Examining Separable Equations for Solution Methods
Explore the concept of separable equations in differential equations, including examples of solving first-order nonlinear equations through variable separation and calculus techniques. Implicit and explicit solutions are discussed, along with handling initial value problems. Visual representations aid in understanding the solutions of these mathematical equations.
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http://bcs.wiley.com/he-bcs/Books?action=resource &bcsId=2026&itemId=047143339X&resourceId=4140 Ch 2.2: Separable Equations In this section we examine a subclass of linear and nonlinear first order equations. Consider the first order equation dy= ( , ) f x y dx We can rewrite this in the form dy + = ( , ) ( , ) 0 M x y N x y dx For example, let M(x,y) = - f(x,y) and N(x,y) = 1. There may be other ways as well. In differential form, ) , ( ) , ( + dy y x N dx y x M = 0 If M is a function of x only and N is a function of y only, then 0 ) ( ) ( = + dy y N dx x M In this case, the equation is called separable.
http://bcs.wiley.com/he-bcs/Books?action=resource &bcsId=2026&itemId=047143339X&resourceId=4140 Example 1: Solving a Separable Equation Solve the following first order nonlinear equation: 1 2 dx + 2 dy x =y 1 Separating variables, and using calculus, we obtain ( ( dx x dy y + = 1 3 3 3 ) 1 ( ) = + 2 2 1 1 y dy x dx ) ( ) 2 2 1 1 = + + 3 3 y y x x C 3 = + + 3 3 y y x x C The equation above defines the solution y implicitly. A graph showing the direction field and implicit plots of several integral curves for the differential equation is given above.
http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=2026&itemId=047143339X&resourceId=4140http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=2026&itemId=047143339X&resourceId=4140 Example 2: Implicit and Explicit Solutions (1 of 4) Solve the following first order nonlinear equation: 2 4 3 y dx + + 2 dy x x = ( ) 1 2 Separating variables and using calculus, we obtain ( ) ( ) x x dy y + + = 2 2 2 ( ) = + + 2 2 1 3 4 2 y dy x x dx ( ) 2 2 1 3 4 2 dx = + + + 2 3 2 y y x x x C The equation above defines the solution y implicitly. An explicit expression for the solution can be found in this case: ( C x x x y y = + + + 0 2 2 2 ( ) + + 2 + + 3 2 ) 2 4 4 2 2 x x x C = 2 3 2 y = + + + 3 2 1 2 2 y x x x C
http://bcs.wiley.com/he-bcs/Books?action=resource &bcsId=2026&itemId=047143339X&resourceId=4140 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 2 2 2 = = C C = + + + 2 3 2 y y x x x C 2 ( ) 1 ( 2 ) 1 3 Thus the implicit equation defining y is = + + + 2 3 2 2 2 2 3 y y x x x Using explicit expression of y, = + + + 3 2 1 2 2 = y x x x C = 1 1 4 C C It follows that = + + + 3 2 1 2 2 4 y x x x
http://bcs.wiley.com/he-bcs/Books?action=resource &bcsId=2026&itemId=047143339X&resourceId=4140 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 square root term: = + + + + 3 2 1 2 2 4 y x x x
http://bcs.wiley.com/he-bcs/Books?action=resource &bcsId=2026&itemId=047143339X&resourceId=4140 Example 2: Domain (4 of 4) Thus the solutions to the initial value problem 2 4 3 y dx + + 2 dy x x = = , ) 0 ( y 1 ( ) 2 1 are given by 2 y = + + + 3 2 2 2 2 (implicit) 3 y x x x = + + + 3 2 1 2 2 4 (explicit) y x x x From explicit representation of y, it follows that ( ) ( 2 2 1 + + + = x x x y )( ) 2 ) ( = + + 2 2 2 1 2 x x and hence domain of y is (-2, ). Note x = -2 yields y = 1, which makes denominator of dy/dx zero (vertical tangent). Conversely, domain of y can be estimated by locating vertical tangents on graph (useful for implicitly defined solutions).
http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=2026&itemId=047143339X&resourceId=4140http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=2026&itemId=047143339X&resourceId=4140 Example 3: Implicit Solution of Initial Value Problem (1 of 2) Consider the following initial value problem: cos 3 = + y y x = , ) 0 ( y 1 y 1 3 Separating variables and using calculus, we obtain y = cos + 3 1 3 dy xdx y 1 + = 2 3 cos y dy xdx y + = + 3 ln sin y y x C Using the initial condition, it follows that + = + 3 ln sin 1 y y x
http://bcs.wiley.com/he-bcs/Books?action=resource &bcsId=2026&itemId=047143339X&resourceId=4140 Example 3: Graph of Solutions (2 of 2) Thus cos + y x = = + = + 3 , ) 0 ( y 1 ln sin 1 y y y x 3 1 3 y The graph of this solution (black), along with the graphs of the direction field and several integral curves (blue) for this differential equation, is given below.
