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Unformatted text preview: Final Exam Practice Problems Math 4027 The final exam will be comprehensive. You should review both the statements of all existence and uniqueness theorems and the various techniques employed for finding solution of differential equations. As with the second exam, you will be allowed to refer to your textbook. 1. Consider the differential equation M ( x, y ) + N ( x, y ) y prime = 0 , (1) where M , N , M y = ∂M/∂y , and N x = ∂N/∂x are continuous functions on the entire plane. (a) What does it mean for equation (1) to be exact ? What does this tell you about solutions to the differential equation? trianglerightsld Solution. Equation (1) is exact provided that there is a differentiable function F ( x, y ) on the plane R 2 such that M = ∂F/∂x and N = ∂F/∂y . If (1) is exact, then the solutions are defined implicitly by the level curves F ( x, y ) = C of the function F ( x, y ) guaranteed by the exactness. triangleleftsld (b) What is the condition on M and N that is equivalent to exactness of equation (1)? trianglerightsld Solution. Equation (1) is exact if and only if M y = N x . triangleleftsld (c) If equation (1) is not exact show that there is an integrating factor μ ( x ) consisting of a function of x alone if and only if the function P = M y N x N depends only on x . If this is the case, then show that P satisfies the ordinary differential equation dμ dx = Pμ. trianglerightsld Solution. Suppose that μ ( x ) (depending only on x ) is a function such that the equation μ ( x ) M ( x, y ) + μ ( x ) N ( x, y ) = 0 ( * ) is exact. By the criterion for exactness in part (b), equation ( * ) is exact if and only if ∂ ( μM ) ∂y = ∂ ( μN ) ∂x ⇐⇒ μ ∂M ∂y = dμ dx N + μ ∂N ∂x ⇐⇒ μ parenleftbigg ∂M ∂y ∂N ∂x parenrightbigg = dμ dx N. 1 Final Exam Practice Problems Math 4027 This last equality is equivalent to P = M y N x N = dμ dx /μ. Since the function on the right depends only on x , the same must be true for P . Thus, if ( * ) is exact, then P depends only on x , and conversely if P depends only on x , we can find an integrating factor μ ( x ) depending only on x since we can solve the first order linear differential equation dμ dx = Pμ for μ . In particular, we find that if P is a function only of x , then μ ( x ) = e P ( x ) , dx . triangleleftsld (d) Apply the technique of part (c) to show that any solution to the differential equation (3 xy + y 2 ) + ( x 2 + xy ) y prime = 0 lies on a curve of the form x 3 y + x 2 y 2 / 2 = C, where C is a constant. trianglerightsld Solution. Compute P = M y N x N = (3 x + 2 y ) (2 x + y ) x 2 + xy = 1 x . This provides an integrating factor of μ ( x ) = x so that (3 x 2 y + xy 2 ) + ( x 3 + x 2 y ) y prime = 0 is exact. Then, we get F ( x, y ) = integraldisplay (3 x 2 y + xy 2 ) dx = x 3 y + 1 2 x 2 y 2 + h ( y ) and the function h ( y ) is determined by ( x 3 + x 2 y ) = ∂F ∂y = x 3 + x 2 y + dh dy ....
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 Spring '06
 Adkins
 Differential Equations, Equations, ORDINARY DIFFERENTIAL EQUATIONS, Frobenius method, Regular singular point, Frobenius, H F J E G, √ tan−1

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