1.06.DiffEq - Clear@y, x, starterx, starteryD; starterx =...

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Growth Authors: Bill Davis, Horacio Porta and Jerry Uhl ©1996-2007 Publisher: Math Everywhere, Inc. Version 6.0 1.06 The Differential Equations of Calculus BASICS B.1) The most important of all differential equations: y £ @ x D = r y @ x D and why you already know how to solve it · B.1.a) Given a number r , to solve the differential equation y' @ x D = r y @ x D , you want a function y @ x D with the property that y' @ x D = r y @ x D for all x 's. This means 100 y £ @ x D y @ x D = 100 r no matter what x is. How does this give away a formula for y @ x D ? · Answer: The equation is y' @ x D = r y @ x D for all x 's. This means 100 y £ @ x D y @ x D = 100 r no matter what x is. This tells you that y @ x D grows with a steady instantaneous percentage growth rate of 100 r percent . And this tells you that y @ x D is the exponential function y @ x D = k e r x for some number k . Try it out: Clear @ x, y, k, r D ; y @ x _ D = k E r x ; y £ @ x D == r y @ x D True No problem-o. You can get Mathematica to spit this out directly: Clear @ y, x D ; DSolve @ y' @ x D == r y @ x D , y @ x D , x D 88 y @ x D Ø ‰ r x C @ 1 D<< This means C @ 1 D can be any constant. You can reconcile the solution y @ x D = k e r x above with the Mathematica solution y @ x D = C @ 1 D e r x by taking k = C @ 1 D . · B.1.b) Now that you know that y @ x D = k e 3 x solves the differential equation y' @ x D = 3 y @ x D , what data do you need to nail down what the value of k is? · Answer: You know y @ x D = k e 3 x ; so you have one unknown (namely k ) to determine. You need one data point on y @ x D to do this. For instance, if you know y @ 1.71 D is equal to 18.06 , then you determine k by solving for k as follows: Clear @ x, k D ; Solve A k I E 3 x ê . x Ø 1.71 M == 18.06, k E 88 k Ø 0.106853 << This gives you the specific solution y @ x D = 0.106853 e 3 x . You can also get an equivalent solution directly from Mathematica : Clear @ y, x, starterx, startery D ; starterx = 1.71; startery = 18.06; sol = DSolve @8 y' @ x D == 3 y @ x D , y @ starterx D == startery < , y @ x D , x D ; y @ x D ê . sol @@ 1 DD 0.106853 3 x Good. B.2) The logistic differential equation y £ @ x D = r y @ x D I 1 - y @ x D b M and how you get a formula for its solution · B.2.a) Given numbers r and b , you want a function y @ x D with the property that y £ @ x D = r y @ x D J 1 - y @ x D b N . How do you find a formula for y @ x D ? · Answer: This one is harder than the others. Let's see what mighty Mathematica thinks about this one: Clear @ y, x, r, b D ; DSolve B y £ @ x D == r y @ x D 1 - y @ x D b , y @ x D , x F :: y @ x D Ø b r x + b C @ 1 D - 1 + ‰ r x + b C @ 1 D >> Here C @ 1 D can be any constant. Call it k and check Mathematica 's solution: Clear @ y, x, k, r, b D ; y @ x _ D = b E r x E r x - E k ; Expand @ y £ @ x DD == Expand B r y @ x D 1 - y @ x D b F True Good. Later in the course, you'll learn how to pull this formula out of a hat with your own hands.
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1.06.DiffEq - Clear@y, x, starterx, starteryD; starterx =...

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