MATH 54, QUIZ 8 (UNGRADED)
(1) Show that ex and e2x are linearly independent functions (in the vector space of all differentiable
functions on R). Do 1, x, and x2 form a linearly independent set? Why or why not?
Solution:
To see that ex and e2x are linearly independent, suppose aex + be2x = 0 for some scalars a and b.
We must show that a and b are zero. Plugging in x = 0 gives the equation a + b = 0. Plugging in
x = 1 gives the equation ae + be2 = 0. Subtracting e times the first equation from the second gives
b(e2 − e) = 0, so b = 0, hence also a = 0.
To see that 1, x, and x2 are independent, suppose that some linear combination of these three
functions is zero, i.e., a + bx + cx2 = 0 for some scalars a, b, and c. We have to show that a, b, and
c are all zero. We plug in different values of x to solve for a, b, c. Taking x = 0 gives a = 0. Taking
x = 1 then gives b + c = 0. Taking x = −1 gives −b + c = 0. Adding thesse two equations shows
that c = 0, and then we get that b = 0. So the equation has only the trivial solution, hence the three
functions are independent.
(2) Find all values of a and b for which the function
f (t) = eat + b
is a solution to the differential equation
f 00 (t) + 2f 0 (t) + f (t) = 0.
Solution:
Substituting f (t) = eat + b into the equation yields (a + 1)2 eat + b = 0, and rearranging, we have
−(a + 1)2 eat = b
This equation is supposed to hold for all t, but the right hand side does not depend on t (i.e., it’s
a constant function of t). Therefore the left hand side must also be a constant function. The onoly
way this can happen is if a = 0 (so the part involving t is zero) or if a = −1 (so the whole left hand
side is zero). Thus the only values of (a, b) such that f is a solution are (0, −1) and (−1, 0).
(3) Find the general solution to the homogeneous linear differential equation
y 00 + 4y 0 + 4y = 0.
Then find a particular solution which satisfies the initial condition y(0) = 1, y 0 (0) = 0.
Solution: We guess that the solution has the form y = erx . Plugging this function into the
equation gives the auxiliary equation r2 + 4r + 4 = 0, which has a double root r = −2. Therefore
two linearly independent solutions are y1 = e−2t and y2 = te−2t , and the general solution is just an
arbitrary linear combination of these two:
y = c1 e−2t + c2 te−2t
If we impose the condition that y(0) = 1, y 0 (0) = 0, we find that c1 = 1 and c0 = 2, so the solutino
to the initial value problem is y = e−2t + 2te−2t .
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