4.5. LINEAR NON-HOMOGENEOUS SECOND ORDER DIFFERENTIAL EQUATIONS179
4.5Linear Non-Homogeneous Second Order Di erential Equations
Theorem 4.5.1 (Variation of Parameters) Consider the equations
y00 |
+ p(x)y0 |
+ q(x)y = r(x) |
(1) |
y00 |
+ p(x)y0 |
+ q(x)y = 0. |
(2) |
Suppose that y1 and y2 are any two linearly independent solutions of (2).
Then the general solution of (1) is |
W (y1, y2) dx − y1(x) Z |
W (y1, y2) dx . |
y(x) = c1y1(x) + c2y2(x) + y2(x) Z |
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y1(x)r(x) |
y2(x)r(x) |
Proof. It is already shown that c1y1(x)+c2y2(x) is the most general solution of the homogeneous equation (2), where c1 and c2 are arbitrary constants. We observe that the di erence of any two solutions of (1) is a solution of (2).
Suppose that y (x) is any solution of (1). We wish to find two functions,
u1 and u2, such that |
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y (x) = u1(x)y1(x) + u2(x)y2(x). |
(3) |
By di erentiation of (3), we get |
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y 0(x) = (u10 y1 + u20 y2) + (u1y10 + u2y20 ). |
(4) |
We impose the following condition (5) on u1 and u2: |
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u10 y1 + u20 y2 = 0. |
(5) |
Then |
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y 0(x) = u1y10 + u2y20
y 00(x) = (u1y100 + u2y200) + u01y10 + u02y20 .
Since y (x) is a solution of (1), we get r(x) = y 00 + p(x)y 0 + q(x)y
=(u1y100 + u2y200) + (u01y10 + u02y20 ) + p(x)[u1y10 + u2y20 ]
+q(x)(u1y1 + u2y2)
=u1[y100 + p(x)y10 + q(x)y1] + u2[y200 + p(x)y20 + q(x)y2]
+(u01y1 + u02y20 )
=u01y10 + u02y20 .
4.5. LINEAR NON-HOMOGENEOUS SECOND ORDER DIFFERENTIAL EQUATIONS181
Example 4.5.1 Solve the di erential equation y00 + 8y0 + 12y = e−3x.
We find the general solution of the homogeneous equation y00 + 8y0 + 12y = 0.
We let y = emx be a solution. Then y0 = memx, y00 = m2emx and
m2emx + 8memx + 12emx = 0
m2 + 8m + 12 = 0 m = −6, −2.
So,
yc(x) = Ae−6x + Be−2x
is the complementary solution. We compute the Wronskian
W(e−6x, e−2x) = e−6x(−2)e−2x − e−2x(−6)e−6x
=e−8x(−2 + 6)
=4e−8x
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6= 0. |
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By Theorem 4.4.1, the particular solution is given by |
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yp = e−2x Z |
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e−6x |
e−3x |
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e−2x |
e−3x |
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dx − e−6x Z |
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dx |
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4e− |
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4 e−xdx − e−6x |
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4 e3xdx |
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4 e−x |
4 e3x |
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=−14 e−3x − 121 e−3x
=−13 e−3x.
The complete solution is the sum of the complementary solution yc and the particular solution yp.
y(t) = Ae−6x + Be−2x − 31 e−3x.
182 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION
Exercises 4.5 Find the complementary, particular and the complete solution for each of the following. Use tables of integrals or computer algebra to do the integrations, if necessary.
1. |
y00 |
+ 4y = sin(3x) |
2. |
y00 |
− 9y = e2x |
3. |
y00 |
+ 9y = cos 2x |
4. |
y00 |
− 4y = e−x |
5. |
y00 |
− y = xex |
6. |
y00 |
− 5y0 + 6y = 3e4x |
7. |
y00 − 4y0 + 4y = e−x |
8. |
y00 + 5y0 + 4y = 2ex |
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9. |
my00 − py0 = mg |
10. |
y00 + 5y0 + 6y = x2e2x |
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In exercises 11–20, compute the complete solution for y.
11.y00 + y = 4x, y(0) = 2, y0(0) = 1
12.y00 − 9y = ex, y(0) = 1, y0(0) = 5
13.y00 − 2y0 − 3y = 4, y(0) = 2, y0(0) = −1
14.y00 − 3y0 + 2y = 4x
15.y00 + 4y = sin 2x
16.y00 − 4y = e2x
17.y00 − 4y = e−2x
18.y00 + 4y = cos 2x
19.y00 + 9y = 2 sin 3x + 4 cos 3x
20.y00 + 4y0 + 5y = sin x − 2 cos x