MAE675 Fall 2020 Homework #2
Note: This homework is given Wednesday, September 2. It is due at the start of class on Wednesday, September 9. Points for each problem are given in parentheses beside problem statement.
Problem 1. A mass-spring-damper system without forcing is described by the linear second-order differential equation with initial conditions:
ÿ + 1.2ẏ + 0.36y = 0, y(0) = 0.3, ẏ(0) = −0.15.
Find a fundamental set of solutions to the homogeneous equation, using the method of reduction of order. Then find the solution to this initial value problem. (5)
Problem 2. A second order equation P (x)y′′ +Q(x)y′ +R(x)y = 0 is said to be exact if it can be written in the form
[ P (x)y′
]′ + [ f(x)y
]′ = 0 where f is in terms of P, Q, R.
a. Show that a necessary condition for exactness is: P ′′(x)−Q′(x) + R(x) = 0. Note that this is also sufficient for exactness.
b. If an equation is not exact it can be made exact by multiplying by an integrating factor µ(x). Derive the differential equation (adjoint equation) governing µ(x).
c. Determine the adjoint equation for the Airy equation: y′′ − xy = 0.
d. A second order linear equation P (x)y′′+Q(x)y′+R(x)y = 0 is said to be self-adjoint if it is identical to its adjoint. Show that the necessary condition for this to be true is for P ′(x) = Q(x). Is the Airy equation self-adjoint? (2.5+4+3.5+4=14)
Problem 3. Consider the following initial value problem,
y′′ + y = g(x), y(x0) = 0, y ′(x0) = 0, x ∈ [0,∞).
a. Show that the general solution of y′′ + y = g(x) is given by
y = φ(x) =
( c1 −
∫ x α
g(t) sin t dt
( c2 +
∫ x β
g(t) cos t dt
where c1, c2 are arbitrary constants and α, β are any conveniently chosen points.
b. Using the result of (a) show that y(x0) = 0 and y ′(x0) = 0 if,
∫ x0 α
g(t) sin t dt, c2 = − ∫ x0 β
g(t) cos t dt,
and hence the solution of the above initial value problem for arbitrary g(x) is,
y = φ(x) =
∫ x x0
g(t) sin(x− t) dt.
Notice that this equation gives a formula for computing the solution of the original initial value problem for any given nonhomogeneous term g(x). The function φ(x) will not only satisfy the differential equation but will also automatically satisfy the initial conditions. If we think of x as time, the formula also shows the relation between the input g(x) and the output φ(x). Further, we see that the output at time x depends only on the behavior of the input from the initial time x0 to the time of interest. This integral is often referred to as the convolution of sin x and g(x).
c. Now that we have the solution of the linear nonhomogeneous differential equa- tion satisfying homogeneous initial conditions, we can solve the same problem with nonhomogeneous initial conditions by superimposing a solution of the homogeneous equations satisfying nonhomogeneous initial conditions. Show that the solution of
y′′ + y = g(x), y(x0 = 0) = y0, y ′(x0 = 0) = y
y = φ(x) =
∫ x x0
g(t) sin(x− t) dt+ y0 cosx+ y′0 sinx. (4 + 8 + 5 = 17)
Problem 4. The Tchebycheff (1821-1894) differential equation is
(1− x2)y′′ − xy′ + α2y = 0, α constant.
a. Determine two linearly independent solutions in powers of x for |x| < 1. For α = 1 graph 5 and 10 terms of both solutions.
b. Resolve this problem using (1) the dsolve command, and (2) the dsolve command with “series” option in MAPLE (or equivalent commands in Mathematica or Matlab). Plot these results on your curves obtained in part a.
c. Show that if α is a non-negative integer n, then there is a polynomial solution of degree n. These polynomials, when properly normalized, are called Tchebycheff polynomials. They are very useful in problems requiring a polynomial approximation to a function defined on −1 ≤ x ≤ 1.
d. Find polynomial solutions for each of the cases n = 0, 1, 2, 3. (2.5+4+3.5+4=14)