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Diff Equ Cheat Sheet-1

# Diff Equ Cheat Sheet-1 - Final Review Separable...

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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 = Nx. 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 Reduction of Order plug in the given y1(x) into y = v(t)y1(x) and take its derivatives until you can plug it into the differential equation and simplify. You then get all the v’s on one side and the y’s on the other and integrate to get v which is y2(x) 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) Mixing Problems dA/dt = (Amount In) – (Amount Out) Mass on a Spring problems are solved using mu’’ + ¥u’ + ku = 0 where k = mg/L and ¥ is the damper constant. 12/08/10 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 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. Y’’(s) = s2Y(s) – sy(0) – y’(0) Y’(s) = sY(s) – y(0) 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 Separation of Variables u(x,t) = X(x)T(t) so uxx = X’’(x)T(t) and ut = X(x)T’(t) plug this into the differential equation and solve for lambda = –T’/T. Newton’s Law of Cooling where u(t) is the temperature at time, t. 12/08/10 Fourier Series is calculated using 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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