Page 20 - Schaum's Outline of Differential Equations

CHAP. 1] BASIC CONCEPTS 3

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Example 1.7. The problem y" + 2y' = e ;y(n) = 1, y'(Tt) = 2 is an initial-value problem, because the two subsidiary con-
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ditions are both given at x = n. The problem y" + 2y' = e ; y(0) = 1, y(l)=l is a boundary-value problem, because the two
subsidiary conditions are given at the different values x = 0 and x = 1.
A solution to an initial-value or boundary-value problem is a function y(x) that both solves the differential
equation and satisfies all given subsidiary conditions.

Solved Problems

1.1. Determine the order, unknown function, and the independent variable in each of the following differential
equations:

(a) Third-order, because the highest-order derivative is the third. The unknown function is y; the independent
variable is x.
(b) Second-order, because the highest-order derivative is the second. The unknown function is y; the independent
variable is t.
(c) Second-order, because the highest-order derivative is the second. The unknown function is t; the independent
variable is s.
(d) Fourth-order, because the highest-order derivative is the fourth. Raising derivatives to various powers does not
alter the number of derivatives involved. The unknown function is b; the independent variable is p.

1.2. Determine the order, unknown function, and the independent variable in each of the following
differential equations:

(a) Second-order. The unknown function is x; the independent variable is y.
(b) First-order, because the highest-order derivative is the first even though it is raised to the second power. The
unknown function is x; the independent variable is y.
(c) Third-order. The unknown function is x; the independent variable is t.
(d) Fourth-order. The unknown function is y; the independent variable is t. Note the difference in notation between
(
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the fourth derivative y *\ with parentheses, and the fifth power y , without parentheses.
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1.3. Determine whether y(x) = 2e + xe is a solution of y" + 2y' + y = 0.
Differentiating y(x), it follows that

Substituting these values into the differential equation, we obtain

Thus, y(x) is a solution.