Answer:
y"-6y'+18y=0
Second order
Step-by-step explanation:
Since there are 2 constants, the order of the differential equation will be 2. This means we will need to differentiate twice.
y = e^(3x) (acos3x +bsin3x)
y'=3e^(3x) (acos3x+bsin3x)
+e^(3x) (-3asin3x+3bcos3x)
Simplifying a bit by reordering and regrouping:
y'=e^(3x) cos3x (3a+3b)+e^(3x) sin3x (3b-3a)
y"=
3e^(3x) cos3x (3a+3b)+-3e^(3x) sin(3x) (3a+3b)
+3e^(3x) sin3x (3b-3a)+3e^(3x) cos(3x) (3b-3a)
Simplifying a bit by reordering and regrouping:
y"=
e^(3x) cos3x (9a+9b+9b-9a)
+e^(3x) sin3x (-9a-9b+9b-9a)
Combining like terms:
y"=
e^(3x) cos3x (18b)
+e^(3x) sin3x (-18a)
Let's reorder y like we did y' and y".
y = e^(3x) (acos3x +bsin3x)
y=e^(3x) cos3x (a) + e^(3x) sin3x (b)
Objective is to find a way to combine or combine constant multiples of y, y', and y" so that a and b are not appearing.
Let's start with the highest order derivative and work down
y"=
e^(3x) cos3x (18b)
+e^(3x) sin3x (-18a)
We need to get rid of the 18b and 18a.
This is what we had for y':
y'=e^(3x) cos3x (3a+3b)+e^(3x) sin3x (3b-3a)
Multiplying this by -6 would get rid of the 18b and 18a in y" if we add them.
So we have y"-6y'=
e^(3x) cos3x (-18a)+e^(3x) sin3x (-18b)
Now multiplying
y=e^(3x) cos3x (a) + e^(3x) sin3x (b)
by 18 and then adding that result to the y"-6y' will eliminate the -18a and -18b
y"-6y'+18y=0
Also the characteristic equation is:
r^2-6r+18=0
This can be solved with completing square or quadratic formula.
I will do completing the square:
r^2-6r+18=0
Subtract 9 on both sides:
r^2-6r+9=-9
Factor left side:
(r-3)^2=-9
Take square root of both sides:
r-3=-3i or r-3=3i
Add 3 on both sides for each:
r=3-3i or r=3+3i
This confirms our solution.
Another way to think about the problem:
Any differential equation whose solution winds up in the form y=e^(px) (acos(qx)+bsin(qx)) will be second order and you can go to trying to figure out the quadratic to solve that leads to solution r=p +/- qi
Note: +/- means plus or minus
So we would be looking for a quadratic equation whose solution was r=3 ×/- 3i
Subtracting 3 on both sides gives:
r-3= +/- 3i
Squaring both sides gives:
(r-3)^2=-9
Applying the exponent on the binomial gives:
r^2-6r+9=-9
Adding 9 on both sides gives:
r^2-6r+18=0
