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# 3410 - Math 3410 Fall 2009 Second semester dierential...

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Math 3410, Fall 2009 Second semester differential equations 1 Some review 1.1 First order equations We will need to know how to do separable equations and linear equations. A separable equation is one like dy dt = ty 2 . We can rewrite this as dy y 2 = t dt. Integrating both sides, - 1 y = 1 2 t 2 + c, so y = - 1 1 2 t 2 + c . A linear equation is one like y + 2 t y = t. One multiplies by an integrating factor p = e R 2 t = e 2 ln t = t 2 to get t 2 y + 2 ty = t 3 , or ( t 2 y ) = t 3 , 1

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which leads to t 2 y = 1 4 t 4 + c, and then y = 1 4 t 2 + c t 2 . When the linear equation has constant coefficients and is homogeneous (i.e., the right hand side is 0), things are much easier. To solve y - 4 y = 0 , we guess a solutions of the form y = e rt , so y = re rt . Then re rt - 4 e rt = 0 , or r = 4, and therefore the solution is y = ce 4 t . To identify c , one needs an initial condition, e.g., y (0) = 2. Then 2 = ce 4 · 0 = c, so we then have y = 2 e 4 t . For non-homogeneous equations, such as y - 4 y = e 3 t , one way to solve it is to solve the homogeneous equation y - 4 y = 0, and then use y = ce 4 t + y p , where y p is a particular solution. One way to find a particular solution is to make an educated guess. If we guess y p = Ae 3 t , then we have y p - 4 y p = 3 Ae 3 t - 4 Ae 3 t , and this will equal e 3 t if A = - 1. We conclude the solution to the non- homogeneous equation is y = ce 4 t - e 3 t . 2
1.2 Series From calculus we have the Taylor series e x = 1 + x + x 2 2! + x 3 3! + · · · , cos x = 1 - x 2 2! + x 4 4! - · · · , and sin x = x - x 3 3! + x 5 5! - · · · . If i = - 1, substituting and doing some algebra shows that e ix = cos x + i sin x. 2 Second order linear 2.1 Applications First consider a spring hung from the ceiling with a weight hanging from it. Let u be the distance the weight is below equilibrium. There is a restoring force upwards of amount ku by Hooke’s law. There is damping resistance against the motion, which is - Ru . And the net force is related to accelera- tion by Newton’s laws, so ku - Ru = F = mu . This leads to mu + Ru - ku = 0 . If there is an external force acting on the spring, then the right hand side is replaced by F ( t ). The second example is that of a circuit with a resistor, inductance coil, and capacitor hooked up in series. Let I be the current, Q the charge, R the resistance, L the inductance, and C the capacitance. We know that I = dQ/dt . The voltage drop across the resistor is IR , across the capacitor 3

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Q/C , and across the inductance coil L dI dt . So if E ( t ) is the potential put into the current, E ( t ) = LQ + RQ + 1 C Q. Sometimes this is differentiated to give E ( t ) = LI + RI + 1 C I. 2.2 Linear, constant coefficients, homogeneous Let’s look at an example: y - 5 y + 4 y = 0 . From Math 211, we know a way of solving this. Let v = y , and this one equation becomes a system y = v ; v = 5 v - 4 y. We then set up matrices, where X = y v , A = - 4 5 0 1 , and the equation is X = AX. We assume W = w 1 w 2 and that our solution is of the form X = We rt for some r , w 1 , and w 2 . We will review this method later when we want to generalize it, but let’s look at an easier method.
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