Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 20 Linear Homogeneous Equations
Example. Consider the diﬀerential equation
y + 5 y + 6 y = 0.
We discuss the Wronskian associated with this equation. We have the characteristic equation
r2 + 5 r + 6 = 0.
Since the polynomial factors as r2 = 5 r + 6 = (r + 2) (r + 3), the characteristic equation has the two distinct
real roots r1 = −2 and r2 = −3. Hence two solutions to the diﬀerential equation are
y1 (t) = e−2t and y2 (t) = e−3t . The Wronskian of y1 and y2 is the function
W (t) = y1 y2 − y1 y2 = e−2t −3 e−3t − −2 e−2t e−3t = −3 e−5t + 2 e−5t = −e−5t .
Example. Consider the more general constant coeﬃcient equation
a y + b y + c y = 0.
We have the characteristic equation a r2 + b r + c = 0.
Say that this has roots r1 and r2 , and consider the solutions
y1 (t) = er1 t and y2 (t) = er2 t . The Wronskian of y1 and y2 is the function
W (t) = y1 y2 − y1 y2 = er1 t r2 er2 t − r1 er1 t er2 t = (r2 − r1 ) e(r1 +r2 )t .
In particular, we have the value W0 = W (t0 ) = (r2 − r1 ) e(r1 +r2 )t0 .
Hence W0 = 0 precisely when r1 = r2 . In fact, if r1 = r2 then W (t) = 0 for any t. Fundamental Set of Solutions. We spend the remainder of the lecture discussing when solutions y1 and
y2 form the general solution y = c1 y1 + c2 y2 . Consider the diﬀerential equation
a(t) y + b(t) y + c(t) y = 0.
Say that we can ﬁnd two solutions y1 = y1 (t) and y2 = y2 (t) satisfying the initial conditions
y1 (t0 ) = 1
y2 (t0 ) = 0
and
y1 (t0 ) = 0
y2 (t0 ) = 1
The set {y1 , y2 } is called a fundamental set of solutions. We will show that the general solution to the
diﬀerential equation is
y (t) = c1 y1 (t) + c2 y2 (t)
for constants c1 and c2 . To see why this is true, consider the initial conditions
y (t0 ) = y0 and
y (t0 ) = y0 . We must construct constants c1 and c2 which solve this initial value problem. By the Principle of Superposition, we know that y = c1 y1 + c2 y2 is a solution to the diﬀerential equation. We have the system of
equations
y0 = c1 y1 (t0 ) + c2 y2 (t0 ) = c1
y0 = c1 y1 (t0 ) + c2 y2 (t0 ) = c2
Hence y (t) = y0 · y1 (t) + y0 · y2 (t) is the solution to the initial value problem.
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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
 Equations

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