Math 334 Assignment 5: Solutions to Differential Equations, Exercises of Mathematics

The solutions to assignment 5 of math 334, which includes finding the solutions to initial value problems of second order differential equations, the general solution to a system of two second order differential equations, and the general solution to a fourth order differential equation. The document also covers finding the general solution to the homogeneous equation, a particular solution, and the simplest form of the general solution in the case of g(x) ≡ 1.

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Uploaded on 01/10/2013

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Math 334
Assignment 5
Due: 12 Noon on Thursday, October 19, 2006.
1. Find the solution to the following initial value problems.
(a) y′′′ y′′ 4y+ 4y= 0; y(0) = 4, y(0) = 1, y′′(0) = 19.
(b) y′′′ 4y′′ + 7y6y= 0; y(0) = 1, y(0) = 0, y′′(0) = 0.
2. Find the general solution to the following system of two second order differential equations:
d2x
dt2x+y= 0, x +d2y
dt2y=e3t.
3. Consider the fourth order differential equation:
y(iv)(x)k2y′′(x) = g(x).(1)
(a) Find the general solution to the homogeneous equation.
(b) Show that a particular solution to Eq. (1) can be written in the form
φp(x) = 1
k2Zxg(x)dx
x
k2Zg(x)dx +ekx
2k3Zg(x)ekx dx
ekx
2k3Zg(x)ekx dx.
(c) Show that the general solution in part (b) can be rewritten in the form
φp(x) = Zx
0
g(s)G(xs)ds,
where
G(ξ) = 1
k3(sinh ).
(d) Determine the simplest form of the general solution to Eq. (1) in the case g(x)1.

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Math 334

Assignment 5

Due: 12 Noon on Thursday, October 19, 2006.

  1. Find the solution to the following initial value problems.

(a) y′′′^ − y′′^ − 4 y′^ + 4y = 0; y(0) = − 4 , y′(0) = − 1 , y′′(0) = −19. (b) y′′′^ − 4 y′′^ + 7y′^ − 6 y = 0; y(0) = 1, y′(0) = 0, y′′(0) = 0.

  1. Find the general solution to the following system of two second order differential equations:

d^2 x dt^2 −^ x^ +^ y^ = 0,^ x^ +^

d^2 y dt^2 −^ y^ =^ e

3 t.

  1. Consider the fourth order differential equation:

y(iv)(x) − k^2 y′′(x) = g(x). (1) (a) Find the general solution to the homogeneous equation. (b) Show that a particular solution to Eq. (1) can be written in the form

φp(x) = (^) k^12

xg(x) dx − (^) kx 2

g(x) dx + e

kx 2 k^3

g(x)e−kx^ dx − e

−kx 2 k^3

g(x)ekx^ dx.

(c) Show that the general solution in part (b) can be rewritten in the form

φp(x) =

∫ (^) x 0

g(s)G(x − s) ds,

where G(ξ) = (^) k^13 (sinh kξ − kξ). (d) Determine the simplest form of the general solution to Eq. (1) in the case g(x) ≡ 1.