Unformatted text preview: 2 MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 18 Example. Consider the initial value problem
2
t − 3 t y + t y − (t + 3) y = 0; y (1) = 2, y (1) = 1. We do not have enough techniques yet to solve this diﬀerential equation, but we can discuss whether a
solution exists. To this end, we compute the largest interval I where there exists a unique solution – without
solving the diﬀerential equation.
Deﬁne the functions
a(t) = t2 − 3 t, b(t) = t, c(t) = − (t + 3) , and f (t) = 0. These are polynomials, so these functions are all continuous on the entire real line R = (−∞, ∞). The
polynomial a(t) = t (t − 3) vanishes when t = 0, 3 so 0, 3 ∈ I . This means we have three possibilities for I :
(−∞, 0), (0, 3), or (3, ∞). This interval must contain t0 = 1, so the longest interval is I = (0, 3).
Linear Homogeneous Equations
Principle of Superposition. Recall in the last lecture that we showed that given the constant coeﬃcient
equation
a y + b y + c y = 0
once we know two solutions y1 = er1 t and y2 = er2 t , then more solutions are in the form
y (t) = c1 y1 (t) + c2 y2 (t)
for any constants c1 and c2 . We generalize this idea.
Consider the more general equation L[y ] = 0 in terms of
d
d2
L[y ] = a(t) y + b(t) y + c(t) = a(t) 2 + b(t)
+ c(t) y.
dt
dt
If y1 = y1 (t) and y2 = y2 (t) are two solutions, then the linear combination
y (t) = c1 y1 (t) + c2 y2 (t)
is also a solution for any constants c1 and c2 . We prove this in the same way we proved this in the case of
constant coeﬃcients. We have the derivatives
y = c1 y1 + c2 y2
y = c1 y1 + c2 y2
y = c1 y1 + c2 y2 which gives the expression
L[y ] = a(t) y + b(t) y + c(t) y
= a(t) c1 y1 + c2 y2 + b(t) c1 y1 + c2 y2 + c(t) c1 y1 + c2 y2
= c1 a(1) y1 + b(t) y1 + c(t) y + c2 a(t) y2 + b(2) y2 + c(t) y2
= c1 L[y1 ] + c2 L[y2 ]. Since L[y1 ] = L[y2 ] = 0 we must have L[y ] = 0 as well. Hence y = c1 y1 + c2 y2 is indeed a solution to the
diﬀerential equation. ...
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 Spring '09
 EdrayGoins
 y2, constants c1, linear homogeneous equations

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