Diff Equ Test 1 Review

Diff Equ Test 1 Review - Diff Equ Test 1 Review...

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Unformatted text preview: 9/18/10 Diff Equ Test 1 Review First Order Differential Equations are solved by one of the following methods to find the unknown function whose derivative is the given equation. Given Equation: df = 0 General Solution: f = c Separable Equations are solved by getting both variables on a side by themselves so that the equation is in the form: F(y)dy = G(x)dx. In this form both sides can be easily integrated. Do not forget to check for solutions that were lost. Exact Equations can be solved by integrating M(x,y) with respect to x which results in an equation, Ψ, that includes a constant h(y). Setting the partial of Ψ with respect to y equal to N(x,y) allows you to solve for h(y) by using integration. h(y) is then plugged into Ψ and then entire equation is set equal to c to produce a general solution. The equation M(x,y)dx + N(x,y)dy = 0 is exact if My = Mx. Integrating Factors turn non ­exact equations into exact equations as long as the equations have a solution. Case 1: My  ­ Nx N is a function of only x then: μ = e∫$(x)dx Case 2: My d Nx  ­  ­M is a function of only y then: μ = e∫$(y)dy d Distribute the integrating factor and then ∫ μM(x,y)dx = ∫ μN(x,y)dy = c. First Order Homogeneous Equations can be rewritten in a way that makes them separable by substituting y = xv and dy = xdv + vdv for all y and dy if M(x,y) and N(x,y) are both homogenous functions of the same degree. Once rewritten the equation can be solved using the method of separable equations followed by a simple substitution back of the original variables: v = y/x. First Order Linear Equations are of the form y’ + P(x)y = Q(x) and require an integrating factor μ to solve. The integrating factor is always μ = e∫P dx. After distributing the integrating factor the left side quickly simplifies by recognizing a reverse product rule to an equation that is easy to integrate. 9/18/10 Variation of Parameters is accomplished by finding a general solution for the homogeneous equation of a second order linear equation and then varying the parameters c1 and c2 with functions v1(x) and v2(x)………………………………………… Second Order Homogeneous Equations are found by finding two linearly independent equations y1 and y2 Constant Coefficient distributes an auxiliary polynomial, y = emx, into a second order homogeneous equation in order to find the general solution. After solving for m, there are three cases to combine these two solutions into a general solution. The cases are determined by the discriminant b2 – 4ac. Case 2: The discriminant is positive Case 2: The discriminant is zero y = c1em x + c2em x 1 2 y = c1emx + c2xemx y = erx(c1cos sx + c2sin sx) m = r ± si Case 2: The discriminant is negative Reduction of Order Partial Fractions Integration by Parts ...
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This note was uploaded on 02/01/2011 for the course MATH 427K taught by Professor Delallave during the Spring '11 term at University of Texas.

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