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1.3 Initial Conditions; Initial-Value Problems
As we noted in the preceding section, we can obtain a particular solution of an nth order differential equation simply by assigning specific values to the n constants in the general solution. However, in typical applications of differential equations you will be asked to find a solution of a given equation that satisfies certain preassigned conditions.
Example 1. Find a solution of y′^ = 3x^2 − 2 x
that passes through the point (1, 3).
SOLUTION In this case, we can find the general solution by integrating:
y =
3 x^2 − 2 x
dx = x^3 − x^2 + C.
The general solution is y = x^3 − x^2 + C.
To find a solution that passes through the point (2, 6), we set x = 2 and y = 6 in the general solution and solve for C:
6 = 2^3 − 22 + C = 8 − 4 + C which implies C = 2.
Thus, y = x^3 − x^2 + 2 is a solution of the differential equation that satisfies the given condition. In fact, it is the only solution that satisfies the condition since the general solution represented all solutions of the equation and the constant C was uniquely determined.
Example 2. Find a solution of
x^2 y′′^ − 2 xy′^ + 2y = 4x^3
which passes through the point (1, 4) with slope 2.
SOLUTION As shown in Example 4 in the preceding section, the general solution of the differential equation is y = C 1 x^2 + C 2 x + 2x^3.
Setting x = 1 and y = 4 in the general solution yields the equation
C 1 + C 2 + 2 = 4 which implies C 1 + C 2 = 2.
The second condition, slope 2 at x = 1, is a condition on y′; we want y′(1) = 2. We calculate y′: y′^ = 2C 1 x + C 2 + 6x^2 ,
and then set x = 1 and y′^ = 2. This yields the equation
2 C 1 + C 2 + 6 = 2 which implies 2 C 1 + C 2 = − 4.
Now we solve the two equations simultaneously:
C 1 + C 2 = 2
2 C 1 + C 2 = − 4
We get: C 1 = − 6 , C 2 = 8. A solution of the differential equation satisfying the two conditions is y(x) = − 6 x^2 + 8x + 2x^3.
It will follow from our work in Chapter 3 that this is the only solution of the differential equation that satisfies the given conditions.
INITIAL CONDITIONS Conditions such as those imposed on the solutions in Examples 1 and 2 are called initial conditions. This term originated with applications where processes are usually observed over time, starting with some initial state at time t = 0.
Example 3. The position y(t) of a weight suspended on a spring and oscillating up and down is governed by the differential equation
y′′^ + 9y = 0.
(a) Show that the general solution of the differential equation is:
y(t) = C 1 sin 3t + C 2 cos 3t.
(b) Find a solution that satisfies the initial conditions y(0) = 1, y′(0) = −2.
SOLUTION
(a) y = C 1 sin 3t + C 2 cos 3t y′^ = 3C 1 cos 3t − 3 C 2 sin 3t y′′^ = − 9 C 1 sin 3t − 9 C 2 cos 3t
Substituting into the differential equation, we get
y′′^ + 9y = (− 9 C 1 sin 3t − 9 C 2 cos 3t) + 9 (C 1 sin 3t + C 2 cos 3t) = 0.
Thus y(t) = C 1 sin 3t + C 2 cos 3t is the general solution.
Similarly, the problem:
- Find a solution of the differential equation
y′′′^ − 3 y′′^ + 3y′^ − y = 0
satisfying the conditions y(0) = 1, y′(0) = 2 is not an initial-value problem; a third order equation requires three conditions: y(c) = k 0 , y′(c) = k 1 , y′′(c) = k 2.
EXISTENCE AND UNIQUENESS The fundamental questions in any course on dif- ferential equations are:
(1) Does a given initial-value problem have a solution? That is, do solutions to the problem exist? (2) If a solution does exist, is it unique? That is, is there exactly one solution to the problem or is there more than one solution.
The initial-value problems in Examples 1, 2, and 3 each had a unique solution; values for the arbitrary constants in the general solution were uniquely determined.
Example 4. The function y = x^2 is a solution of the differential equation y′^ = 2√y and y(0) = 0. Thus the initial-value problem
y′^ = 2
y; y(0) = 0.
has a solution. However, y ≡ 0 also satisfies the differential equation and y(0) = 0. Thus, the initial-value problem does not have a unique solution. In fact, for any positive number a, the function
ya(x) =
0 , x ≤ a (x − a)^2 , x > a
is a solution of the initial-value problem.
a x
y
Example 5. The one-parameter family of functions y = Cx is the general solution of
y′^ = y x.
There is no solution that satisfies y(0) = 1; the initial-value problem
y′^ = y x , y(0) = 1
does not have a solution.
The questions of existence and uniqueness of solutions will be addressed in the specific cases of interest to us. A general treatment of existence and uniqueness of solutions of initial-value problems is beyond the scope of this course.
Exercises 1.
- (a) Show that each member of the one-parameter family of functions
y = Ce^5 x
is a solution of the differential equation y′^ − 5 y = 0. (b) Find a solution of the initial-value problem y′^ − 5 y = 0, y(0) = 2.
- (a) Show that each member of the two-parameter family of functions
y = C 1 e^2 x^ + C 2 e−x
is a solution of the differential equation y′′^ − y′^ − 2 y = 0. (b) Find a solution of the initial-value problem
y′′^ − y′^ − 2 y = 0; y(0) = 2, y′(0) = 1.
- (a) Show that each member of the one-parameter family of functions
y =
Cex^ + 1 is a solution of the differential equation y′^ + y = y^2. (b) Find a solution of the initial-value problem y′^ + y = y^2 ; y(1) = − 1.
- (a) Show that each member of the three-parameter family of functions
y = C 2 x^2 + C 1 x + C 0
is a solution of the differential equation y′′′^ = 0. (b) Find a solution of the initial-value problem
y′′′^ = 0; y(1) = 1, y′(1) = 4, y′′(1) = 2.
(a) Determine whether there are one or more members of this family that satisfy the conditions y(0) = 0, y(π) = 0. (b) Show that the zero function, y ≡ 0, is the only member of the family that satisfies the conditions y(0) = 0, y(π/2) = 0.
- Given the differential equation (y′)^2 − xy′^ + y = 0. (a) Show that the family of straight lines y = Cx − C^2 is the general solution of the equation (b) Show that y = 14 x^2 is a solution of the equation. Note that this function is not included in the general solution of the equation; it is a singular solution of the equation.
- Given the differential equation x(y′)^2 − 2 yy′^ + 4x = 0.
(a) Show that the one-parameter family y = x
2 + C 2
C
is the general solution of the equation (b) Show that each of y = 2x and y = − 2 x is a solution of the equation. Note that these functions are not included in the general solution of the equation; they are singular solutions of the equation.
Find the differential equation of the given family.
- y = Cx^3 + 1.
- y = Cx^2 + 3
- y^3 = Cx^2 + 3.
- y^2 = Cx^4 − 2
- y = Ce^2 x^ + e−^2 x
- y = Cex^ + sin x.
- y = C 1 x + C 2.
- y = C 1 ex^ + C 2 e−^2 x.
- y = (C 1 + C 2 x)e^2 x
- y = C 1 x + C 2 x^2.
- y = C 1 x + C 2 x−^1.
- y = C 1 cos 3x + C 2 sin 3x.
- y = C 1 sin (3x + C 2 ).
- y = C 1 e^2 x^ cos 3x + C 2 e^2 x^ sin 3x.
- y = C 1 + C 2 x + C 3 x^2.
- y = C 1 x + C 2 x^2 + C 3 x^3.
- Refer to Example 4. Verify that the function ya(x) is a solution of the initial-value problem y′^ = 2√y, y(0) = 0.
