Find the general solution of y"+9y=(t^2)(e^3t)+6.
r^2+9=0
r=±3i
Answer: c_{1}cos 3t+c_{2}sin 3t+(1/162)(9t^2-6t+1)e^(3t)+2/3
Please show the work step by step.
Find the general solution of y"+9y=(t^2)(e^3t)+6.
r^2+9=0
r=±3i
Answer: c_{1}cos 3t+c_{2}sin 3t+(1/162)(9t^2-6t+1)e^(3t)+2/3
Please show the work step by step.
The general solution(y_{0}) of your homogeneous equation y" + 9y = 0 is
y_{0} = C_{1}sin3t + C_{2} 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 = t^{2}e^{3t} (2)
If we introduce a new variable z = y - 2/3 we can write
z" + 9z = t^{2}e^{3t} (3)
We are going to look for a particular solution for (3). Then a particular solution for y will be
y_{1} = z + 2/3 (4)
Because the right side of the equation (3) is t^{2} e^{3t} we can look for the solution in the form
z = e^{3t} (at^{2} +bt + c) (5) (quadratic expression times exponenet)
Substitute (5) into (3). You will obtain:
e^{3t}[ 9at^{2} + (6a + 9b)t + 3b +9c +6at +2a +3b] + 9e^{3t} (at^{2} +bt +c) = t^{2} e^{3t}
Now cancel alll terms by e^{3t}. It gives you
18at^{2} + (12a + 18b(t + (2a +6b + 18c) = t^{2} (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 = e^{3t}( t^{2}/18 - t/27 +1/162) = e^{3t}(9t^{2} - 6t +1)/162
and
y = y_{0} + z + 2/3 = C_{1} sin(3t) + C_{2} cos (3t) + e^{3t}(9t^{2} - 6t +1)/162 + 2/3 (general solution).
If you need more details let me know. Good luck!
r=±3i
Homogenerous solution: y_{o} = 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 = y_{o} + y* = c1cos 3t+c2sin 3t+(1/162)(9t^2-6t+1)e^(3t)+2/3 <==Answer