By Wei-Chau Xie
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Additional info for Differential equations for engineers
Combining Cases 1 and 2, the solutions of the differential equation are y 2 + 2C x + C 2 = 0, y = ±x. 3 Exact Differential Equations and Integrating Factors Consider differential equations of the form M(x, y)dx + N(x, y)dy = 0, or M(x, y) dy =− , N(x, y) = 0, dx N(x, y) (1) ∂N ∂M and are continuous. Suppose the solution of equation (1) is ∂y ∂x u(x, y) = C, C = constant. Taking the differential yields where du = ∂u ∂u ∂u ∂u dx + dy = dC = 0 =⇒ dx + dy = 0. , the coefﬁcients of dx and dy in equations (1) and (2) are proportional ∂u ∂u ∂x = ∂y = μ(x, y) =⇒ ∂u = μM, M(x, y) ∂x N(x, y) ∂u = μN.
Equation (4) is variable separable, which can be solved easily by integration ln μ = ∂N 1 ∂M − dx =⇒ μ(x) = exp N ∂y ∂x ∂N 1 ∂M − dx . N ∂y ∂x (5) 40 2 ﬁrst-order and simple higher-order differential equations Note that, since only one integrating factor is sought, there is no need to include a constant of integration C. Interchanging M and N, and x and y in equation (5), one obtains an integrating factor for another special case μ( y) = exp ∂M 1 ∂N − dy . M ∂x ∂y function of y only Integrating Factors Consider the differential equation M(x, y)dx + N(x, y)dy = 0.
Dx 2v +1 2v +1 2v +1 Case 1. , (v −3)(v +2) = 0 y 3 = −2x or =⇒ v = −2 or v = 3, which gives y 3 = 3x. Case 2. v 2 −v −6 = 0, separating the variables gives 2v +1 v 2 −v −6 dv = 1 dx. x Variable separable Integrating both sides yields 2v +1 v 2 −v −6 dv = 1 dx + C. 2 method of transformation of variables 29 The ﬁrst integral can be evaluated using partial fractions (see pages 259–261 for a brief review on partial fractions) 2v +1 v 2 −v −6 = 2v +1 A B = + . (v −3)(v +2) v −3 v +2 Using the cover-up method, the coefﬁcients A and B can be easily determined A= 2v +1 v 2 −v −6 2v +1 v +2 dv = v=3 1 5 = 75 , B= 2v +1 v −3 v= − 2 = 35 , 7 3 + dv = 75 ln v −3 + 53 ln v +2 .