Chapter 2 First-Order Differential Equations i
Let M(x, y) + N(x, y) = 0 Some equations can be written in the form A(x) + B(y) = 0 DEFINITION 2.2. (Separable Equation) A first-order differential equation of the form = f(x, y) = g(x). h(y) is said to be separable or to have separable variables. For example, the equations = y2 xe 3x+4y and = y + sin x are separable and nonseparable, respectively. In the first equation we can factor g(x) h(y) f(x, y) = y 2 xe 3x+4y = (xe 3x )(y 2 e 4y ), but in the second equation there is no way of expressing y + sin x as a product of a function of x times a function of y. Note: There is no need to use two constants in the integration of a separable equation, because if we write H(y) + c = G(x) + c 2, then the difference c 2 c can be replaced by a single constant c. In many instances throughout the chapters that follow, we will relabel constants in a manner convenient to a given equation. For example, multiples of constants or combinations of constants can sometimes be replaced by a single constant. EXAMPLE (Solving a Separable DE) ( + x) y = 0. Dividing by ( + x)y, we can write y = /( + x), from which it follows that y = + x ln y = ln + x + c y = e ln +x +c = e ln +x. e c laws of exponents = + x e c + x = + x, x = ±e c { ( + x). + x = ( + x), x < Relabeling ±e c as c then gives y = c( + x). 6
ALTERNATIVE (2.2) Separable Variables Because each integral results in a logarithm, a judicious choice for the constant of integration is ln c rather than c. Rewriting the second line of the solution as ln y = ln + x + ln c enables us to combine the terms on the right-hand side by the properties of logarithms. From ln y = ln c( + x) we immediately get y = c( + x). Even if the indefinite integrals are not all logarithms, it may still be advantageous to use ln c. However, no firm rule can be given. LOSING A EXAMPLE 2 (Losing a Solution) We put the equation in the form y 2 4 = or [ 4 y 2 = y2 4. 4 ] =. () y + 2 The second equation in () is the result of using partial fractions on the left-hand side of the first equation. Integrating and using the laws of logarithms gives 4 ln y 2 4 ln y + 2 = x + c ln y 2 y + 2 = 4x + c 2 y 2 y + 2 = ±e4x+c 2 Here we have replaced 4c by c 2. Finally, after replacing ±e c 2 by c and solving the last equation for y, we get the one-parameter family of solutions y = 2 + ce4x. (2) ce4x Now if we factor the right-hand side of the differential equation as = (y 2)(y + 2), we know from the discussion of critical points that y = 2 and y = 2 are two constant (equilibrium) solutions. The solution y = 2 is a member of the family of solutions defined by (2) corresponding to the value c = 0. However, y = 2 is a singular solution; it cannot be obtained from (2) for any choice of the parameter c. EXAMPLE 3 (An Initial-Value Problem) (e 2y y) cos x = ey sin 2x, y(0) = 0. Dividing the equation by e y cos x gives e 2y y e y = 7 sin 2x cos x. Before integrating, we use termwise division on the left-hand side and the trigonometric identity
on the right-hand side. Then yields (2.2) Separable Variables integration by parts sin 2x = 2 sin x cos x (e y ye y ) = 2 sin x e y + ye y + e y = 2 cos x + c. The initial condition y = 0 when x = 0 implies c = 4. Thus a solution of the initial-value problem is e y + ye y + e y = 4 2 cos x. Extra Examples: EXAMPLE 4 sin x sin y + cos x cos y = 0. Dividing the equation by sin y cos x gives sin x sin y cos x cos y + sin y cos x sin y cos x = 0 By integrating, yields sin x cos y + cos x sin y = 0 tan x + cot y = 0 tan x + cot y = c ln sec x + ln sin y = c ln sec x. sin y = c ln. sin y = c cos x sin y ln cos x = c Take the exponential for both sides, we have sin y cos x = c, where c = e c Hence, sin y = c cos x Another way: We have tan x + cot y = ln c ln sec x + ln sin y = ln c ln sec x. sin y = ln c 8
ln. sin y = ln c cos x sin y ln = ln c cos x Take the exponential for both sides, we have sin y cos x = c Hence, sin y = c cos x EXAMPLE 5 Dividing the equation by (x ) gives x 2 x 2 + y(x ) = 0. x + y(x ) x = 0 By integrating, yields Hence, x 2 + y = 0 x x2 + y = c x (x + + ) + y = c x x 2 2 + x + ln x + y2 2 = c x 2 + 2x + 2 ln x + y 2 = 2c x 2 + 2x + ln(x ) 2 + y 2 = c, where c = 2c EXAMPLE 6 Dividing the equation by (xy) gives 2y = 3x when x = 2, y = 2y 3x = xy xy By integrating, 2 x = 3 y 2 x = 3 + ln c y 9
yields 2 ln x = 3 ln y + ln c ln x 2 = ln y 3 + ln c Hence, ln x 2 = ln cy 3 x 2 = cy 3 It s called a general solution Use x = 2, y = or y(2) = Substitute into x 2 = cy 3, we obtain (2) 2 = c() 3 c = 4 Therefore, x 2 = 4y 3 It s called a particular solution 20
Exercises 2.2: Page 50 (2.2) Separable Variables In the following problems, solve the given differential equation by separation of variables. (4) (y ) 2 = 0 Solution: (y ) 2 = 0 = (y ) 2 Dividing the equation by ( y) 2 gives (y ) 2 = = ( y) 2 Integrate both sides, = ( y) 2 ( y) 2 = ( y) 2 = ( y) = x + c = x + c y y = x + c y = x + c =. x + c x + c --------------------------------------------------------------------------------- xy + 3x y 3 (9) = xy 2x + 4y 8 Solution: xy + 3x y 3 = xy 2x + 4y 8 Multiply both sides into (xy 2x + 4y 8), we obtain (xy 2x + 4y 8) = (xy + 3x y 3) Multiply the equation by [y(x + 4) 2(x + 4)] = [y(x ) + 3(x )] (x + 4)(y 2) = (x )(y + 3) (x+4)(y+3), gives y 2 y + 3 Before integrating, we use the long division to have = x x + 4 ( 5 5 ) = ( y + 3 x + 4 ) 2
Integrate both sides, ( 5 5 ) = ( y + 3 x + 4 ) y 5 ln y + 3 = x 5 ln x + 4 + c. --------------------------------------------------------------------------------- (2) = x y2 Solution: = x y2 Multiply both sides into, we have y 2 Rewrite the equation as = x y2 = x y2 Integrate both sides, or y 2 = x sin y = 2 x2 + c y = sin ( 2 x2 + c). 22
Exercises 2.2: Page 50 (Homework) In the following problems, solve the given differential equation by separation of variables. (8) e x y = e y + e 2x y (2) sin 3x + 2y cos 3 3x = 0 (4) x( + y 2 ) /2 = y( + x 2 ) /2 (20) xy + 2y x 2 = xy 3y + x 3 (22) (e x + e x ) = y2 ------------------------------------------------------------------------------------------------------------------- In the following problems, find an explicit solution of the given initial-value problem. (25) x 2 = y xy, y( ) = (28) ( + x 4 ) + x( + 4y 2 ) = 0, y() = 0 Exercises 2.2: Page 50 (Assignment) In the following problem, solve the given differential equation by separation of variables. (22) (e x + e x ) = y2 23