Question 1115567
Using Laplace transforms: <br>

 L(y'') = {{{ s^2 }}}Y(s) - sy(0) - y'(0) 
 L(y') =  sY(s) - y(0)
 L(c) =  c/s      (c=a constant) <br>

 Noting y'(0)=y(0)=0, the Laplace transform is:

  {{{  s^2Y(s) - 3sY(s) = 9/s }}}<br>

Now we "just" need to isolate Y(s) and then take the inverse Laplace transform.
 
   {{{ Y(s) =  (9/(s*(s^2-3s))) }}}
   {{{  Y(s) =  (9/(s^2*(s-3))) }}} <br>

Use partial fraction expansion to get the right hand side into a form in which the inverse Laplace can be taken:
    {{{  (9/(s^2*(s-3))) = A/s + B/s^2 + C/(s-3) }}}   (1)

Multiply both sides by {{{ s^2(s-3) }}} :

    {{{  9 =  A*(s*(s-3)) + B*(s-3) + C*s^2 }}}
    {{{  9 =  (As^2 - 3As) + (Bs - 3B) + Cs^2 }}}
    
From this you get three equations in three unknowns:
           A+C = 0     (from the {{{s^2}}} terms)
           -3A+B = 0    (from the {{{s^1}}} terms)
           -3B = 9        (from the {{{s^0}}} terms)

—>  B = -3  —> A= -1   —>  C = 1  <br>
            
So we can write (1) as:

   {{{  Y(s) = -1/s + -3/s^2 + 1/(s-3)  }}}

Taking the inverse Laplace gives:
   {{{ highlight(  y(x)  =  -1 -3x + e^(3x) ) }}}

——

Check:  
Initial conditions:
y(0) = -1-3*0+e^(0) = -1 + 1 = 0 (ok)
y'(x) = -3 + 3e^(3x) and y'(0) = -3 + 3e^(0) = -3+3 = 0  (also ok)

Entire equation:
  y''(x) = 9e^(3x)

   y'' - 3y' = 9e^(3x) - 3(-3+3e^(3x)) = 9e^(3x) + 9 -9e^(3x) = 9  (ok)

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My answer is the general solution.   A particular solution is often guessed at the start, and then combined with the homogeneous solution (i.e. particular solution would be a function y(x) that satisfies y''-3y' = 9 while the homogenous ("complementary") solution would satisfy y''-3y' = 0 and you  add the two solutions together to get the general solution.  I don't know how to guess a proper particular solution for this problem.  One could guess y(x) = Ae^(kx) + Bxe^(mx) + C,  I suppose, but I wouldn't know to guess that without seeing the general solution first.