Download Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)

NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
NonHomogeneous Linear Equations (Section 17.2)
The solution of a second order nonhomogeneous linear differential equation of
the form
ay 00 + by 0 + cy = G (x)
where a, b, c are constants, a 6= 0 and G (x) is a continuous function of x on a
given interval is of the form
y (x) = yp (x) + yc (x)
where yp (x) is a particular solution of ay 00 + by 0 + cy = G (x) and yc (x) is the
general solution of the complementary equation/ corresponding
homogeneous equation ay 00 + by 0 + cy = 0.
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
NonHomogeneous Linear Equations (Section 17.2)
The solution of a second order nonhomogeneous linear differential equation of
the form
ay 00 + by 0 + cy = G (x)
where a, b, c are constants, a 6= 0 and G (x) is a continuous function of x on a
given interval is of the form
y (x) = yp (x) + yc (x)
where yp (x) is a particular solution of ay 00 + by 0 + cy = G (x) and yc (x) is the
general solution of the complementary equation/ corresponding
homogeneous equation ay 00 + by 0 + cy = 0.
I
Since we already know how to find yc , the general solution to the
corresponding homogeneous equation, we need a method to
find a particular solution, yp , to the equation. One such methods is
described below. This method may not always work. A second method
which is always applicable is demonstrated in the extra examples in your
notes.
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
The method of Undetermined Coefficients
We wish to search for a particular solution to ay 00 + by 0 + cy = G (x).
If G (x) is a polynomial it is reasonable to guess that there is a particular
solution, yp (x) which is a polynomial in x of the same degree as G (x) (because
if y is such a polynomial, then ay 00 + by 0 + c is also a polynomial of the same
degree.)
Method to find a particular solution: Substitute yp (x) = a polynomial of
the same degree as G into the differential equation and determine the
coefficients.
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
The method of Undetermined Coefficients
We wish to search for a particular solution to ay 00 + by 0 + cy = G (x).
If G (x) is a polynomial it is reasonable to guess that there is a particular
solution, yp (x) which is a polynomial in x of the same degree as G (x) (because
if y is such a polynomial, then ay 00 + by 0 + c is also a polynomial of the same
degree.)
Method to find a particular solution: Substitute yp (x) = a polynomial of
the same degree as G into the differential equation and determine the
coefficients.
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
I
We first find the
solution of the complementary/ corresponding homogeneous equation,
y 00 + 3y 0 + 2y = 0:
Auxiliary equation: r 2 + 3r + 2 = 0
Roots: (r + 1)(r + 2) = 0 → r1 = −1, r2 = −2. Distinct real roots.
Solution to corresponding homogeneous equation:
yc = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Undetermined coefficients Example (polynomial)
y (x) = yp (x) + yc (x)
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x
We now need a particular solution yp (x).
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Undetermined coefficients Example (polynomial)
y (x) = yp (x) + yc (x)
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x
We now need a particular solution yp (x).
I
We consider a trial solution of the form yp (x) = Ax 2 + Bx + C .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Undetermined coefficients Example (polynomial)
y (x) = yp (x) + yc (x)
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x
We now need a particular solution yp (x).
I
We consider a trial solution of the form yp (x) = Ax 2 + Bx + C .
I
Then yp0 (x) = 2Ax + B,
yp00 (x) = 2A.
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Undetermined coefficients Example (polynomial)
y (x) = yp (x) + yc (x)
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x
We now need a particular solution yp (x).
I
We consider a trial solution of the form yp (x) = Ax 2 + Bx + C .
I
Then yp0 (x) = 2Ax + B,
I
yp00 ,
yp00 (x) = 2A.
yp0
We plug
and yp into the equation to get
2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2
→ 2Ax 2 + (6A + 2B)x + (2A + 3B + 2C ) = x 2 .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Undetermined coefficients Example (polynomial)
y (x) = yp (x) + yc (x)
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x
We now need a particular solution yp (x).
I
We consider a trial solution of the form yp (x) = Ax 2 + Bx + C .
I
Then yp0 (x) = 2Ax + B,
I
yp00 ,
yp00 (x) = 2A.
yp0
We plug
and yp into the equation to get
2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2
→ 2Ax 2 + (6A + 2B)x + (2A + 3B + 2C ) = x 2 .
I
Equating coefficients, we get
1
2A = 1 → A = , 6A + 2B = 0, 2A + 3B + 2C = 0.
2
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Undetermined coefficients Example (polynomial)
y (x) = yp (x) + yc (x)
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x
We now need a particular solution yp (x).
I
We consider a trial solution of the form yp (x) = Ax 2 + Bx + C .
I
Then yp0 (x) = 2Ax + B,
I
yp00 ,
yp00 (x) = 2A.
yp0
We plug
and yp into the equation to get
2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2
→ 2Ax 2 + (6A + 2B)x + (2A + 3B + 2C ) = x 2 .
I
I
Equating coefficients, we get
1
2A = 1 → A = , 6A + 2B = 0, 2A + 3B + 2C = 0.
2
Plugging the value for A into the middle equation, we get
3
3 + 2B = 0 → B = − .
2
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Undetermined coefficients Example (polynomial)
y (x) = yp (x) + yc (x)
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x
We now need a particular solution yp (x).
I Let us assume yp (x) = Ax 2 + Bx + C .
I
Then yp0 (x) = 2Ax + B,
I
We plug yp00 , yp0 and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2
yp00 (x) = 2A.
→
2Ax
2
2
+ (6A + 2B)x + (2A + 3B + 2C ) = x .
2A = 1 → A =
I
Equating coefficients, we get
I
Plugging the value for A into the middle equation, we get
Annette Pilkington
1
, 6A + 2B = 0, 2A + 3B + 2C = 0.
2
3 + 2B = 0 → B = −
3
.
2
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Undetermined coefficients Example (polynomial)
y (x) = yp (x) + yc (x)
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x
We now need a particular solution yp (x).
I Let us assume yp (x) = Ax 2 + Bx + C .
I
Then yp0 (x) = 2Ax + B,
I
We plug yp00 , yp0 and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2
yp00 (x) = 2A.
→
2Ax
2
2
+ (6A + 2B)x + (2A + 3B + 2C ) = x .
2A = 1 → A =
I
Equating coefficients, we get
I
Plugging the value for A into the middle equation, we get
I
1
, 6A + 2B = 0, 2A + 3B + 2C = 0.
2
3 + 2B = 0 → B = −
3
.
2
Using values for A and B in the third equation, we get
7
1 − 92 + 2C = 0 → C = .
4
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Undetermined coefficients Example (polynomial)
y (x) = yp (x) + yc (x)
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x
We now need a particular solution yp (x).
I Trial Solution: yp (x) = Ax 2 + Bx + C .
I
I
Then yp0 (x) = 2Ax + B, yp00 (x) = 2A.
We plug yp00 , yp0 and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2
→
I
A=
1
,
2
B=−
3
,
2
2Ax
2
2
+ (6A + 2B)x + (2A + 3B + 2C ) = x .
C =
7
4
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Undetermined coefficients Example (polynomial)
y (x) = yp (x) + yc (x)
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x
We now need a particular solution yp (x).
I Trial Solution: yp (x) = Ax 2 + Bx + C .
I
I
Then yp0 (x) = 2Ax + B, yp00 (x) = 2A.
We plug yp00 , yp0 and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2
→
I
I
A=
1
,
2
B=−
3
,
2
2Ax
2
2
+ (6A + 2B)x + (2A + 3B + 2C ) = x .
C =
7
4
Hence a particular solution is given by yp =
Annette Pilkington
7i
1h 2
x − 3x +
,
2
2
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Undetermined coefficients Example (polynomial)
y (x) = yp (x) + yc (x)
Example Solve the differential equation: y 00 + 3y 0 + 2y = x 2 .
yc (x) = c1 e r1 x + c2 e r2 x = c1 e −x + c2 e −2x
We now need a particular solution yp (x).
I Trial Solution: yp (x) = Ax 2 + Bx + C .
I
I
Then yp0 (x) = 2Ax + B, yp00 (x) = 2A.
We plug yp00 , yp0 and yp into the equation to get 2A + 3(2Ax + B) + 2(Ax 2 + Bx + C ) = x 2
→
I
A=
1
,
2
B=−
3
,
2
2Ax
2
2
+ (6A + 2B)x + (2A + 3B + 2C ) = x .
C =
7
4
7i
1h 2
x − 3x +
,
2
2
I
Hence a particular solution is given by yp =
I
and the general solution is given by
7i
1h 2
y = yp + yc =
x − 3x +
+ c1 e −x + c2 e −2x .
2
2
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = Ce kx . Example (Exponential)
We wish to search for a particular solution to ay 00 + by 0 + cy = G (x).
If G (x) is of the form Ce kx , where C and k are constants, then we use a trial
solution of the form yp (x) = Ae kx and solve for A if possible.
Example Solve y 00 + 9y = e −4x .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = Ce kx . Example (Exponential)
We wish to search for a particular solution to ay 00 + by 0 + cy = G (x).
If G (x) is of the form Ce kx , where C and k are constants, then we use a trial
solution of the form yp (x) = Ae kx and solve for A if possible.
Example Solve y 00 + 9y = e −4x .
I
We first find the solution of the corresponding homogeneous equation,
y 00 + 9y 0 = 0:
Auxiliary equation: r 2 + 9 = 0
Roots: r 2 = −9 → r1 = 3i, r2 = −3i. Complex roots.
Solution to corresponding homogeneous equation :
yc = c1 cos(3x) + c2 sin(3x) .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = Ce kx . Example (Exponential)
We wish to search for a particular solution to ay 00 + by 0 + cy = G (x).
If G (x) is of the form Ce kx , where C and k are constants, then we use a trial
solution of the form yp (x) = Ae kx and solve for A if possible.
Example Solve y 00 + 9y = e −4x .
I
We first find the solution of the corresponding homogeneous equation,
y 00 + 9y 0 = 0:
Auxiliary equation: r 2 + 9 = 0
Roots: r 2 = −9 → r1 = 3i, r2 = −3i. Complex roots.
Solution to corresponding homogeneous equation :
yc = c1 cos(3x) + c2 sin(3x) .
I
Next we find a particular solution. Our trial solution is of the form
yp (x) = Ae −4x .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = Ce kx . Example (Exponential)
We wish to search for a particular solution to ay 00 + by 0 + cy = G (x).
If G (x) is of the form Ce kx , where C and k are constants, then we use a trial
solution of the form yp (x) = Ae kx and solve for A if possible.
Example Solve y 00 + 9y = e −4x .
I
We first find the solution of the corresponding homogeneous equation,
y 00 + 9y 0 = 0:
Auxiliary equation: r 2 + 9 = 0
Roots: r 2 = −9 → r1 = 3i, r2 = −3i. Complex roots.
Solution to corresponding homogeneous equation :
yc = c1 cos(3x) + c2 sin(3x) .
I
Next we find a particular solution. Our trial solution is of the form
yp (x) = Ae −4x .
yp0 (x) = −4Ae −4x , yp00 (x) = 16Ae −4x .
I
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = Ce kx . Example (Exponential)
We wish to search for a particular solution to ay 00 + by 0 + cy = G (x).
If G (x) is of the form Ce kx , where C and k are constants, then we use a trial
solution of the form yp (x) = Ae kx and solve for A if possible.
Example Solve y 00 + 9y = e −4x .
I
We first find the solution of the corresponding homogeneous equation,
y 00 + 9y 0 = 0:
Auxiliary equation: r 2 + 9 = 0
Roots: r 2 = −9 → r1 = 3i, r2 = −3i. Complex roots.
Solution to corresponding homogeneous equation :
yc = c1 cos(3x) + c2 sin(3x) .
I
Next we find a particular solution. Our trial solution is of the form
yp (x) = Ae −4x .
yp0 (x) = −4Ae −4x , yp00 (x) = 16Ae −4x .
We plug yp00 , yp0 and yp into the equation to get
I
I
16Ae −4x + 9Ae −4x = e −4x
Annette Pilkington
→
e −4x [16A + 9A] = e −4x .
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = Ce kx . Example (Exponential)
We wish to search for a particular solution to ay 00 + by 0 + cy = G (x).
If G (x) is of the form Ce kx , where C and k are constants, then we use a trial
solution of the form yp (x) = Ae kx and solve for A if possible.
Example Solve y 00 + 9y = e −4x .
I
We first find the solution of the corresponding homogeneous equation,
y 00 + 9y 0 = 0:
Auxiliary equation: r 2 + 9 = 0
Roots: r 2 = −9 → r1 = 3i, r2 = −3i. Complex roots.
Solution to corresponding homogeneous equation :
yc = c1 cos(3x) + c2 sin(3x) .
I
Next we find a particular solution. Our trial solution is of the form
yp (x) = Ae −4x .
yp0 (x) = −4Ae −4x , yp00 (x) = 16Ae −4x .
We plug yp00 , yp0 and yp into the equation to get
I
I
16Ae −4x + 9Ae −4x = e −4x
I
→
e −4x [16A + 9A] = e −4x .
Since e −4x > 0, we can cancel to get 25A = 1
Annette Pilkington
A=
1
.
25
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = Ce kx . Example (Exponential)
We wish to search for a particular solution to ay 00 + by 0 + cy = G (x).
If G (x) is of the form Ce kx , where C and k are constants, then we use a trial
solution of the form yp (x) = Ae kx and solve for A if possible.
Example Solve y 00 + 9y = e −4x .
I
yc = c1 cos(3x) + c2 sin(3x) .
I
Next we find a particular solution. Our trial solution is of the form
yp (x) = Ae −4x .
I
yp0 (x) = −4Ae −4x ,
I
yp00 (x) = 16Ae −4x .
00
0
We plug yp , yp and yp into the equation to get
16Ae
I
−4x
+ 9Ae
Since e −4x > 0, we can cancel to get
−4x
= e
25A = 1
Annette Pilkington
−4x
→
A=
e
−4x
[16A + 9A] = e
−4x
.
1
.
25
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = Ce kx . Example (Exponential)
We wish to search for a particular solution to ay 00 + by 0 + cy = G (x).
If G (x) is of the form Ce kx , where C and k are constants, then we use a trial
solution of the form yp (x) = Ae kx and solve for A if possible.
Example Solve y 00 + 9y = e −4x .
I
yc = c1 cos(3x) + c2 sin(3x) .
I
Next we find a particular solution. Our trial solution is of the form
yp (x) = Ae −4x .
I
yp0 (x) = −4Ae −4x ,
I
yp00 (x) = 16Ae −4x .
00
0
We plug yp , yp and yp into the equation to get
16Ae
−4x
+ 9Ae
−4x
= e
25A = 1
−4x
→
A=
e
−4x
[16A + 9A] = e
.
1
.
25
I
Since e −4x > 0, we can cancel to get
I
Hence a particular solution is given by : yp =
Annette Pilkington
−4x
1 −4x
e
,
25
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = Ce kx . Example (Exponential)
We wish to search for a particular solution to ay 00 + by 0 + cy = G (x).
If G (x) is of the form Ce kx , where C and k are constants, then we use a trial
solution of the form yp (x) = Ae kx and solve for A if possible.
Example Solve y 00 + 9y = e −4x .
I
yc = c1 cos(3x) + c2 sin(3x) .
I
Next we find a particular solution. Our trial solution is of the form
yp (x) = Ae −4x .
I
yp0 (x) = −4Ae −4x ,
I
yp00 (x) = 16Ae −4x .
00
0
We plug yp , yp and yp into the equation to get
16Ae
−4x
+ 9Ae
−4x
= e
−4x
25A = 1
→
A=
e
−4x
[16A + 9A] = e
.
1
.
25
I
Since e −4x > 0, we can cancel to get
I
Hence a particular solution is given by : yp =
I
General solution: y = yp + yc =
Annette Pilkington
−4x
1 −4x
e
,
25
1 −4x
e
+ c1 cos(3x) + c2 sin(3x)
25
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = C cos kx or C sin kx. Example (Trigonometric)
If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then
we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for
A and B if possible.
Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants
K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0.
Example Solve y 00 − 4y 0 − 5y = cos(2x).
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = C cos kx or C sin kx. Example (Trigonometric)
If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then
we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for
A and B if possible.
Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants
K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0.
Example Solve y 00 − 4y 0 − 5y = cos(2x).
I
We first find the solution of the corresponding homogeneous equation,
y 00 − 4y 0 − 5y = 0:
Auxiliary equation: r 2 − 4r − 5 = 0
Roots: (r + 1)(r − 5) = 0 → r1 = −1, r2 = 5. Distinct real roots.
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = C cos kx or C sin kx. Example (Trigonometric)
If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then
we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for
A and B if possible.
Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants
K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0.
Example Solve y 00 − 4y 0 − 5y = cos(2x).
I
We first find the solution of the corresponding homogeneous equation,
y 00 − 4y 0 − 5y = 0:
Auxiliary equation: r 2 − 4r − 5 = 0
Roots: (r + 1)(r − 5) = 0 → r1 = −1, r2 = 5. Distinct real roots.
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x .
I
Next we find a particular solution. Our trial solution takes the form:
yp (x) = A cos(2x) + B sin(2x).
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = C cos kx or C sin kx. Example (Trigonometric)
If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then
we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for
A and B if possible.
Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants
K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0.
Example Solve y 00 − 4y 0 − 5y = cos(2x).
I
We first find the solution of the corresponding homogeneous equation,
y 00 − 4y 0 − 5y = 0:
Auxiliary equation: r 2 − 4r − 5 = 0
Roots: (r + 1)(r − 5) = 0 → r1 = −1, r2 = 5. Distinct real roots.
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x .
I
I
Next we find a particular solution. Our trial solution takes the form:
yp (x) = A cos(2x) + B sin(2x).
We plug yp00 , yp0 and yp into the equation to get
−4A cos(2x) − 4B sin(2x) − 4[−2A sin(2x) + 2B cos(2x)] − 5[A cos(2x) +
B sin(2x)] = cos(2x)
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = C cos kx or C sin kx. Example (Trigonometric)
If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then
we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for
A and B if possible.
Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants
K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0.
Example Solve y 00 − 4y 0 − 5y = cos(2x).
I
We first find the solution of the corresponding homogeneous equation,
y 00 − 4y 0 − 5y = 0:
Auxiliary equation: r 2 − 4r − 5 = 0
Roots: (r + 1)(r − 5) = 0 → r1 = −1, r2 = 5. Distinct real roots.
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x .
I
I
I
Next we find a particular solution. Our trial solution takes the form:
yp (x) = A cos(2x) + B sin(2x).
We plug yp00 , yp0 and yp into the equation to get
−4A cos(2x) − 4B sin(2x) − 4[−2A sin(2x) + 2B cos(2x)] − 5[A cos(2x) +
B sin(2x)] = cos(2x)
Tidying up, we get
(−4A − 8B − 5A − 1) cos(2x) + (−4B + 8A − 5B) sin(2x) = 0 →
(−9A − 8B − 1) cos(2x) + (−9B + 8A) sin(2x) = 0.
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = C cos kx or C sin kx. Example (Trigonometric)
If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then
we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for
A and B if possible.
Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants
K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0.
Example Solve y 00 − 4y 0 − 5y = cos(2x).
I
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x .
I
Next we find a particular solution. Our trial solution takes the form:
I
We plug yp00 , yp0 and yp into the equation to get
yp (x) = A cos(2x) + B sin(2x).
−4A cos(2x) − 4B sin(2x) − 4[−2A sin(2x) + 2B cos(2x)] − 5[A cos(2x) + B sin(2x)] = cos(2x)
I
Tidying up, we get (−4A − 8B − 5A − 1) cos(2x) + (−4B + 8A − 5B) sin(2x) = 0
→ (−9A − 8B − 1) cos(2x) + (−9B + 8A) sin(2x) = 0.
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = C cos kx or C sin kx. Example (Trigonometric)
If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then
we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for
A and B if possible.
Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants
K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0.
Example Solve y 00 − 4y 0 − 5y = cos(2x).
I
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x .
I
Next we find a particular solution. Our trial solution takes the form:
I
We plug yp00 , yp0 and yp into the equation to get
yp (x) = A cos(2x) + B sin(2x).
−4A cos(2x) − 4B sin(2x) − 4[−2A sin(2x) + 2B cos(2x)] − 5[A cos(2x) + B sin(2x)] = cos(2x)
I
Tidying up, we get (−4A − 8B − 5A − 1) cos(2x) + (−4B + 8A − 5B) sin(2x) = 0
I
→ (−9A − 8B − 1) cos(2x) + (−9B + 8A) sin(2x) = 0.
We must have −9A − 8B = 1,
8A − 9B = 0.
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = C cos kx or C sin kx. Example (Trigonometric)
If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then
we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for
A and B if possible.
Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants
K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0.
Example Solve y 00 − 4y 0 − 5y = cos(2x).
I
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x .
I
Next we find a particular solution. Our trial solution takes the form:
I
We plug yp00 , yp0 and yp into the equation to get
yp (x) = A cos(2x) + B sin(2x).
−4A cos(2x) − 4B sin(2x) − 4[−2A sin(2x) + 2B cos(2x)] − 5[A cos(2x) + B sin(2x)] = cos(2x)
I
I
I
Tidying up, we get (−4A − 8B − 5A − 1) cos(2x) + (−4B + 8A − 5B) sin(2x) = 0
→ (−9A − 8B − 1) cos(2x) + (−9B + 8A) sin(2x) = 0.
We must have −9A − 8B = 1,
8A − 9B = 0.
From the second equation, we get B = 89 A and substituting this into the
first equation, we get
−9A −
64
A=1
9
→
Annette Pilkington
−
145
A=1 →
9
A=−
9
.
145
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = C cos kx or C sin kx. Example (Trigonometric)
If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then
we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for
A and B if possible.
Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants
K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0.
Example Solve y 00 − 4y 0 − 5y = cos(2x).
I
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x .
I
Next we find a particular solution. Our trial solution takes the form:
I
From the second equation, we get
−9A − 64 A = 1
9
→
B = 98 A
− 145 A = 1
9
→
yp (x) = A cos(2x) + B sin(2x).
and substituting this into the first equation, we get
A=−
Annette Pilkington
9
.
145
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = C cos kx or C sin kx. Example (Trigonometric)
If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then
we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for
A and B if possible.
Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants
K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0.
Example Solve y 00 − 4y 0 − 5y = cos(2x).
I
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x .
I
Next we find a particular solution. Our trial solution takes the form:
I
From the second equation, we get
−9A − 64 A = 1
9
I
→
B = 98 A
− 145 A = 1
9
→
yp (x) = A cos(2x) + B sin(2x).
and substituting this into the first equation, we get
A=−
Substituting we get B = 89 A →
Annette Pilkington
9
.
145
B=−
8
.
145
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = C cos kx or C sin kx. Example (Trigonometric)
If G (x) is of the form C cos kx or C sin kx, where C and k are constants, then
we use a trial solution of the form yp (x) = A cos(kx) + B sin(kx) and solve for
A and B if possible.
Below we use the fact that if K1 cos(αx) + K2 sin(αx) = 0 for constants
K1 , K2 , α, where α 6= 0, Then we must have K1 = K2 = 0.
Example Solve y 00 − 4y 0 − 5y = cos(2x).
I
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e 5x .
I
Next we find a particular solution. Our trial solution takes the form:
I
From the second equation, we get
−9A − 64 A = 1
9
→
B = 98 A
− 145 A = 1
9
→
yp (x) = A cos(2x) + B sin(2x).
and substituting this into the first equation, we get
A=−
9
.
145
8
.
145
I
Substituting we get B = 89 A →
I
9
8
Hence a particular solution is given by yp = − 145
cos(2x) − 145
sin(2x),
and the general solution is given by
9
8
y = yp + yc = −
cos(2x) −
sin(2x) + c1 e −x + c2 e 5x .
145
145
Annette Pilkington
B=−
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Trouble: Trial solution fits corresponding hom. eqn.
Troubleshooting
If the trial solution yp is a solution of the corresponding homogeneous equation,
then it cannot be a solution to the non-homogeneous equation. In this case, we
multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial
solution that does not satisfy the corresponding homogeneous equation. Then
proceed as above.
Example Solve y 00 − y = e x .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Trouble: Trial solution fits corresponding hom. eqn.
Troubleshooting
If the trial solution yp is a solution of the corresponding homogeneous equation,
then it cannot be a solution to the non-homogeneous equation. In this case, we
multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial
solution that does not satisfy the corresponding homogeneous equation. Then
proceed as above.
Example Solve y 00 − y = e x .
I
We first find the solution of the corresponding homogeneous equation,
y 00 − y = 0:
Auxiliary equation: r 2 − 1 = 0
Roots: (r + 1)(r − 1) = 0 → r1 = −1, r2 = 1. Distinct real roots.
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e x .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Trouble: Trial solution fits corresponding hom. eqn.
Troubleshooting
If the trial solution yp is a solution of the corresponding homogeneous equation,
then it cannot be a solution to the non-homogeneous equation. In this case, we
multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial
solution that does not satisfy the corresponding homogeneous equation. Then
proceed as above.
Example Solve y 00 − y = e x .
I
We first find the solution of the corresponding homogeneous equation,
y 00 − y = 0:
Auxiliary equation: r 2 − 1 = 0
Roots: (r + 1)(r − 1) = 0 → r1 = −1, r2 = 1. Distinct real roots.
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e x .
I
Next we find a particular solution. Note that Ae x , is a solution to the
corresponding homogeneous equation and therefore cannot be a solution
to the non-homogeneous equation.
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Trouble: Trial solution fits corresponding hom. eqn.
Troubleshooting
If the trial solution yp is a solution of the corresponding homogeneous equation,
then it cannot be a solution to the non-homogeneous equation. In this case, we
multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial
solution that does not satisfy the corresponding homogeneous equation. Then
proceed as above.
Example Solve y 00 − y = e x .
I
We first find the solution of the corresponding homogeneous equation,
y 00 − y = 0:
Auxiliary equation: r 2 − 1 = 0
Roots: (r + 1)(r − 1) = 0 → r1 = −1, r2 = 1. Distinct real roots.
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e x .
I
Next we find a particular solution. Note that Ae x , is a solution to the
corresponding homogeneous equation and therefore cannot be a solution
to the non-homogeneous equation.
I
As recommended, we use yp (x) = Axe x .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Trouble: Trial solution fits corresponding hom. eqn.
Troubleshooting
If the trial solution yp is a solution of the corresponding homogeneous equation,
then it cannot be a solution to the non-homogeneous equation. In this case, we
multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial
solution that does not satisfy the corresponding homogeneous equation. Then
proceed as above.
Example Solve y 00 − y = e x .
I
We first find the solution of the corresponding homogeneous equation,
y 00 − y = 0:
Auxiliary equation: r 2 − 1 = 0
Roots: (r + 1)(r − 1) = 0 → r1 = −1, r2 = 1. Distinct real roots.
Solution to corresponding homogeneous equation : yc = c1 e −x + c2 e x .
I
Next we find a particular solution. Note that Ae x , is a solution to the
corresponding homogeneous equation and therefore cannot be a solution
to the non-homogeneous equation.
I
As recommended, we use yp (x) = Axe x .
I
yp0 (x) = Ae x + Axe x ,
yp00 (x) = 2Ae x + Axe x .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Trouble: Trial solution fits corresponding hom. eqn.
Troubleshooting
If the trial solution yp is a solution of the corresponding homogeneous equation,
then it cannot be a solution to the non-homogeneous equation. In this case, we
multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial
solution that does not satisfy the corresponding homogeneous equation. Then
proceed as above.
Example Solve y 00 − y = e x .
I
I
yc = c1 e −x + c2 e x .
Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore
cannot be a solution to the non-homogeneous equation.
I
As recommended, we use yp (x) = Axe x .
I
yp0 (x) = Ae x + Axe x ,
yp00 (x) = 2Ae x + Axe x .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Trouble: Trial solution fits corresponding hom. eqn.
Troubleshooting
If the trial solution yp is a solution of the corresponding homogeneous equation,
then it cannot be a solution to the non-homogeneous equation. In this case, we
multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial
solution that does not satisfy the corresponding homogeneous equation. Then
proceed as above.
Example Solve y 00 − y = e x .
I
I
yc = c1 e −x + c2 e x .
Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore
cannot be a solution to the non-homogeneous equation.
I
As recommended, we use yp (x) = Axe x .
I
yp0 (x) = Ae x + Axe x ,
I
yp00
We plug
yp00 (x) = 2Ae x + Axe x .
and yp into the equation to get
2Ae x + Axe x − Axe x = e x
→ 2Ae x = e x
Annette Pilkington
→ 2A = 1 →
A = 1/2
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Trouble: Trial solution fits corresponding hom. eqn.
Troubleshooting
If the trial solution yp is a solution of the corresponding homogeneous equation,
then it cannot be a solution to the non-homogeneous equation. In this case, we
multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial
solution that does not satisfy the corresponding homogeneous equation. Then
proceed as above.
Example Solve y 00 − y = e x .
I
I
yc = c1 e −x + c2 e x .
Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore
cannot be a solution to the non-homogeneous equation.
I
As recommended, we use yp (x) = Axe x .
I
yp0 (x) = Ae x + Axe x ,
I
yp00
We plug
yp00 (x) = 2Ae x + Axe x .
and yp into the equation to get
2Ae x + Axe x − Axe x = e x
I
→ 2Ae x = e x
Hence a particular solution is given by yp =
Annette Pilkington
→ 2A = 1 →
A = 1/2
1 x
xe
2
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Trouble: Trial solution fits corresponding hom. eqn.
Troubleshooting
If the trial solution yp is a solution of the corresponding homogeneous equation,
then it cannot be a solution to the non-homogeneous equation. In this case, we
multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial
solution that does not satisfy the corresponding homogeneous equation. Then
proceed as above.
Example Solve y 00 − y = e x .
I
I
yc = c1 e −x + c2 e x .
Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore
cannot be a solution to the non-homogeneous equation.
I
As recommended, we use yp (x) = Axe x .
I
Particular solution: yp =
1 x
xe
2
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
Trouble: Trial solution fits corresponding hom. eqn.
Troubleshooting
If the trial solution yp is a solution of the corresponding homogeneous equation,
then it cannot be a solution to the non-homogeneous equation. In this case, we
multiply the trial solution by x (or x 2 or x 3 ... as necessary) to get a new trial
solution that does not satisfy the corresponding homogeneous equation. Then
proceed as above.
Example Solve y 00 − y = e x .
I
I
yc = c1 e −x + c2 e x .
Next we find a particular solution. Note that Ae x , is a solution to the corresponding homogeneous equation and therefore
cannot be a solution to the non-homogeneous equation.
I
As recommended, we use yp (x) = Axe x .
I
Particular solution: yp =
I
general solution: y = yp + yc =
1 x
xe
2
Annette Pilkington
1 x
xe + c1 e −x + c2 e x
2
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)
NonHomogeneous Second Order Linear Equations (Section 17.2) Example Polynomial Example Exponentiall Example Trigonometric Troubleshooting
G (x) = G1 (x) + G2 (x).
To solve the equation ay 00 + by 0 + cy = G1 (x) + G2 (x), we can find particular
solutions, yp1 and yp2 to of the equations
ay 00 + by 0 + cy = G1 (x),
ay 00 + by 0 + cy = G2 (x)
separately. The general solution of the equation
ay 00 + by 0 + cy = G1 (x) + G2 (x) is yp1 + yp2 + yc , where yc is the solution of
the corresponding homogeneous equation ay 00 + by 0 + cy = 0.
Example For the equation y 00 + y 0 + y = x 2 + e x , we use a trial solution of the
form
yp = (Ax 2 + Bx + C ) + De x .
Annette Pilkington
Lecture 22 : NonHomogeneous Linear Equations (Section 17.2)