfrac{2}{5}\cos(x)-\dfrac{1}{5}\sin(x)\)

Answer 19

so that is the homogeneous solution to this question, from here what are the steps to work it out? I can try it

Answer 20

No, the solution to the homogeneous equation is the monstrosity with the \(\sqrt{5}\) wild. The sine and cosine just shown is one solution to the nonhomogeneous equation. However, this is not all. I used only sine and cosine. With two derivatives, we should also include x*sin(x) and x^2(sin(x)) along with x*cos(x) and x^2(cos(x)). After that, it is a very tedious simultaneous solution of 6 equations.

Answer 21

....6 equations :O
:(
I've been studying straight for seven hours, a question with 6 equations is going to hurt >.<

Answer 22

if I had the solution from this I could work it out and how I got to the solution from there, but there is no solutions for this >.<

Answer 23

Well, it may turn out that the last four I mentioned are all zero, so don't lose hope.

Answer 24

I shall try as much as I can, what are the equations for this large question?

Answer 25

I used \(f(x) = A\cos(x) + B\sin(x) + Cx\cos(x) + Dx\sin(x) + Ex^{2}\cos(x) + Fx^{2}\sin(x)\)

Constructing \(\dfrac{d^{2}}{dx^{2}}f(x) - 4\dfrac{d}{dx}f(x) + 9f(x) = 4\cos(x)\), leads to the six equations. Like I said, it is tedious.

Answer 26

I see, hmm how do we get the six equations?

Answer 27

After expanding the massive expression on the left, group by the six function types:

cos(x)
sin(x)
x*cos(x)
x*sin(x)
x^2 * cos(x)
x^2 * sin(x)

Each of these will have a coefficient made up of various combinations of A, B, C, D, E, and F. All but the coefficient on cos(x) are set to zero (0), since they do not appear int he particular solution. The one attached to cos(x) is 4.

Answer 28

so in use, the five equation below cox(x) are equal to zero

Answer 29

that means we are just dealing with cos(x) ?

Answer 30

No, the coefficients on the types are all zero. We are NOT just dealing with cos(x). We are dealing with anything that can result in a cos(x) up to the 2nd derivative.

Briefly, if the particular solution is f(x) = Acos(x) + Bsin(x), we have

f"(x) = -Acos(x) - Bsin(x)
f'(x) = -Asin(x) + Bcos(x)
f(x) = Acos(x) + Bsin(x)

Giving:

f"(x) - 4f'(x) + 9f(x) =
(-Acos(x) - Bsin(x))-4(-Asin(x) + Bcos(x))+9(Acos(x) + Bsin(x)) = 4cos(x)

Collecting: cos(x)[-A - 4B + 9A] + sin(x)[-B + 4A + 9B] = 4cos(x)

For equality of the expressions, we must have:

[-A - 4B + 9A] = 8A - 4B = 4 ==> 4A - 2B = 2
[-B + 4A + 9B] =4A + 8B = 0 ==> A + 2B = 0

This leads to 5A = 2 ==> A = 2/5 and finally B = -1/5, which is the solution I gave above.

Answer 31

oh my gosh, I totally understand that!!! :D

Answer 32

The trouble is wondering if that f(x) is sufficiently general.

Do we need \(h(x) = f(x) + Cx\cos(x) + Dx\sin(x)\)?
Do we need \(r(x) = h(x) + Ex^{2}\cos(x) + Fx^{2}\sin(x)\)?

Note that the 1st and 2nd derivatives of h(x) and r(x) result in additional terms consisting of only cos(x). This may contribute to a more general solution.

Answer 33

ah and that's where I get down :/

Answer 34

and also, isn't \[\frac{2}{5}cos(x)-\frac{1}{5}sin(x)\] the final answer?? Because I just arrived at that step from my calculations

Answer 35

No worries. It's not massively difficult to demonstrate that you do not actually need these additional complications.

IF we use just \(h(x) - f(x) = Cx\cos(x) + Dx\sin(x)\), because of symmetry we get the lovely result \(C = D = 0\). The same thing happens with \(r(x) - h(x) = Ex^{2}\cos(x) + Fx^{2}\sin(x)\). We get \(E = F = 0\) and I think we're done!

We do NOT need to try anything with greater exponents, \(Gx^{3}\cos(x)\), for example. You might want to ponder why this is so.

Now, go get some sleep!

Answer 36

oh my gosh, do you mean the final answer is \[\large \frac{2}{5}cos(x)-\frac{1}{5}sin(x)\] ???

Answer 37

I think you left of the solution to the homogeneous equation. Other than that, yes.

Answer 38

thank you soo very much @tkhunny what a great help you are :D yes I shall sleep now :D