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Unformatted text preview: 12/08/10 Final Review 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 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. 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 1: 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 3: The discriminant is negative Variation of Parameters is done using the Wronskian. The particular solution is found by the equation yp = u1(x)y1(x) + u2(x)y2(x) where u1’ =
y2(x)g(x) / W u2’ = y1(x)g(x) / W Wronskian shows that two functions are linearly independent if the following is not equal to zero: W = y1(x)y2’(x) – y1’(x)y2(x) Mass on a Spring problems are solved using mu’’ + yu’ + ku = 0 where k = mg/u0 Series Circuits problems are solved using LQ’’ + RQ’ + Q/C = 0 Undetermined Coefficients if g(x) is in the form of: • erx, try a particular solution of the form Aerx, where A is a constant. Because 12/08/10 derivatives of erx reproduce erx, you have a good chance of finding a particular solution this way. • a polynomial of order n, try a polynomial of order n. For example, if g(x) = x2 + 1, try a polynomial of the form Ax2 + B. The particular solution is not allowed to have the same powers of x as the homogeneous solution so multiple by x until the particular does not have the same powers before solving for the coefficients. • a combination of sines and cosines, sin αx + cos βx, try a combination of sines and cosines with undetermined coefficients, A sin αx + B cos βx. Then plug into the differential equation and solve for A and B. Laplace Transform is done by transforming the given equation from a function of x to a function of s, manipulating this equation until Y(s) is by itself, and then transforming the equation back to a function of x using reverse Laplace Transform. Partial Fractions allow you to find the anti
derivative of the form ∫N(x)/D(x) First check that the degree of numerator < degree of the denominator. If it isn’t do division. Second factor the denominator and write the form of the partial fraction. Then solve for the coefficients. x2 + 2x + 1 = a + b + cx + d (x
2)2(x2 + 2x + 5) (x
2)2 (x
2) (x2 + 2x + 5) Integration by Parts simplifies integrals by rewriting them like: ∫u dv = uv
∫v du Newton’s Law of Cooling where u(t) is the temperature at time, t. Fourier Series is calculated using 12/08/10 where π can be substituted for any L where L = T / 2 Heat Conduction in a Rod Temperature according to position and time is where ...
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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 at Austin.
 Spring '11
 DELALLAVE
 Differential Equations, Equations

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