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Unformatted text preview: 2 MA 36600 LECTURE NOTES: MONDAY, MARCH 2 It makes sense to choose A and B such that
−10 A
2A + −2 B
+ −10 B = −8
=
0 We can solve for A and B by multiplying the ﬁrst equation by −5; or multiplying the second by 5:
50 A
2A
52 A +
10 B
+ −10 B = 40
=0
= 40 −10 A + −2 B
10 A + −50 B
−52 B = −8
=
0
= −8 This gives A = 40/52 = 10/13 and B = 8/52 = 2/13, so that one solution to the diﬀerential equation is
10 t
2t
Y (t) =
e cos 2t +
e sin 2t.
13
13
Hence the general solution to the nonhomogeneous diﬀerential equation
y − 3 y − 4 y = −8 et cos 2t must be the function y (t) = c1 y1 (t) + c2 y2 (t) + Y (t) = c1 e −t + c2 e +
4t for some constants c1 and c2 . 10 t
2t
e cos 2t +
e sin 2t
13
13 Example 2. Finally, consider the diﬀerential equation
y − 3 y − 4 y = 2 e−t . Again, we ﬁnd the general solution. It suﬃces to ﬁnd one solution Y = Y (t).
If we attempt to make a guess in the form Y (t) = A eαt , we have
Y − 3 Y − 4 Y = α2 − 3 α − 4 A eαt = 2 e−t .
Upon choosing α = −1 would ﬁnd that α2 − 3 α − 4 A eαt = 0 for all t. Hence a guess in this form would
not work! This fails because the function f (t) = 2 e−t involves a polynomial of degree d = 0, and we must
guess a solution which involves a polynomial of degree (d + 2) = 2.
Make the guess that the solution is in the form
Y (t) = A0 + A1 t + A2 t2 eαt
where α = −1; and the Aj are coeﬃcients yet to be determined. We have the derivatives
Y
Y
Y =
A0 e−t
=
(−A0 + A1 ) e−t
= (A0 − 2 A1 + 2 A2 ) e−t This gives the expression +
A1 t e−t
+ (−A1 + 2 A2 ) t e−t
+
(A1 − 4 A2 ) t e−t +
A2 t2 e−t
+ (−A2 ) t2 e−t
+
A2 t2 e−t
−t
−t
Y − 3 Y − 4 Y = −5 A1 + 2 A2 e + −10 A2 t e + 0 t2 e−t = 2 e−t .
It makes sense to choose A0 , A1 , and A2 such that
−5 A1 + 2 A2
−10 A2 =2
=0 This gives A1 = −2/5 and A2 = 0. Since A0 does not appear in these equations, there is no harm in choosing
A0 = 0, so that one solution to the diﬀerential equation is
2
Y = − t e−t .
5
Hence the general solution to the nonhomogeneous diﬀerential equation
must be the function y − 3 y − 4 y = 2 e−t
y (t) = c1 y1 (t) + c2 y2 (t) + Y (t) = c1 e −t 2
+ c2 e + − t e−t
5
4t ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.
 Spring '09
 EdrayGoins

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