Ch2-Sec( 2.2): Separable Equations

In this section we examine a subclass of linear and nonlinear first order equations. Consider the first order equation

We can rewrite this in the form

For example, let M(x,y) = - f (x,y) and N (x,y) = 1. There may be other ways as well. In differential form,

If M is a function of x only and N is a function of y only, then

In this case, the equation is called separable.

0),(),( dx

dyyxNyxM

0),(),( dyyxNdxyxM

0)()( dyyNdxxM

),( yxfdx

dy

Example 1: Solving a Separable Equation

Solve the following first order nonlinear equation:

Separating variables, and using calculus, we obtain

The equation above defines the solution y implicitly. A graph showing the direction field and implicit plots of several solution curves for the differential equation is given above.

2

2

1 y

x

dx

dy

Cxyy

Cxyy

dxxdyy

dxxdyy

33

33

22

22

3

3

1

3

1

1

1

4 2 2 4xt

4

2

2

4yt

Example 2: Implicit and Explicit Solutions (1 of 4)

Solve the following first order nonlinear equation:

Separating variables and using calculus, we obtain

The equation above defines the solution y implicitly. An explicit expression for the solution can be found in this case:

12

243 2

y

xx

dx

dy

Cxxxyy

dxxxdyy

dxxxdyy

222

24312

24312

232

2

2

Cxxxy

CxxxyCxxxyy

221

2

224420222

23

23232

Example 2: Initial Value Problem (2 of 4)

Suppose we seek a solution satisfying y(0) = -1. Using the implicit expression of y, we obtain

Thus the implicit equation defining y is

Using an explicit expression of y,

It follows that

411

221 23

CC

Cxxxy

3)1(2)1(

2222

232

CC

Cxxxyy

3222 232 xxxyy

4221 23 xxxy

12

243 2

y

xx

dx

dy

Example 2: Initial Condition y(0) = 3 (3 of 4)

Note that if initial condition is y(0) = 3, then we choose the positive sign, instead of negative sign, on the square root term:

4221 23 xxxy

12

243 2

y

xx

dx

dy

Example 2: Domain (4 of 4)

Thus the solutions to the initial value problem

are given by

From explicit representation of y, it follows that

and hence the domain of y is (-2, ). Note x = -2 yields y = 1, which makes the denominator of dy/dx zero (vertical tangent).

Conversely, the domain of y can be estimated by locating vertical tangents on the graph (useful for implicitly defined solutions).

(explicit) 4221

(implicit) 3222

23

232

xxxy

xxxyy

2212221 22 xxxxxy

1)0(,12

243 2

y

y

xx

dx

dy

Example 3: Implicit Solution of an Initial Value Problem (1 of 2)

Consider the following initial value problem:

Separating variables and using calculus, we obtain

Using the initial condition, y(0)=1, it follows that C = 17.

1)0(,4

43

3

yy

xxy

cCCxxyy

cxxyy

dxxxdyy

dxxxdyy

4 where816

24

)4(4

)4()4(

244

424

33

33

4

1

4

1

17816 244 xxyy

Example 3: Graph of Solutions (2 of 2)

The graph of this particular solution through (0, 2) is shown in red along with the graphs of the direction field and several other solution curves for this differential equation, are shown: The points identified with blue dots correspond to the points onthe red curve where the tangentline is vertical:

but at allpoints where the line connecting the blue points intersects solution curvesthe tangent line is vertical.

17816 is (0,2)hrough solution t theand

816 issolution general theThus244

244

xxyy

Cxxyy

1)0(,4

43

3

yy

xxy

4 2 2 4x

3

2

1

1

2y

-1.5874 43 ycurve, red on the 3488.3 x

Parametric Equations

The differential equation:

is sometimes easier to solve if x and y are thought of as dependent variables of the independent variable t and rewriting the single differential equation as the system of differential equations:

Chapter 9 is devoted to the solution of systems such as these.

),(

),(

yxG

yxF

dx

dy

),( and ),( yxGdt

dxyxF

dt

dy