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Part III: The Power Stuff: Advanced Techniques
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After assuming the correct form of your solution, you re ready to do the math In the following sections, I explain how to calculate all your missing coefficients Remember that A is the following: J 1 1N O A=K K 4 1O L P
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Finding a
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When you start calculating the missing coefficients, why not start with finding a In this case, Aa = a becomes: Ja1N J 1 1N J a 1 N K OK O = - K O K 4 1OK a 2 O K a2O L PL P L P Or, in simpler terms: a1 + a2 = a1 and: 4a1 + a2 = a2 These can be rewritten even more simply as: 2a1 = a2 and: 4a1 = 2a2 These equations differ by a factor of 2 Because you re only interested in a single particular solution, go with a1 = 1 and a2 = 2, which gives you: J-1N a=K O K2O L P
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Finding b
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Now find b You already know the following: J- 2 N Ab = a - b - K O K 0O L P
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12: Tackling Systems of First Order Linear Differential Equations
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So after plugging in your solution for a, you get: J b 1 + b 2 N J-1N J b 1 N J- 2 N K O K O K O K O K 4b 1 + b 2 O = K 2 O - K b 2 O - K 0 O L P L P L P L P Or, more simply: 2b1 + b2 = 1 and 4b1 + 2b2 = 2 A solution to these equations (which only differ by a factor of 2) is: J 1N b=K O K -1O L P
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Finding c
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Ready to find c You already know the following: J0 N Ac = - K O K 3O L P So in other words, you get: J 1 1N J c 1 N J0 N K O K O = -K O K 4 1O K c 2 O K 3O L PL P L P Here s what you get after simplifying: c1 + c2 = 0 and: 4c1 + c2 = 3 The solution to these equations, after a little number crunching, is: J-1N c=K O K 1O L P
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Part III: The Power Stuff: Advanced Techniques Finding d
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Finally you just have to find d You know that Ad = c In other words, by plugging in your solution for c you get: d1 + d2 = 1 and: 4d1 + d2 = 1 So the solution to this pair of equations is: J 23 N d=K 5 O K - 3O L P
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After going through the previous sections and assuming a form of the solution and crunching the numbers, you just have to put the work together to get a particular solution to your original system That particular solution is: J x 1 N J-1N J 1N J-1N J 2 3 N x = K O = K O te - t + K O e - t + K O t + K 5 O K 1O K - 3 O K -1O K x 2 O K 2O L P L L P L P L P P
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13
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Discovering Three Fail-Proof Numerical Methods
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In This
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Brushing up on the Euler method Taking it to the next level with the improved Euler method Getting great results with the Runge-Kutta method
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group of elite computer scientists shuffles into your office with a problem: We re not comfortable writing software that solves differential equations Isn t there an easier, more computer-friendly way of handling differential equations Well, you say, you can use the method of undetermined coefficients and then solve for Solve for the group asks You mean using variables No, no we want something numeric Ah, you tell them You want methods like Euler s method and the RungeKutta method You show them some programming code That s it! the group cries, grabbing your code and running off My bill, you call after them, will be in the mail Better use e-mail, cries a junior member, disappearing around the corner Differential equations can stump even the best and brightest, but I have a secret weapon for you: numerical methods This chapter is all about the computer-based methods that you can use to solve differential equations when everything else fails Do remember that you ll get numbers out of these techniques, not elegant, finished formulas But sometimes, numbers are just what you want, as is often the case in engineering
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