# # a dt in Java Generating QR in Java # a dt

# a dt
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In this equation, exp(x) means ex As an example, try solving the following differential equation with the shortcut: dy 1 + y=4+t dt 2 Assume that the initial condition is y = 8, when t = 0 This equation is an example of the general equation solved in the previous section In this case, g(t) = 4 + t, and a = 1 2 Using a, you find that the integrating factor is et/2, so multiply both sides of equation by that factor: e t/2 dy e t/2 + y = 4e t/2 + te t/2 dt 2
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Now you can combine the two terms on the left to give you this equation: d _ e t/2 y i = 4e t/2 + te t/2 dt All you have to do now is integrate this result The term on the left and the first term on the right are no problem The last term on the right is another story You can use integration by parts to integrate this term Integration by parts works like this:
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# f ^ x h g l ^ x h dx = f ^ b h g ^ b h
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# f l^ x h g ^ x h dx
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2: Looking at Linear First Order Differential Equations
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Applying integration by parts to the last term on the right, and integrating the others, gives you: et/2 y = 8et/2 + 2 t et/2 4et/2 + c where c is an arbitrary constant, set by the initial conditions Dividing by eat gives you this equation: y = 4 + 2t + ce at By applying the initial condition, y(0) = 8, you get y(0) = 8 8=4+c Or c = 4 So the general solution of the differential equation is: y = 4 + 2t + 4e t/2 In 1, I explain that direction fields are great tools for visualizing differential equations You can see a direction field for the previously noted general solution in Figure 2-1
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Figure 2-1: The direction field of the general solution
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Part I: Focusing on First Order Differential Equations
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Connecting the slanting lines in a direction field gives you a graph of the solution You can see a graph of this solution in Figure 2-2
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Figure 2-2: The graph of the general solution
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I think you re ready for another, somewhat more advanced, example Try solving this differential equation to show that you can have different integrating factors: t dy + 2y = 4t 2 dt
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where y(1) = 4 To solve, first you have to find an integrating factor for the equation To get it into the form: dy + ay = g ^ t h dt you have to divide both sides by t, which gives you this equation: dy 2 + t y = 4t dt
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2: Looking at Linear First Order Differential Equations
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To find the integrating factor, use the shortcut equation from the previous section, like this: ^ t h = exp
# a dt = exp # 2 dt t # 2 dt = e t
2 ln t
Performing the integral gives you this equation: ^ t h = exp = t2
So the integrating factor here is t2, which is a new one Multiplying both sides of the equation by the integrating factor, (t) = t2, gives you: t2 dy + 2ty = 4t 3 dt
Because the left side is a readily apparent derivative, you can also write it in this form: d _ yt 2 i = 4t 3 dt Now simply integrate both sides to get: yt2 = t4 + c Finally you get: y = t 2 + c2 t where c is an arbitrary constant of integration Now you can plug in the initial condition y(1) = 4, which allows you to see that c = 3 And that helps you come to this solution: y = t 2 + 32 t And there you have it You can see a direction field for the many general solutions to this differential equation in Figure 2-3