The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
III Second Order DE
§3.5 Nonhomogeneous LSODEs
Method of Undetermined Coefficients
Satya Mandal, KU
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
SODEs
◮
Recall, second order DE (SODE) has the form
d 2y
dy
= f t, y ,
dt 2
dt
(1)
This is also written as
y ” = f (t, y , y ′ )
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Nonhomogeneous LSODE
Now, we consider and solve nonhomogeneous LSODEs.
◮ A nonhomogeneous linear SODE (LSODE), can be
written as:
L(y ) = y ” + p(t)y ′ + q(t)y = g (t)
◮
(2)
where p(t), q(t), g (t) are functions of t.
The corresponding homogeneous DE
L(y ) = y ” + p(t)y ′ + q(t)y = 0
(3)
would play a significant role in solution of (2).
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
◮
Another form of LSODE (2) is:
L(y ) = P(t)y ” + Q(t)y ′ + R(t)y = G (t)
◮
where P(t), Q(t), R(t), G (t) are functions of t.
As in the last frame, corresponding homogeneous DE:
L(y ) = P(t)y ” + Q(t)y ′ + R(t)y = 0
◮
(4)
(5)
Target: We consider nonhomogeneous linear DEs, with
constant coefficients only. We use §3.1, 3.3, 3.4.
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Role of the Homogeneous DE
The role of the corresponding homogeneous equation:
Theorem 3.5.1
◮ Suppose Y1 , Y2 are two solutions of the nonhomogeneous
LSODE (2) L(y ) = g (t).
◮ Then, the difference Y1 − Y2 is a solution of
corresponding homogeneous equation (3) L(y ) = 0.
◮ Proof. L(Y1 ) = g (t), L(Y2 ) = g (t) =⇒
L(Y1 − Y2 ) = L(Y1 ) − L(Y2 ) = g (t) − g (t) = 0.
◮
Same is true for the other form of (4) L(y ) = G (t).
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
The General Solution
Theorem 3.5.2
◮ Suppose Y is a solution of the equation nonhomogeneous
LSODE (2) L(y ) = g (t) [likewise the other form (4)].
◮ Suppose y1 , y2 is a fundamental set of solutions of the
corresponding homogeneous equation (3) L(y ) = 0.
◮ Then, the general solution of (2) [likewise of (4)] is:
y = ϕ(t) = c1 y1 (t) + c2 y2 (t) + Y (t)
(6)
where c1 , c2 are arbitrary constants. Textbook uses the
notation yc = c1 y1 (t) + c2 y2 (t).
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Method(s) of undetermined coefficients
◮
◮
◮
We solve some nonhomogeneous DEs (2).
We consider only those, for which the homogeneous part
L(y ) = ay ” + by ′ + cy = 0 has constant coefficients.
By theorem 3.5.2, to solve DE L(y ) = g (t), we need:
◮
◮
A pair of fundamental solutions of the homogeneous
equation L(y ) = 0.
We also need a particular solution Y (t).
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Continued
◮
◮
◮
The form of a particular solution Y (t) is known, when
g (t) has certain convenient form. In this §3.5, we use
these forms and determine the "coefficients". It is called
the methods of undetermined coefficients.
I will deal with variety of forms of g (t).
We only consider equations of the form:
L(y ) = ay ” + by ′ + cy = g (t)
Satya Mandal, KU
(7)
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
When g (t) is a polynomial
◮
◮
Consider the DE (7) and suppose
g (t) = Pn (t) = a0 t n + a1 t n−1 + · · · + an is a polynomial
of degree n (i.e. a0 6= 0).
So, (7) has the form
L(y ) = ay ” + by ′ + cy = a0 t n + a1 t n−1 + · · · + an (8)
◮
Then, if (c 6= 0), a particular solution of (8) has the form
Y (t) = A0 t n + A1 t n−1 + · · · + An
◮
(9)
where A0 , . . . , An can be determined, as follows.
Minor adjustment is needed, when c = 0.
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
Sample PI: Ex 3 (edited)
Consider the IVP:

 y ” − y ′ − 2y = −2t + 4t 2
y (0) = 0
 ′
y (0) = 0
Solve this IVP and give the graph of the solution.
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
Solution: Compute Y (t)
Major Step I: Compute Y (t)
◮ Here g (t) = −2t + 4t 2 is a polynomial of degree 2.
◮ By (9), we have
Y (t) = At 2 + Bt + C . So, Y ′ = 2At + B, Y ” = 2A
◮
Substituting in the DE:
2A − (2At + B) − 2(At 2 + Bt + C ) = −2t + 4t 2 =⇒
(−2A)t 2 + (−2A − 2B)t + (2A − B − 2C ) = −2t + 4t 2
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
Continued
◮
Equating coefficients of t 2 , t, t 0 :
−2A = 4,
◮
◮
− 2A − 2B = −2,
2A − B − 2C = 0
They solve easily. However, I am giving the matrix form:

 


4
A
−2 0
0
 −2 −2 0   B  =  −2 
0
C
2 −1 −2
So, A = −2, B = 3, C = −3.5
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
Continued
◮
So, a particular solution is
Y (t) = At 2 + Bt + C = −2t 2 + 3t − 3.5
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
Solve the homogeneous DE
Second Major Step: Solve the homogeneous DE
L(y ) = y ” − y ′ − 2y = 0
◮ The CE: r 2 − r − 2 = 0
√
◮ So, r = 1± 1+8 , r1 = −1, r2 = 2
2
◮ So, two solutions of L(y ) = 0 are y1 = e −t , y2 = e 2t
◮ So, a general solution of L(y ) = 0 is yc = c1 e −t + c2 e 2t
◮ Remark. As in §3.3 3.4, in this case of distinct roots also,
y1 , y2 form a fundamental set. One can check directly,
that the Wronskian W (y1 , y2 ) 6= 0.
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
The general solution
◮
The general solution of the nonhomogeneous equation is:
y = yc + Y = c1 e −t + c2 e 2t − 2t 2 + 3t − 3.5
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
Major Step III: Determine c1, c2
◮
First, we compute y ′ :
y ′ = −c1 e −t + 2c2 e 2t − 4t + 3
◮
The initial conditions:
y (0) = c1 + c2 − 3.5 = 0
c1 =
=⇒
′
c2 =
y (0) = −c1 + 2c2 + 3 = 0
Satya Mandal, KU
20
6
1
6
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
The Answer
◮
The general of the nonhomogeneous equation is:
y=
20 −t 1 2t
e + e − 2t 2 + 3t − 3.5
6
6
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
Case: g (t) = e αt Pn (t)
◮
Suppose in DE (7), g (t) = e αt Pn (t) is product of an
exponential function and a polynomial, of degree n. So,
ay ” + by ′ + cy = e αt (a0 t n + a1 t n−1 + · · · + an )
◮
Assume α is not a root of th CE (i.e. aα2 + bα + c 6= 0).
Then, a particular solution of (10) has the form
Y (t) = e αt (A0 t n + A1 t n−1 + · · · + An )
◮
(10)
(11)
Minor adjustment is needed, in case aα2 + bα + c = 0.
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
◮
◮
◮
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
To establish (11), write Y (t) = e αt u(t). We will see that
DE (10) reduces to a DE in u, of form in case 1.
So, Y ′ = e αt (αu + u ′ ), Y ” = e αt (u” + 2αu ′ + α2 u).
So, (10) reduces to: L(Y ) = L (e αt u) =
a[e αt (u”+2αu+α2 u)]+b[e αt (αu+u ′ )]+ce αt u(t) = e αt Pn (t)
◮
So,
au” + (2α + b)u ′ + (aα2 + bα + c)u = Pn (t),
(12)
which is, as in case 1.
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
With Trig Functions
◮
Suppose
g (t) =
e αt Pn (t) cos βt
e αt Pn (t) sin βt
So, we have:
′
ay ” + by + cy =
e αt (a0 t n + a1 t n−1 + · · · + an ) cos βt
e αt (a0 t n + a1 t n−1 + · · · + an ) sin βt
(13)
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
Case 3: Continued
◮
Assume z := α + iβ is not a root of th CE (i.e.
az 2 + bz + c 6= 0). Then, a particular solution of (13) has
the form
Y (t) = e αt [(A0 t n + A1 t n−1 + · · · + An ) cos βt]
e αt [+(B0 t n + B1 t n−1 + · · · + Bn ) sin βt]
◮
(14)
Minor adjustment is needed, in case az 2 + bz + c = 0.
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Case 1: g (t) = Pn (t) is a polynomial
Sample PI
Case 2: g (t) = e αt Pn (t)
Case 3: g (t) = e αt Pn (t) cos βt or g (t) = e αt Pn (t) sin βt
Final Summary: The Tool Box
Summary
◮
◮
In 3-cases, the form of a particular solution: (9), (11),
(14).
Add Solutions: Assume g (t) = g1 (t) + g2 (t). So, (7) is:
L(y ) = ay ” + by ′ + cy = g1 (t) + g2 (t)
(15)
Suppose Y1 (t) is a particular solution
L(y ) = ay ” + by ′ + cy = g1 (t)
(16)
And Y2 (t) is a particular solution
L(y ) = ay ” + by ′ + cy = g2 (t)
(17)
Then, Y (t) = Y1 (t) + Y2 (t) is a solution of (15).
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
General Sample: g (t) = ste αt
The following is slightly general than Ex. 5.
Consider the DE
ay ” + by ′ + cy = ste αt
(18)
Assume the CE has two distinct real roots. Find the general
solution.
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Step I: The solution of L(y ) = 0
◮
◮
◮
◮
First, we solve the homogeneous equation
L(y ) = ay ” + by ′ + cy = 0
The CE ζ(r ) = ar 2 + br + c = 0.
We assumed, r1 6= r2 are two real roots of the EC
ζ(r ) = 0.
So, a general solution of L(y ) = 0 is
yc = c1 e r1 t + c2 e r2 t
Satya Mandal, KU
(19)
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Step II: The particular solution
◮
By (11), a particular solution has the form
Y = e αt (A0 t + A1 ),
So,
Y ′ = αe αt (A0 t + A1 ) + e αt A0 = e αt [αA0 t + (αA1 + A0 )]
Y ” = αe αt [αA0 t + (αA1 + A0 )] + e αt αA0
= e αt [α2 A0 t + (α2 A1 + 2αA0 )]
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
◮
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Substituting in (18)
e αt [a(α2 A0 t + (α2 A1 + 2αA0 )) + b(αA0 t + (αA1 + A0 ))
+c(A0 t + A1 )] = ste αt
=⇒
t((aα2 +bα+c)A0 +[a(α2 A1 +2αA0 )+b(αA1 +A0 )+cA1 ]
= st
◮
So,
2
(aα + bα + c)A1
Satya Mandal, KU
(aα2 + bα + c)A0 = s
+(2aα + b)A0
=0
(20)
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
◮
◮
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
System (20) has Echelon form. We rewrite System (20):
(
s
A0 = ζ(α)
ζ(α)A0
=s
′ (α)
=⇒
ζ(α)A1 +ζ ′ (α)A0 = 0
A1 = − sζ
ζ(α)2
if ζ(α) 6= 0.
So, a particular solution is:
αt
Y = e (A0 t + A1 ) = e
Satya Mandal, KU
αt
sζ ′ (α)
s
t−
ζ(α)
ζ(α)2
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
◮
◮
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Combining with (19), the general solutions is
s
sζ ′ (α)
αt
y = Y + yc = e
+ c1 e r1 t + c2 e r2 t
t−
ζ(α)
ζ(α)2
(21)
The Point: ζ(α) plays an important role.
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Exercise 5 (edited)
Consider the DE
y ” − 3y ′ − 4y = 36te −2t
(22)
Give a general solution.
◮ This is clearly a particular case of the above.
◮ We repeat the steps above, instead of using the solution
(21).
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Step I: The solution of L(y ) = 0
◮
◮
◮
◮
First, we solve the homogeneous equation
L(y ) = y ” − 3y ′ − 4y = 0
The CE: r 2 − 3r − 4 = 0
We have r1 = −1, r2 = 4
So, a general solution of L(y ) = 0 is
yc = c1 e −t + c2 e 4t
Satya Mandal, KU
(23)
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Step II: The particular solution
◮
Here g (t) = 36te −2t . By (11), a particular solution has
the form
Y = e −2t (A0 t + A1 ),
So,
Y ′ = −2e −2t (A0 t+A1 )+e −2t A0 = e −2t [−2A0 t + (−2A1 + A0 )]
Y ” = −2e −2t [−2A0 t + (−2A1 + A0 )] − 2e −2t A0
= e −2t [4A0 t + (4A1 − 4A0 )]
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
◮
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Substituting in (22)
e −2t [(4A0 t + (4A1 − 4A0 )) − 3(−2A0 t + (−2A1 + A0 ))
−4(A0 t + A1 )] = 36te −2t
=⇒
t((4 + 6 − 4)A0 + [(4A1 − 4A0 ) − 3(−2A1 + A0 ) − 4A1 ]
= 36t
◮
So,
6A1
6A0
= 36
−7A0 = 0
Satya Mandal, KU
=⇒
A0 = 6
A1 = 7
(24)
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
◮
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
So, a particular solution is:
Y = e −2t (A0 t + A1 ) = e −2t (6t + 7)
◮
Combining with (23), the general solutions is
y = Y + yc = e −2t (6t + 7) + (c1 e −t + c2 e 4t )
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Exercise 2 (edited)
Consider the DE
y ” − 2y ′ + 10y = 37 sin 3t
(25)
Give a general solution.
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Step I: The solution of L(y ) = 0
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◮
◮
◮
First, we solve the homogeneous equation
L(y ) = y ” − 2y ′ + 10y = 0
The CE: r 2 − 2r + 10 = 0
We have r1 = 1 + 3i, r2 = 1 − 3i
So, a general solution of L(y ) = 0 is (See §3.3)
yc = e t (c1 cos 3t + c2 sin 3t)
Satya Mandal, KU
(26)
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Step II: The particular solution
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◮
◮
Here g (t) = 37 sin 3t.
We use (14), with α = 0, β = 3. We need to ensure
ζ(α + iβ) = ζ(3i) 6= 0.
So, by (14), a particular solution has the form
Y = A0 cos 3t + B0 sin 3t.
Y ′ = −3A0 sin 3t + 3B0 cos 3t
Y ” = −9A0 cos 3t − 9B0 sin 3t
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
◮
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Substituting in (25)
(−9A0 cos 3t − 9B0 sin 3t) − 2(−3A0 sin 3t + 3B0 cos 3t)
+10(A0 cos 3t + B0 sin 3t) = 37 sin 3t =⇒
cos 3t(−9A0 − 6B0 + 10A0 ) + sin 3t(−9B0 + 6A0 + 10B0 )
= 37 sin 3t
◮
So,
A0 −6B0 = 0
6A0 +B0 = 37
Satya Mandal, KU
=⇒
A0 = 6
B0 = 1
(27)
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
◮
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
So,
Y = A0 cos 3t + B0 sin 3t = 6 cos 3t + sin 3t
◮
Combining with (26), the general solutions is
y = Y + yc = (6 cos 3t + sin 3t) + e t (c1 cos 3t + c2 sin 3t)
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Exercise 9
Consider the DE
2y ” + 3y ′ + y = t 2 + 3 sin t
(28)
Give a general solution.
◮ To solve this we solve the following two DEs:
2y ” + 3y ′ + y = t 2
(29)
2y ” + 3y ′ + y = 3 sin t
(30)
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Step I: The solution of L(y ) = 0
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◮
◮
◮
In either case, the homogeneous equation is
L(y ) = 2y ” + 3y ′ + y = 0
The CE: ζ(r ) := 2r 2 + 3r + 1 = 0
We have r1 = −1, r2 = −.5
So, a general solution of L(y ) = 0 is
yc = c1 e −t + c2 e −.5t
Satya Mandal, KU
(31)
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Step IIa: The particular solution of (29)
We first find a particular of (29)).
◮ Here g (t) = t 2 . By (9), a particular solution has the form
Y = A0 t 2 + A1 t + A2
Y ′ = 2A0 t + A1 ,
Satya Mandal, KU
Y ” = 2A0
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Particular Solution of (29)
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Substituting in (29)
2(2A0 ) + 3(2A0 t + A1 ) + (A0 t 2 + A1 t + A2 ) = t 2
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Comparing coefficients of

 A0
6A0 +A1

4A0 +3A1 +A2
Satya Mandal, KU
T 2 , t, t 0 , we have,

=1
 A0 = 1
= 0 =⇒
A1 = −6

A2 = 14
=0
(32)
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Particular Solution of (29)
◮
a particular Solution of (29):
Y1 = A0 t 2 + A1 t + A2 = t 2 − 6t + 14
Satya Mandal, KU
(33)
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Step IIb: The particular solution of (30)
Now, we find a particular solution of (30).
◮ Here g (t) = 3 sin t.
◮ We use (14), with α = 0, β = 1. We need to ensure
ζ(α + iβ) = ζ(i) 6= 0 or α + iβ 6= r1 , r2 .
◮ So, by (14), a particular solution has the form
Y = A0 cos t + B0 sin t.
Y ′ = −A0 sin t + B0 cos t
Y ” = −A0 cos t − B0 sin t
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
◮
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
Substituting in (30)
2(−A0 cos t − B0 sin t) + 3(−A0 sin t + B0 cos t)
+(A0 cos t + B0 sin t) = 3 sin t =⇒
cos t(−2A0 + 3B0 + A0 ) + sin t(−2B0 − 3A0 + B0 )
= 3 sin t
◮
So,
−A0 +3B0 = 0
−3A0 −B0 = 3
Satya Mandal, KU
=⇒
9
A0 = − 10
3
B0 = − 10
(34)
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
◮
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
So, a particular solution of (30):
Y2 = A0 cos t + B0 sin t = −
◮
9
3
cos t −
sin t
10
10
(35)
Combining with (33), (35), (31) the general solutions is
9
3
2
y = Y1 +Y2 +yc = (t −6t +14)+ − cos t −
sin t
10
10
+ c1 e −t + c2 e −.5t
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod
The Homogeneous DE and the General Solutions
Different forms of g (t)
Steps by Step: g (t) = ste αt
Sample IIEP: Exercise 5 (edited)
Sample IIITrig: Exercise 2 (edited)
Sample IV Sum: Exercise 9
§3.5 Assignments and Homework
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Most Importantly, be familiar with the chart in page 182
Read Example 1-5 (They are helpful).
Homework: §3.5 See the Homework Site!
Satya Mandal, KU
III Second Order DE §3.5 Nonhomogeneous LSODEsMethod