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Unformatted text preview: MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 18 Second Order Differential Equations
Example. Consider the initial value problem
y + 5 y + 6 y = 0; where y (0) = 2, y (0) = 3. To solve this equation, we proceed in two steps: First, we ﬁnd the general solution to the diﬀerential equation.
Second, we ﬁnd the particular solution to the initial value problem by taking in account the initial conditions.
We found the general solution to this diﬀerential equation above:
y (t) = c1 e−2t + c2 e−3t
for some constants c1 and c2 . We must ﬁnd which constants satisfy the initial conditions. We have the
y (t) =
c1 e−2t +
3 = (−2) c1 + (−3) c2
y (t) = (−2) c1 e
+ (−3) c2 e
Upon multiplying the ﬁrst equation by 3 then by 2, we have the following systems of equations:
+ −3 c2 =6
=9 2 c1
−2 c1 +
+ −3 c2
=7 The constants are c1 = 9 and c2 = −7. Hence the solution to the initial value problem is
y (t) = 9 e−2t − 7 e−3t . Non-Example. Now say that we wish to ﬁnd the general solution to the diﬀerential equation
y + 2 y + y = 0.
Again, we consider the characteristic equation:
r2 + 2 r + 1 = 0.
This polynomial factors as
r2 + 2 r + 1 = (r + 1) 2 =⇒ r1 = r2 = −1. In this case, the two roots are not distinct. Hence the trick we have introduced does not give the general
solution to the diﬀerential equation. We will introduce in a future lecture a trick which will solve such
equations – even when the roots of the characteristic equation are not distinct.
Existence and Uniqueness Theorems. We are motivated by two questions:
#1. Does one solution exist?
#2. When do we have all solutions?
We attempt to answer these questions by stating the “Existence and Uniqueness Theorem” for second order
equations. To this end, we recall the corresponding result for ﬁrst order diﬀerential equations.
Say we have the initial value problem
a(t) y + b(t) y + c(t) y = 0; y (t0 ) = y0 , If we have an interval I = (α, β ) such that
i. f (t) is continuous on I ,
ii. b(t) and c(t) are continuous on I ,
iii. a(t) is continuous yet a(t) = 0 on I , and
iv. t0 ∈ I ;
then the initial value problem has a unique solution y = y (t).
y (t0 ) = y0 ...
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