Math 345A - Fall 2008 - Midterm Exam 2
This is a no-book, no-notes test. If you forget a process or method, such as
“reduction of order”, you may re-derive it. You MAY use an integral table. You
may have unlimited time, but if you’ve kept up on the material, this test is doable
in one to two hours.
1. Consider the differential equation: d2y
dx2+ 3dy
dx −10y= 0. Determine a fundamental
set of solutions, y1(x) and y2(x). Prove that these ARE a fundamental set of solutions.
Write the general solution for this DE. Find the unique solution given initial conditions
y(0) = 3, y0(0) = 0.
2. Find the general solution of: i) 2x2y00(x)+4xy0(x)−264y= 0 ii) y00(x)+4y0(x)+7y= 0,
iii) x2y00(x)−5xy0(x)+9y= 0
3. Consider the linear DE: y00 +3
xy0+q(x)y= 0, defined on 0 < x < ∞, for which q(x) is
unknown. It IS known that y(x) = ln(x)
xsolves this DE. Find the general solution.
4. State what Abel’s theorem says about the Wronskian Determinant of solutions of certain
differential equatoins. Consider now a second order DE with solutions y1(x) = x4and
y2(x) = e3x. Find the roots of the Wronskain Determinant, w(x). What must be true
in order for solutions y1and y2to not violate Abel’s theorem.
5. Find the general solution to i) y00 +y= (x+ 2)ex+ 6 and ii) y00 −y= (x+ 2)exusing
the method of undetermined coefficients.
6. Find a particular solution to y00 −2y0=e2xusing the method of variation of parameters.
7. Use the annihilator method to find the general solution to y00 + 2y0= 4.
8. Prove that the operator L≡D2+ 3 sin(x) is linear.
9. Let’s create a new method! Consider the general first order linear DE: y0(x)+p(x)y(x) =
q(x). Suppose we know a homogenous solution, yh(x). We suspect that we can determine
a particular solution, yp(x) of the form yp(x) = u(x)yh(x). Proceed in an analogous
manner to the derivation of the method of variation of parameters, and determine the
function u(x), and therefore, yp(x). Verify your method on the differential equation:
y0(x) + 1
xy=x.
10. Prove the following superposition principle: Given a linear DE L[y] = g(x), with ho-
mogenous solution yh(x) and particular solution yp(x), that c·yh(x) + yp(x) is also a
particular solution, for a scalar constant, c.
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