Separable Di fferential Equations

Separable Differential Equations

S. F. Ellermeyer October 1, 2003

The differential equation

dy = ky dt

(1)

(where k is a given constant) is extremely important in applications and in the

general theory of differential equations and, hence, we should make it a point

to become comfortable with its family of solutions. We can find this family

of solutions by using the method of separation of variables. Solving the

differential equation (1) will be our first illustration of this method.

1 General Solution of dy/dt = ky

Our first observation about the differential equation (1) is that the constant function y = 0 is a solution. This is easily checked, for if y (t) = 0 for all t then

dy d dt = dt (0) = 0 = k ? 0 = ky.

Next, we would like to find all solutions, y, of the differential equation (1) such that y (t) 6= 0 for any t. If y is a solution of (1) and y (t) 6= 0 for any t, then we can divide both sides of (1) by y to obtain

1 ? dy = k.

(2)

y dt

Next, we integrate both sides of equation (2) with respect to the independent

variable t to obtain

Z?

?

Z

1 y

?

dy dt

dt =

k dt.

(3)

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To evaluate the indefinite integral on the left hand side of equation (3), we make the substitution u = y. This substitution implies that

dy du = dt dt.

We thus have

Z

? 1

? dy

Z1

? dt = du = ln |u| + A = ln |y| + A

(4)

y dt

u

(where A is an arbitrary constant).

Evaluation of the integral on the right hand side of equation (3) is straight-

forward. We obtain

Z

k dt = kt + B

(5)

(where B is an arbitrary constant). Substituting the results of equations (4) and (5) into equation (3), we

obtain ln |y| + A = kt + B

or ln |y| = kt + (B - A) .

Since A and B are both arbitrary constants, we can simply set C = B - A

and write

ln |y| = kt + C

(6)

(where C is an arbitrary constant). The only thing that remains is to solve the equation (6) for y. To to this,

we exponentiate both sides of equation (6) to obtain

eln|y| = e(kt+C)

which gives us

|y| = e(kt+C).

Using a property of exponents, we can write the above equation as

|y| = eC ? ekt

or, by setting D = eC, we can simply write

|y| = Dekt.

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Since C is an arbitrary constant, D = eC is an arbitrary positive constant. (Recall that eC is always positive, no matter what the sign of C.)

Finally, recalling that |p| = q implies that p = ?q, we obtain

y = ?Dekt.

(7)

Since D is an arbitrary positive constant, we can set E = ?D and rewrite

equation (7) as

y = Eekt

(8)

where E is an arbitrary positive or negative constant. To check that each

member of the family of functions (8) is a solution of the differential equation (1), we note that for y = Eekt, we have

dy

=

d

? E

ekt?

=

Ekekt

=

kEekt

=

ky.

dt dt

Finally, note that if E = 0, then equation (8) simply reduces to y = 0, which we know is also a solution of the differential equation (1). Our conclusion is that the family of solutions of the differential equation (1) is

y = Cekt

where C is an arbitrary constant.

Example 1 Find the family of solutions of the differential equation

dy

dt = 3y.

(9)

and draw a picture of some members of this family.

Solution: By the work done above, the family of solutions of the differ-

ential equation (9) is

y = Ce3t

(where C is an arbitrary constant). Some members of this family are pictured in Figure 1.

Example 2 Find the family of solutions of the differential equation

dy = -2y.

(10)

dt

3

4 3 y2 1

-2

-1

0

-1

-2

-3

-4

1t

2

Figure 1: Family of Solutions of dy/dt = 3y

and draw a picture of some members of this family.

Solution: By the work done above, the family of solutions of the differ-

ential equation (10) is

y = Ce-2t

(where C is an arbitrary constant). Some members of this family are pictured in Figure 2.

4 3 y2 1

-2

-1

0

-1

-2

-3

-4

1t

2

Figure 2: Family of Solutions of dy/dt = -2y

In-class Exercise 1: Find the family of solutions of the differential 4

equation

dy = 0.5y dt

and sketch a picture of this family.

In-class Exercise 2: Find the family of solutions of the differential

equation

dy dt = -5y

and sketch a picture of this family.

Remark 3 If k is a positive constant, then limt ekt = and limt- ekt = 0. Thus, if k > 0, then the family of solutions of dy/dt = ky will be similar (qualitatively) to the family shown in Figure 1 of Example 1; whereas, if k is a negative constant, then limt ekt = 0 and limt- ekt = . Thus, if k < 0, then the family of solutions of dy/dt = ky will be similar (qualitatively) to

the family shown in Figure 2 of Example 2. What about the case k = 0? This question is addressed in the exercises.

1.1 Solution of Initial Value Problem dy/dt = ky, y (t0) = y0

Since we know that the general solution of

dy = ky dt is y = Cekt,

we can easily solve initial value problems for dy/dt = ky.

Example 4 Find the solution of the initial value problem

dy dt = 3y y (0) = 0.5.

Solution: Every solution of the differential equation dy/dt = 3y has the form y = Ce3t. In order that the initial condition y (0) = 0.5 also be satisfied, we must choose the constant C such that Ce3?0 = 0.5; i.e., we must choose C = 0.5. Thus, the solution of the above initial value problem is y = 0.5e3t. This solution is pictured in Figure 3.

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