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Unformatted text preview: MATH 152 L1 & L2 Applied Linear Algebra and Differential Equations Spring 200405 ¥ Week 02 Worksheet Solutions : Firstorder differential equations Q1. (Change of variables) Consider the firstorder differential equation 1 + x y sin y ¶ dy dx = 0 , which is neither linear nor separable. The purpose of this problem is to show you a trick. (a) Instead of thinking of the variable y as a function of x , suppose we think of the variable x as a function of y . The derivative of x with respect to y is dx dy , the reciprocal of dy dx . Rewrite the given equation so that it is an equation for dx dy . Solution x y sin y ¶ = dx dy , dx dy + x y = sin y. (1) (b) Is your rewritten equation linear, separable or both? Solution The equation (1) is a firstorder linear ordinary differential equation for x ( y ). However, it is a nonseparable equation. (c) Solve your differential equation for x as a function of y . This gives you an implicit function for y in terms of x . Solution Consider the equation (1) for x ( y ). The integrating factor is exp •Z 1 y dy ‚ = e ln y = y. Then consider d dy ( yx ) = y dx dy + x y ¶ = y sin y. Integrating with respect to y gives yx = Z y sin y dy = Z y d (cos y ) = y cos y + Z cos y dy = y cos y + sin y + C. The implicit solution for y in terms of x is given by xy + y cos y sin y = C . Q2. (Homogeneous equation) In the previous problem, we showed how one sometimes converts nonlinear differential equations into linear differential equations. This problem shows you another trick to convert a certain type of differential equation into a separable equation. Let us consider the equationa certain type of differential equation into a separable equation....
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This note was uploaded on 09/30/2010 for the course MATH MATH152 taught by Professor Kcc during the Spring '10 term at HKUST.
 Spring '10
 KCC
 Math, Linear Algebra, Algebra, Equations

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