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Date: April 22, 2003
Instructor: Ivan Avramidi
MATH 335: Applied Analysis I
Spring 2003
TEST 2: Solutions
- Find the solution of the initial value problem
y(4)^ + y = 0, y(0) = 0, y′(0) = 0, y′′(0) = − 1 y′′′(0) = 0.
How does the solution behave as t → ∞?
Solution. The characteristic equation is
r^4 + 1 = 0
The roots of this equation are:
rk = e(iπ+i^2 πk)/^4 , k = 1, 2 , 3 , 4
r 1 =
(−1 + i), r 2 =
(− 1 − i), r 3 =
(1 − i), r 4 =
(1 + i)
Denoting α =
, the general solution can be written as
y = c 1 eαt^ cos(αt) + c 2 eαt^ sin(αt) + c 3 e−αt^ cos(αt) + c 4 e−αt^ sin(αt)
We compute the derivatives
y′^ = α(c 1 + c 2 )eαt^ cos(αt) + α(−c 1 + c 2 )eαt^ sin(αt) + α(c 3 + c 4 )e−αt^ cos(αt) +α(c 3 − c 4 )e−αt^ sin(αt)
y′′^ = 2 α^2 c 2 eαt^ cos(αt) − 2 α^2 c 1 eαt^ sin(αt) − 2 α^2 c 4 e−αt^ cos(αt) + 2α^2 c 3 e−αt^ sin(αt)
y′′′^ = 2 α^3 (−c 1 + c 2 )eαt^ cos(αt) + 2α^3 (−c 1 − c 2 )eαt^ sin(αt) + 2α^3 (c 3 + c 4 )e−αt^ cos(αt) +2α^3 (−c 3 + c 4 )e−αt^ sin(αt)
By using the initial conditions we find
y(0) = 0 = c 1 + c 3 y′(0) = 0 = c 1 + c 2 + c 4 − c 3 y′′(0) = − 1 = c 2 − c 4 y′′′(0) = 0 = −c 1 + c 2 + c 3 + c 4
By adding the second and the forth equation we find c 2 + c 4 = 0. This together with the third equation gives
c 2 = −
, c 4 =
Further, from the second equation we have c 1 − c 3 = 0, which together with the first equation means that c 1 = c 3 = 0.
Thus, the solution of the initial value problem is
y(t) = −
eαt^ − e−αt
sin(αt) = − sinh(αt) sin(αt) ,
where α =
As t → ∞ the solution oscillates with the amplitude that grows to ∞.
- Find the general solution of the differential equation
y(4)^ − y = 3t + cos t
So,
Y 2 = −
t sin t
and, finally,
y = c 1 et^ + c 2 e−t^ + c 3 cos t + c 4 sin t − 3 t −
t sin t
- Solve the differential equation
2 y′′^ + xy′^ + 3y = 0
by means of a power series about the point x 0 = 0. a) Find the recurrence relation. b) Find the first three terms in each of two linearly independent solutions.
Solution. The point x = 0 is an ordinary point of the equation. So, we look for a power series solution in the form y =
anxn
where an = 0 for n < 0. We compute
y′^ =
annxn−^1
xy′^ =
annxn
y′′^ =
ann(n − 1)xn−^2 =
an+2(n + 2)(n + 1)xn
Substituting into the differential equation we obtain the recurrence relation
2 an+2(n + 2)(n + 1) + ann + 3an = 0
or an+2 = − n + 3 2(n + 2)(n + 1) an
By using these relations we obtain
a 2 = −
a 0
a 3 = −
a 1
a 4 = (−1)^2
a 0
a 5 = (−1)^2
a 1
a 6 = (−1)^3
a 0
a 7 = (−1)^3
a 1
and, in general,
a 2 k =
(−1)k(2k + 1)!! 2 k(2k)! a 0 , a 2 k+1 =
(−1)k(2k + 2)!! 2 k+1(2k + 1)! a 1
Finally, we get the first three terms in two linearly independent solutions by letting a 0 = 1 and a 1 = 0 for y 1
y 1 = 1 −
x^2 +
x^4 + · · ·
and a 0 = 0 and a 1 = 1 for y 2
y 2 = x −
x^3 +
x^5 + · · ·
- Solve the differential equation
(1 − x^2 )y′′^ − xy′^ + 4y = 0
by means of a power series about the point x 0 = 0. a) Find the recurrence relation. b) Find the first three terms in each of two linearly independent solutions. c) Show that one solution is in fact a quadratic polynomial.
It is easy to see that all subsequent coefficients of even order are equal to zero
a 2 k = 0, k = 2, 3 , 4 ,...
Since the series for the second solution terminates, it is just a quadratic polynomial
y 2 = 1 − 2 x^2
- Show that this differential equation
2 xy′′^ + y′^ + xy = 0
has a regular singular point at x = 0. a) Determine the indicial equation (and its roots). b) Find the recurrence relation. c) Find the first three terms in each of two linearly independent solutions (for x > 0).
Solution. We rewrite the equation in the canonical form
x^2 y′′^ + xφ(x)y′^ + ψ(x)y = 0
where φ(x) =
, ψ(x) =
x^2
Since both φ(x) and ψ(x) are polynomials, they are analytic at x = 0, and, therefore, the point x = 0 is a regular singular point. So, we search for a solution of the form
y =
anxn+r
where an = 0 for n < 0. We compute
xy = =
anxn+r+1^ =
an− 1 xn+r
y′^ =
an(n + r)xn+r−^1 =
an+1(n + r + 1)xn+r
y′′^ =
an(n + r)(n + r − 1)xn+r−^2
xy′′^ =
an(n + r)(n + r − 1)xn+r−^1 =
an+1(n + r + 1)(n + r)xn+r
Substituting into the differential equation we obtain
2 an+1(n + r + 1)(n + r) + an+1(n + r + 1) + an− 1 = 0
or (n + r)(2n + 2r − 1)an + an− 2 = 0
By letting n = 0 we get the indicial equation (recall that a− 2 = 0)
r(2r − 1) = 0
which has the roots r 1 = 0, r 2 =
We consider first the first root r 1. The recurrence relation becomes
an = −
n(2n − 1) an− 2
We obtain from here
a 1 = 0
a 2 = −
a 0
a 3 = 0
a 4 = (−1)^2
a 0
a 5 = 0
a 6 = (−1)^3
a 0
Therefore, the first solution is (by setting a 0 = 1)
y 1 = 1 −
x^2 +
x^4 + · · ·
For the second root r 2 = 12 we get the recurrence relation
an = −
n(2n + 1) an− 2
