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Method of Undetermined Coefficients
by Andrew Binder
February 23, 2012
Problem
...
Solution
...
The first step of the method of undetermined coefficients is to find
the homogeneous solution to the differential equation
...
The homogeneous solution satisfies
′′
′
yh + 6yh + 13yh = 0
...
We can solve this homogeneous equation using either the factoring operator method or the
characteristic equation method
...
The characteristic
equation for this second order differential equation is
r 2 + 6r + 13 = 0
...
The roots
of the characteristic equation are r = − 3 ± 2i
...
The second step in this method is to guess a particular solution of our given problem
...
(4)
Our guess for the particular solution depends on the functions on the right hand side of the
above equation
...
In our
case, the right hand side consists of trigonometric functions
...
In order to determine A and B, we will substitute
y p into Eq
...
Before I do so, let me calculate the derivatives of y p that
appear in Eq
...
Now, substitute y p and its derivatives into Eq
...
(6)
Simplify:
[9B + 12A]cos(2t) + [9A − 12B]sin(2t) = 9cos(2t) − 87sin(2t)
...
To get to that point, I set the coefficient of the sine function on the left hand side equal to the
coefficient of the sine function on the right hand side
...
Solving this sytem of linear equations, you will find that A = − 3 and B = 5
...
1
(8)
You can double check this result by substituting it into the particular solution’s differential
equation
...
(9)
In order to find C1 and C2, we will use our initial conditions
...
You
will get the wrong values for C1 and C2
...
(10)
The second initial condition requires a first derivative, so let’s calculate the first derivative of y:
y ′ = − 3e−3t[C1cos(2t) + C2sin(2t)] + e−3t[ − 2C1sin(2t) + 2C2cos(2t)] − 10sin(2t) −
6cos(2t)
...
(12)
Therefore, the solution to our initial value problem is
y = e−3t[4cos(2t) + 7sin(2t)] + 5cos(2t) − 3sin(2t)