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Find the general solution?

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2 Answers

The general solution(y0) of your homogeneous equation y" + 9y = 0 is

  y0 = C1sin3t + C2 cos 3t  (because, as you have already noticed, r = ±3i )    (1)

(we have discussed it several times in the past).

To find a particular solution for the inhomogeneous equation let' s rewrite it in the following way:

                         y" + 9y - 6 = t2e3t                         (2)     

If we introduce a new variable z = y - 2/3 we can write

                       z" + 9z = t2e3t                                (3)

We are going to look for a particular solution  for (3). Then a particular solution for y will be

                                             y1 = z + 2/3               (4)

Because the right side of the equation (3) is t2 e3t we can look for the solution in the form

             z = e3t (at2 +bt + c)                          (5)    (quadratic  expression times exponenet)  

Substitute (5) into (3). You will obtain:

     e3t[ 9at2 + (6a + 9b)t + 3b +9c +6at +2a +3b]     + 9e3t (at2 +bt +c)  = t2 e3t    

Now cancel alll terms by e3t. It gives you        

                   18at2 + (12a + 18b(t + (2a +6b + 18c) = t2             (6)

To make both sides of the equation (6) equal, we have to put

           18a = 1 

          12a + 18b = 0 

          2a + 6b + 18c = 0   

Solve the system and you will come up with answers: a = 1/18, b = -1/27  and c = 1/162. Thus

     z = e3t( t2/18 - t/27 +1/162)  = e3t(9t2 - 6t +1)/162 

and

     y = y0 + z + 2/3 = C1 sin(3t) + C2 cos (3t) + e3t(9t2 - 6t +1)/162  + 2/3   (general solution).

If you need more details let me know. Good luck!

r=±3i

Homogenerous solution: yo = c1cos 3t+c2sin 3t

Let y* = (At^2+Bt+C)e^3t + 6/9 be a particular solution. Plugging into the original equation,

y*' = [3(At^2+Bt+C) + 2At + B] e^3t

y*'' = [9(At^2+Bt+C) + 3(2At + B) + 3(2At+B) + 2A] e^3t

y*"+9y* = (t^2)(e^3t)+6 becomes

[18(At^2+Bt+C) + 3(2At + B) + 3(2At+B) + 2A] e^3t = (t^2)(e^3t)

Comparing coefficients:

t^2: 18A = 1 => A = 1/18

t: 18B+12A = 0 => B = -(2/3)(1/18) = -1/27

t^0: 18C+6B+2A = 0 => C = -(1/3)B - (1/9)A = 1/81 - 1/162 = 1/162

So, the general solution is

y = yo + y* = c1cos 3t+c2sin 3t+(1/162)(9t^2-6t+1)e^(3t)+2/3 <==Answer