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Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27 3 Then the general solution to the nonhomogeneous equation is the function
y (t) = c1 y1 (t) + c2 y2 (t) + Y (t)
for some constants c1 and c2 . We remark that typically ﬁnding y1 and y2 is easy: when the coeﬃcients a(t),
b(t), and c(t) are constant functions, one can compute y1 and y2 by considering roots of the characteristic
equation a r2 + b r + c = 0. We will spend the next couple of lectures discussing how to ﬁnd a solution Y .
Note that once one solution Y is found, we can compute all solutions y .
We give a proof of the claim above. Say that y = y (t) is any solution to the nonhomogeneous equation,
and let Y = Y (t) be as in assumption (iii). Denote the diﬀerence
u(t) = y (t) − Y (t). We see that u = u(t) is a solution to the homogeneous equation:
a(t) u + b(t) u + c(t) u = a(t) y − Y + b(t) y − Y + c(t) y − Y
= a(t) y + b(t) y + c(t) y − a(t) Y + b(t) Y + c(t) Y = f (t) − f (t)
= 0. By assumption (i), both y1 and y2 are also solutions to a(t) u + b(t) u + c(t) u = 0; and by assumption (ii)
these functions are linearly independent. Hence {y1 , y2 } is a fundamental set of solutions for the homogeneous
equation. That means there exist constants c1 and c2 such that
u(t) = c1 y1 (t) + c2 y2 (t) =⇒ y (t) = u(t) + Y (t) = c1 y1 (t) + c2 y2 (t) + Y (t). Example. Consider the diﬀerential equation
y − 3 y − 4 y = 3 e2t . We will ﬁnd the general solution of this equation. To do so, we compute functions y1 = y1 (t), y2 = y2 (t),
and Y = Y (t) as above.
First we ﬁnd the functions y1 and y2 as solutions to the homogeneous equation
We have the characteristic polynomial
Hence we have the roots y − 3 y − 4 y = 0. r2 − 3 r − 4 = 0 =⇒ (r + 1) (r − 4) = 0 =⇒ r1
r2 = −1
=
4 y1 (t) = e−t
y2 (t) = e4t Since r1 = r2 we see that the functions y1 and y2 are linearly independent. That is, the Wronskian
W y1 , y2 (t) = 0.
Second we seek a function Y = Y (t) as a solution to the nonhomogeneous equation
Y − 3 Y − 4 Y = 3 e2t . It suﬃces just to ﬁnd one solution Y = Y (t). To this end, we make a guess that such a solution is in the
form
Y (t) = A eαt
for some constants A and α. We have the derivatives
Y
=
A eαt
Y
= α A eαt
=⇒
Y − 3 Y − 4 Y = α2 − 3 α − 4 A eαt = 3 e2t .
Y = α2 A eαt
We now choose α = 2, so that α2 − 3 α − 4 A eαt = −6 A e2t . Similarly, we choose A = −1/2. This means
that one solution to the nonhomogeneous equation is the function
1
Y (t) = − e2t .
2
Hence the general solution to the nonhomogeneous diﬀerential equation
y − 3 y − 4 y = 3 e2t ...
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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 University.
 Spring '09
 EdrayGoins

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