Second Order Differential Equations
Ordinary Differential Equations
Dr. Marco A Roque Sol
12/01/2015
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
Example 48
Find a particular solution for the following differential equation.
y 00 − 4y 0 − 12y = sin(2t)
Solution
In this case, since the right hand side has the sin(2t) ( or cos(2t) ),
then we will propose
YP = Acos(2t) + Bsin(2t)
YP0 = −2Asin(2t) + 2Bcos(2t);
Dr. Marco A Roque Sol
YP00 = −4Acos(2t) − 4Bsin(2t)
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
Plugging this into the differential equation and collecting like terms
gives
(−4Acos(2t) − 4Bsin(2t)) − 4(−2Asin(2t) + 2Bcos(2t))...
... − 12(Acos(2t) + Bsin(2t)) = sin(2t)
(−4A − 8B − 12A)cos(2t) + (−4B + 8A − 12B)sin(2t) = sin(2t)
−16A − 8B = 0
8A − 16B = 1
1
1
Solving this system gives us A = , B = −
40
20
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
Thus, a particular solution to the differential equation is then,
YP =
1
1
cos(2t) + − sin(2t)
40
20
Example 49
Find a particular solution for the following differential equation.
y 00 − 4y 0 − 12y = 2t 3 − t + 3
Solution
In this case, since the right hand side is a polynomial of degree 3,
then we will propose a generic third degree polynomial
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
YP = At 3 + Bt 2 + Ct + D
Does that make sense at all ? We have that if YP is a polinomial
of degree three, then its first and second derivatives are
polynomials of degrees two and one respectively. In this way the
combinantion y 00 + p(t)y 0 + y , gives us a third degree polynomial
with 4 unkwon coefficients that will be determined from the right
hand side of the equation.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
So, differentiate and plug into the differential equation
(6At + 2B) − 4(3At 2 + 2Bt + C )...
... − 12(At 3 + Bt 2 + Ct + D) = 2t 3 − t + 3
and the system we have to solve is
1
−12A = 2 =⇒ A = − ;
6
−12A − 12B = 0 =⇒ B =
1
6A − 8B − 12C = −1 =⇒ C = − ;
9
5
D=−
27
Dr. Marco A Roque Sol
1
6
2B − 4C − 12D = 3 =⇒
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
A particular solution for this differential equation is
1
1
1
5
YP = − t 3 + t 2 − t −
6
6
9
27
Now that we’ve gone over the three basic kinds of functions that
we can use undetermined coefficients on let’s summarize.
g (t)
ae βt
acos(βt) + bsin(βt)
an t n + an−1 t n−1 + ... + a0
Dr. Marco A Roque Sol
YP
Ae βt
Acos(βt) + Bsin(βt)
An t n + An−1 t n−1 + ... + A0
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
We now need move on to some more complicated functions. The
more complicated functions arise by taking products and sums of
the basic kinds of functions. Let’s first look at products
Example 50
Find a particular solution for the following differential equation.
y 00 − 4y 0 − 12y = te 4t
Solution
The guess for the polynomial function t would be
at + b
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
while the guess for the exponential function e 4t would be
ce 4t
Now, since we’ve got a product of two functions it seems like
taking a product of the guesses for the individual pieces
YP = ce 4t (at + b) = e 4t (cat + cb) = e 4t (At + B)
YP = e 4t (At + B)
that is, we just need two constants for the particular solution.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
Now that we’ve got our guess, let’s differentiate, plug into the
differential equation and collect like terms.
e 4t (16At + 16B + 8A) − 4e 4t (4At + 4B + A) − 12e 4t (At + B)
= te 4t
e 4t (16At + 16B + 8A) − 4e 4t (4At + 4B + A) − 12e 4t (At + B)
= te 4t
(16A − 16A − 12A)te 4t + (16B + 8A − 16B − 4A − 12B)e 4t = te 4t
−12Ate 4t + (4A − 12B)e 4t = te 4t
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
1
1
;B = −
12
36
A particular solution for this differential equation is then
−12A = 1;
(4A − 12B) = 0 =⇒ A = −
YP = e 4t (−
1
1
t
− ) = − e 4t (3t + 1)
12 36
36
Example 51
Write down the form of the particular solution to
y 00 + p(t)y 0 + q(t)y = g (t)
for the following g (t)0 s
a)g (t) = 16e 7t sin(10t)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
b)g (t) = (9t 2 − 103t)cos(t)
c)g (t) = −e −2t (3 − 5t)cos(9t)
Solution
a)g (t) = 16e 7t sin(10t). In this case we have
YP = ce 7t (asin(10t) + bcos(10t)) = e 7t (acsin(10t) + bccos(10t))
YP = e 7t (Asin(10t) + Bcos(10t))
b)g (t) = (9t 2 − 103t)cos(t). In this case we have
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
YP = (at 2 + bt + c) (dcos(t) + esin(t)) = (adt 2 + bdt + cd)cos(t) + ...
... + (aet 2 + bet + ce)sin(t)
YP = (At 2 + Bt + C )cos(t) + (Dt 2 + Et + F )sin(t)
c)g (t) = −e −2t (3 − 5t)cos(9t) In this case we have
YP = ae −2t (bt + c) (dcos(9t) + esin(9t)) = e −2t (abdt + acd)cos(9t) + .
... + e −2t (abet + ace)sin(9t)
YP = e −2t (At + B)cos(9t) + e −2t (Ct + D)sin(9t)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
Now, let’s take a look at sums of the basic components.
Theorem
If YP1 (t) is a particular solution for
y 00 + p(t)y 0 + q(t)y = g1 (t)
and if YP2 (t) is a particular solution for
y 00 + p(t)y 0 + q(t)y = g2 (t)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
then YP1 (t) + YP2 (t) is a particular solution for
y 00 + p(t)y 0 + q(t)y = g1 (t) + g2 (t)
Example 52
Find a particular solution for the following differential equation
y 00 − 4y 0 − 12y = 3e 5t + sin(2t) + te 4t
Solution
We look for a solution of the form
YP = Ae 5t + (Bsin(2t) + Ccos(2t)) + (Dt + E )e 4t
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
since we know the solution for each term with the same left
solution, we have that the particular solution is
3
1
1
1
YP = − e 5t − sin(2t) + cos(2t) − (3t + 1)e 4t
7
20
40
36
Example 53
Find a particular solution for the following differential equation.
y 00 − 4y 0 − 12y = e 6t
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
Solution
This is a particular case, because one of the solutions of the
associate homogeneous equation is y1 = e 6t since the
characteristic equation is r 2 − 4r − 12 = (r − 6)(r + 2) = 0 !!!!!
Therefore the method doesn’t work ( why? ). Now, following a
previous idea, we propose
YP = Ate 6t
Plugging this into our differential equation gives
y 00 − 4y 0 − 12y = (12Ae 6t + 36Ate 6t ) − 4(Ae 6t + 6Ate 6t ) − 12(Ate 6t ) =
(36A − 24A − 12A)te 6t ) + (12A − 4A)e 6t = e 6t
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
8Ae 6t = e 6t =⇒ A =
1
8
So, the particular solution in this case is
t
YP = e 6t
8
Example 54
Write down the guess for the particular solution to the given
differential equation. Do not find the coefficients.
a) y 00 + 3y 0 − 28y = 7t + e −7t − 1
b)
y 00 − 100y = 9t 2 e 10t + cost − tsint
c)
4y 00 + y = e −2t sin(t/2) + 6tcos(t/2)
d)
4y 00 + 16y 0 + 17y = e −2t sin(t/2) + 6tcos(t/2)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Second Order Differential Equations
Nonhomogeneous Equations; Method of Undetermined Coefficie
Nonhomogeneous Equations; Method of
Undetermined Coefficients
e)
y 00 + 8y 0 + 16y = e −4t + (t 2 + 5)e −4t
Solution
a) y 00 + 3y 0 − 28y = 7t + e −7t − 1 = (7t − 1) + e −7t
The general solution of the associate homogeneus equation
(y 00 + 3y 0 − 28y = 0 =⇒ r 2 + 3r − 28 = (r − 4)(r + 7) = 0
=⇒ r1 = 4, r2 = −7) is
y (t) = c1 e 4t + c2 e −7t
Thus, the form of the particular solution is
YP = (At + B) + Cte −7t
Dr. Marco A Roque Sol
Ordinary Differential Equations