Unformatted text preview: MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 16 3 If the characteristic equation a r2 + b r + c = 0 has two distinct real roots r1 and r2 , then the general solution
to the diﬀerential equation is
y (t) = c1 er1 t + c2 er2 t
for some constants c1 and c2 . Recall that a quadratic equation a r2 + b r + c = 0 has two distinct real roots
if its discriminant is positive: b2 − 4 a c > 0.
Example. Say that we wish to ﬁnd the general solution to the diﬀerential equation
y + 5 y + 6 y = 0.
To solve this problem, we consider the characteristic equation. In this case, it is
r2 + 5 r + 6 = 0.
The polynomial factors as
r2 + 5 r + 6 = (r + 2) (r + 3) =⇒ r1 = −2, r2 = −3. Since these two roots are distinct, the general solution to the diﬀerential equation is
y (t) = c1 e−2t + c2 e−3t
for some constants c1 and c2 . Proof of the General Solution. We return to the claim made earlier about the general solution to the
homogeneous linear second order diﬀerential equation with constant coeﬃcients. In fact, we make a stronger
claim. Consider the initial value problem
a y + b y + c y = 0; where y (t0 ) = y0 ,
y (t0 ) = y0 . If the characteristic equation a r2 + b r + c = 0 has two distinct real roots r1 and r2 , then the solution to the
diﬀerential equation is the function
y (t) = c1 er1 t + c2 er2 t
in terms of the constants
c1 =
y0 − r2 y0 −r1 t0
e
r1 − r 2 and c2 =
y0 − r1 y0 −r2 t0
e
.
r2 − r1 We sketch why this result is true. The proof relies on the “existence” and “uniqueness” of the solution
to an initial value problem. For now, we give existence i.e., we simply explain why the function above is a
solution to the diﬀerential equation. We have the derivatives
y = c1 er1 t + c2 er2 t y = r1 c1 er1 t + r2 c2 er2 t y = r1 2 c1 er1 t + r2 2 c2 er2 t This gives the expression
a y + b y + c y = a r1 2 c1 er1 t + r2 2 c2 er2 t + b r1 c1 er1 t + r2 c2 er2 t + c c1 er1 t + c2 er2 t
= a r1 2 c1 er1 t + b r1 c1 er1 t + c c1 er1 t + a r2 2 c2 er2 t + b r2 c2 er2 t + c c2 er2 t
= a r1 2 + b r1 + c c1 er1 t + a r2 2 + b r2 + c c2 er2 t . Since r1 and r2 are roots of the characteristic equation, we have
a r1 2 + b r1 + c = 0
=⇒
a y + b y + c y = 0.
a r2 2 + b r2 + c = 0 Hence y is a solution to the diﬀerential equation as claimed. It suﬃces then to explain how to ﬁnd the
constants c1 and c2 from the initial conditions. We have the derivatives
y (t) =
c1 er1 t +
c2 er2 t
y0 =
c1 er1 t0 +
c2 er2 t0
=⇒
y (t) = r1 c1 er1 t + r2 c2 er2 t
y0 = r1 c1 er1 t0 + r2 c2 er2 t0 ...
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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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