1. Determine the Green's function and the general solution for the problem
000(x) = f(x) y(0) = y”(0) = y
(1) = 0
2. Let a and B be constants. Under what conditions does the general first order boundary
(x) + f(x)y(x) = g(x), Ay(α) + By(β) = 0
have a unique solution for all g? When is this case, write down a formula for the solution.
3. Consider the problem of determining a solution to the initial value problem
(x) = f(x, y), y(x0) = a
where f is continuous dierentiable function. The problem is equivalent to the integral
equation, labeled as (1)
y(x) = a +
Begins with an initial approximation φ(x) to the solution of (1) we can generate a sequence
of successive approximations, φa, φa, φ3, . . . via
φn+1(x) = a +
f(ξ, φn(ξ))dξ, n = 0, 1, 2, . . .
Follow the same procedure as described above to nd the sequence of solutions φn+1(x)
for the initial value problem
(x) = 2x(1 + y), y(0) = 0.
4. a) Find the Neumann series of (1)
y(x) = x + λ
(x − ξ)y(ξ)dξ
b) Does the integral equation
y(x) = sin(x) + 3 Z p
i(x + ξ)y(ξ)dξ
have a solution?
5. Let B be a Banach space with norm || · ||. Let T : B → B
(a) Show that if T is a contraction mapping, then it is continuous.
(b) Let [a, b] ⊆ R, let K(t, s) be continuous on [a, b] × [a, b] with |K(t, s)| ≤ M for
some positive constant M. Let f : [a, b] × R− > R be continuous and satisfy Lipschitz
condition with Lipschitz constant L, for some constant L > 0. Also, p : [a, b] → R be
continuous. Define T : C[a, b] → C[a, b] by
(T x)(t) = p(t) + Z b
K(t, s)f(s, x(s))dx.
Show that if ML(b − a) ≤ α, for some real number α with α ∈ (0, 1), then T is a contraction on C[a, b].
(c) Let T : [0, 1] → [0, 1]. Show that if there is a real number α with α ∈ (0, 1) such
(x)| ≤ α,
for all x ∈ [0, 1] where T
is the derivative of T, then T is a contraction on [0, 1].
6. Determine a change of variables which reduces the quadratic forms
Q1 = 5x
1 + 4x1x2 + 3x
, Q2 = x
1 + 2x
simultaneously to canonical forms.
(a) Calculate A100. Do not simplify your answer.
(b) Solve the system of dierential equations
0 = 4x − 2y
0 = 7x − 5y
x(0) = 1, y(0) = 2.
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