homogeneous linear differential equations with constant coefficients a lecture in engiana
TRANSCRIPT
![Page 1: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/1.jpg)
Homogeneous Linear Differential Equations
with Constant Coefficients
A Lecture in ENGIANA
![Page 2: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/2.jpg)
Auxillary Equation
• Consider a second order equation
ay’’ + by’ + cy = 0
where a, b, and c are constants.
• If we try to find a solution of the form
y = emx, then after substitution of
y’ = memx and y’’ = m2emx, the equation becomes
am2emx + bmemx + cemx = 0
![Page 3: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/3.jpg)
Auxillary Equation
• Solving am2emx + bmemx + cemx = 0,
emx(am2 + bm + c) = 0
• The quantity in parenthesis, a quadratic equation, is called the auxiliary equation.
• This means that to find the solution y (see previous slide), we must solve for m.
a2
ac4bbm
0cbmam
2
2
![Page 4: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/4.jpg)
Auxillary Equation
There are three possible cases:
• m1 m2; distinct real roots
• m1 = m2; repeated real roots
• m1 m2; conjugate complex roots
![Page 5: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/5.jpg)
Case 1: Distinct Real Roots
For this case, we have
And hence,
Or
xm2
xm1
21 eyandey
21 yyy
xm2
xm1
21 ececy
![Page 6: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/6.jpg)
Example
Find the general solution of
(D2 + D – 6) y = 0
![Page 7: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/7.jpg)
x2
2x3
1
2
2
ececy
,Hence
2m|3m
0)2m)(3m(
06mm
isequationauxiliarythe
,0y)6DD(From
:Solution
![Page 8: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/8.jpg)
Case 2: Real Repeated Roots
• Having two real, repeated roots means
• Now, one solution isxm
11ey
1
2
2
ma2
bm
a2
ac4bbm
0cbmam
![Page 9: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/9.jpg)
Case 2: Real Repeated Roots
Recall that
a2(x)y’’ + a1(x)y’ + a0(x)y = 0
can be written as
y” + P(x)y’ + Q(x)y = 0
where
P(x) = a1(x)/a2(x)
Q(x) = a0(x)/a2(x)
![Page 10: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/10.jpg)
Case 2: Real Repeated Roots
In our case, the coefficients are constants:
ay’’ + by’ + cy = 0
Thus,
y” + Py’ + Qy = 0
where
P = b/a
Q = c/a
![Page 11: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/11.jpg)
Case 2: Real Repeated Roots
Recall also that another solution y2 is
dx)x(y
e)x(yy
21
dx)x(P
12
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Case 2: Real Repeated Roots
Hence,
)x(ey
dx)e(
eey
dx)e(
eey
dx)x(y
e)x(yy
xm2
2xm
xm2xm
2
2xm
dx)m2(xm
2
21
dx)x(P
12
1
1
11
1
11
![Page 13: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/13.jpg)
Case 2: Real Repeated Roots
The general solution is then
xm2
xm1
21
11 xececy
yyy
![Page 14: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/14.jpg)
Example
Find the general solution of
y’’ + 8y’ + 16y = 0
![Page 15: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/15.jpg)
x42
x41
2
2
xececy
,Hence
)twice(4m
0)4m(
016m8m
isequationauxiliarythe
,0y16'y8''yFrom
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Case 3: Conjugate Complex Roots
• If m1 and m2 are complex, then we have
m1 = + i
m2 = - i
where and are real and positive
• Hence, we can write
y = C1e( + i)x + C2e( - i)x
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Case 3: Conjugate Complex Roots
• However, in practice we prefer to work with real functions instead of complex exponentials.
• To this end, we use Euler’s formula:
ei = cos + isin
where is any real number
![Page 18: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/18.jpg)
Case 3: Conjugate Complex Roots
• Thus, we have
e ix = cosx + isinx
e- ix = cosx - isinx
• Note that
e ix + e- ix = 2cosx &
e ix – e- ix = 2isinx
![Page 19: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/19.jpg)
Case 3: Conjugate Complex Roots
• Our solution is then
y = C1e (+i)x + C2e(-i)x
• If we let C1 = 1 and C2 = 1:
y1 = e (+i)x + e(-i)x
y1 = e x(eix + e-ix)
y1 = e x(2cosx)
y1 = 2e xcosx
![Page 20: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/20.jpg)
Case 3: Conjugate Complex Roots
• If we let C1 = 1 and C2 = -1:
y2 = e (+i)x - e(-i)x
y2 = e x(eix - e-ix)
y2 = e x(2isinx)
y2 = 2ie xsinx
![Page 21: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/21.jpg)
Case 3: Conjugate Complex Roots
Thus, the solution to
y = C1e( + i)x + C2e( - i)x
is
y = c1y1 + c2y2
y = c1(e xcosx) + c2(e xsinx)
or
y = e x(c1cosx + c2sinx)
![Page 22: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/22.jpg)
Example
Find the general solution of
(D2 – 4D + 7) y = 0
![Page 23: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/23.jpg)
x3sinecx3cosecy
,Hence
i32m
2
122m
)1(2
)7)(1(4)4()4(m
,Then
07m4m
isy)7D4D(ofequationauxiliaryThe
:Solution
x22
x21
2
2
2
![Page 24: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/24.jpg)
Higher-Order (n>2) Equations: Distinct Roots
Consider the case where the auxiliary equation has distinct roots.
Say we are given
f(D)y = 0.
Then one possible solution is emx,
f(D)emx = 0,
if the auxiliary equation is
f(m) = 0
![Page 25: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/25.jpg)
Higher-Order (n>2) Equations: Distinct Real Roots
In other words, if the distinct roots of the auxiliary equation are m1, m2, …, mn, then the corresponding solutions are exp(m1x), exp(m2x), …, exp(mnx).
The general solution is xm
nxm
2xm
1n21 ec...ececy
![Page 26: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/26.jpg)
Example
Find the general solution of
(D3 + 6D2 + 11D + 6) y = 0
![Page 27: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/27.jpg)
x2
3x3
2x
1
23
ecececy
,Hence
2m|3m|1m
0)2m)(3m)(1m(
,Then
0)6m11m6m(
isequationauxiliaryThe
:Solution
![Page 28: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/28.jpg)
Higher-Order (n>2) Equations: Repeated Real Roots
Consider the case where the auxiliary equation has repeated roots.
Say we are given
f(D)y = 0.
If there are several identically repeated roots m1 = m2 = … = mn = b, then this means
(D - b)n y = 0
![Page 29: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/29.jpg)
Higher-Order (n>2) Equations: Repeated Roots
If we let
y = xkebx [k = 0, 1, 2, …, (n-1)]
Then,
(D – b)n y = (D – b)n [xkebx]
But
(D – b)n [xkebx] = ebxDn[xk] = ebx (0)
Thus,
(D – b)n y = (D – b)n [xkebx] = 0
![Page 30: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/30.jpg)
Higher-Order (n>2) Equations: Repeated Roots
The functions yk = xkebx [e.g., e7x, xe7x, x2e7x, etc.], where k = 0, 1, 2, …, (n – 1) are linearly independent because, aside from the common factor ebx, they contain only the respective powers x0, x1, x2, …, xn-1.
The general solution is thus
y = c1ebx + c2xebx + … + cnxn-1ebx
![Page 31: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/31.jpg)
Example
Find the general solution of
(D4 + 6D3 + 9D2) y = 0
![Page 32: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/32.jpg)
x34321
x34
x33
x02
x01
22
22
234
e)xcc(xccy
or
xececxececy
,Hence
)twice(3m
and)twice(0m
0)3m(m
0)9m6m(m
0m9m6m
:Solution
![Page 33: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/33.jpg)
Higher-Order (n>2) Equations: Repeated Imaginary Roots
• Repeated imaginary roots lead to solutions analogous to those brought in by repeated real roots.
• For instance, if the conjugate pair m = a bi occur three times, the corresponding general solution is
y = (c1 + c2x + c3x2) eaxcosbx +
(c4 + c5x + c6x2) eaxsinbx
![Page 34: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/34.jpg)
Example
Find the general solution of
(D4 + 18D2 + 81) y = 0
![Page 35: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/35.jpg)
x3sin)xcc(x3cos)xcc(y
or
x3sine)xcc(x3cose)xcc(y
,Hence
)twice(i3m
0)9m(
081m18m
:Solution
4321
x043
x021
22
24
![Page 36: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/36.jpg)
Exercises
Find the solution required:
1) (D2 – 2D – 3)y = 0 y(0)=0; y’(0)=-4
2) (D3 – 4D)y = 0 y(0)=0; y’(0)=0; y’’(0)=2
3) (D4 + 2D3 + 10D2)y = 0
4) (D6 + 9D4 + 24D2 + 16)y = 0
5) (D3 + 7D2 + 19D + 13)y = 0 y(0)=0; y’(0)=2;
y’’(0)=-12
6) (4D4 + 4D3 – 3D2 – 2D + 1)y = 0
7) (D4 – 5D2 – 6D – 2)y = 0
![Page 37: Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA](https://reader033.vdocuments.site/reader033/viewer/2022061509/551aacfe550346856e8b4c3b/html5/thumbnails/37.jpg)
Exercises
Find the solution required:
8) (D3 + D2 – D – 1)y = 0
y(0)=1; y(2)=0;
9) Find for x = 2 the y value for the particular solution required:
(D3 + 2D2)y = 0
y(0)=-3; y’(0)=0; y’’(0)=12
0ylimx