Find fourth order differential equation

kk1995
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Homework Statement


Suppose that a fourth order differential equation has a solution y=-9e^(3x)xcos(x).

a. Find such differential equation, assuming that it is homogeneous and has constant coefficients.

b. Find the general solution to this differential equation. x is the independent variable.

The Attempt at a Solution


I tried to find the derivatives of y (y',y'',y''',y'''').
Then, I placed unknown constants (c1, c2, c3, c4, c5) on each order and made a matrix with each row having coefficients from values containing e^(3x)cosx, e^(3x)xcosx, e^(3x)sinx, and e^(3x)xsinx. I made the columns all have the same order (such as all values from y'' on one column). I also made c1=1 since otherwise the matrix becomes 4*5 and that cannot be solved for the unknown constants. However, this did not work out.

I would sincerely appreciate your help. I mean, I can do those functions that have y=e^rx and y=cos(x), but combination of these makes my head ache.
 
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Have you tried writing cos (x) in terms of complex exponentials?
 
Think about what the solution whose DE had this characteristic equation$$
(r-a)^2+b^2=0$$would be.
 
kk1995 said:

Homework Statement


Suppose that a fourth order differential equation has a solution y=-9e^(3x)xcos(x).

a. Find such differential equation, assuming that it is homogeneous and has constant coefficients.

b. Find the general solution to this differential equation. x is the independent variable.

The Attempt at a Solution


I tried to find the derivatives of y (y',y'',y''',y'''').
Then, I placed unknown constants (c1, c2, c3, c4, c5) on each order and made a matrix with each row having coefficients from values containing e^(3x)cosx, e^(3x)xcosx, e^(3x)sinx, and e^(3x)xsinx.
Great! Or course, in order to have e^{3x}cos(x) and e^{3x}sin(x) as solutions, the characteristic equation must have roots 3+ i and 3- i and so must have (x- 3- i) and (x- 3+ i) as factors. In order to have each of those times x as solutions, those must be double roots so the characteristic equation must be of the form (x- 3- i)^2(x- 3+ i)^2= [(x- 3- i)(x- 3+ i)]^2=[(x- 3)^2+ 1]^2= (x^2- 6x+ 10)^2= x^4- 12x^3+ 56x^2- 12x+ 100.

You can get the differential equation itself from that.

I made the columns all have the same order (such as all values from y'' on one column). I also made c1=1 since otherwise the matrix becomes 4*5 and that cannot be solved for the unknown constants. However, this did not work out.

I would sincerely appreciate your help. I mean, I can do those functions that have y=e^rx and y=cos(x), but combination of these makes my head ache.
"columns"? "matrix"? Too advanced for me!
 
Thank you all for your replies. All of your replies really helped. The first two replies led me to Euler's formula, while the third one by HallsofIvy led the rest of the problem from the point of getting the values of r using Euler's formula. Note: the -12x should be -120x for the answer, just a point out at a typo. :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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