Mathematics II. Tutorial 2 First order differential equations. Groups: B03 & B08

Transcription

1 Tutorial 2 First order differential equations Groups: B03 & B08 February 1, 2012 Department of Mathematics National University of Singapore 1/15

2 : First order linear differential equations In this question, all DEs are of the form either or y + p(x)y = q(x) y + p(x)y = q(x)y n, n R. While the cases (a) and (b) are easy to see, the case (c) is just y + y = xy 1. The case (d) is a little bit unclear since the term (x 1)y 2 contains y up to the second order. Fortunately, since 2xyy = x(y 2 ), we may write the given ODE as xz + (x 1)z = x 2 e x, x > 0, where z = y 2. Notice that the condition y(1) = e 1 in the case (b) plays no role but simply that the generic constant c can be calculated explicitly. 2/15

3 : Catenary In this question, the unknown is the shape of the cable given by the function y(x) with y(0) = 0. The aim of the question is to consider the following equivalent forms of ODEs by integrating or differentiating dy = f(x, y), y(x) = dx x 0 f(t, y(t)) + c. Besides, the quantities µ and T play no role since they are constants. Since the given equation representing y is written in the integral form, we have to differentiate it w.r.t. x in order to obtain the original form. Using the FTC, there holds d dx ( ) dy = µ dx T d dx ( x (dy ) 2 + 1). 0 3/15

4 : Catenary Since y is a function of x, the integral on the right of the given equation should be understood as dy dx = µ T x 0 ( ) d 2 y(t) + 1, }{{} Just a function of t where the integrand is a function of the dummy variable t. Thus, ( ) ( d dy = µ ) dy dx dx T dx If we now denote v(x) = dy dx, we then see that v is also a function of x. In addition, v(0) = 0 and v verifies dv dx = µ v(x) 2 + 1, T that is v = µ v T /15

5 : Catenary Since v = µ T v is a separable ODE, we find that dv v = µ T dx. By integrating, it is easy to find that sinh 1 (v) = µ T x + c. Using the initial condition, v(0) = 0, there holds c = 0. In other words, v = sinh ( µ T x). This helps us to conclude that dy ( µ ) dx = sinh T x. By integrating, we obtain y = T ( µ ) µ cosh T x + c where c = T since y(0) = 0. µ 5/15

6 : Performance curve with C At first, let us try to solve the ODE dp = C(M P ) with P (0) = 0 assuming C and M are constants. Indeed, dp by writing the ODE as M P = C and by integrating we arrive at which yields ln M P = Ct + c, P = M e Ct+c. Using P (0) = 0 and the fact that M > 0 we conclude that P (t) = M Me Ct. Since P (t) is a real number, dimensionless, so is Ct. Thus, the constant C has units of 1/time. From the picture above with M = 2, C = 1/3, and C = 1/3, there holds C > 0. 6/15

7 : Performance curve with C In order to understand the constant C, we observe from the following picture In conclusion, the positive constant C measures how rapidly the student is able to learn although her maximum possible performance is never achieved since M Me Ct < M, t 0. 7/15

8 : Performance curve with C(t) Let us now consider the following dp = C(t)(M P ), C(t) = K tanh ( ) t, t 0. T At first, we wish to solve the above ODE. Rewrite to get ( ) dp t M P = K tanh. T By integrating, we get that ln M P = KT ( ) 2 ln t tanh2 1 T + c ( ( )) t = KT ln sech + c. T Thus, ( )) t KT P = M e (sech c. T Using P (0) = 0, sech(0) = 1, and M > 0, we find that c = ln M, thus, P (t) = M Msech KT ( ) t T. 8/15

9 : Performance curve with C(t) Due to the presence of t T, the constant T has units of time. 1 As such, KT is dimensionless, that is, K has units of time. In order to understand the constants K and T, we observe from the following picture In conclusion, K represents the speed of learning and T measures the amount of time required for her to realise the maximum potential. 9/15

10 Again, we aim to solve the following ODE dr = KR(1300 R) where K a positive constant. We simply rewrite the above ODE as dr 1300KR = KR2. This is a Bernoulli equation, thus, by using Z = R 1 2, the ODE transforms into dz It is then easy to find that KZ = K. 1 R(t) = Z(t) = Ce 1300Kt. Since the rumour was started by one student, so R(0) = 1. 10/15

11 Using the condition R(0) = 1 we know that C = Hence 1 R(t) = e 1300Kt. Clearly, R(t) < 1300 for all t 0. From the picture above, the constant K measures the rapidity with which the rumour will spread, the bigger K is, the faster the rumour will be. 11/15

12 Let us now discuss the hint given in the question: spread slowly both... hardly anyone... everyone has heard. To see this, as a function of variable t, dr, evaluated at a point, is nothing but the slope of the tangent line at that point. keep in mind that dr 0 since R is a non-decreasing function. Thus, the slowly the rumour spreads, the smaller is. dr Based on the above discussion, the rumour will spread slowly at ponts those solve KR(1300 R) = 0, that is, at either R = 0 or R = Similarly, in Q3, by solving C(t)(M P ) = 0, her performance will increase slowly when P approaches M. 12/15

13 : Radioactive Decay A radioactive substance such as Uranium-234 decays in such a way that if U(t) is the amount present in a sample at time t, then U(t) satisfies a relation of the form decay rate r(t) is proportional to amount present at time t, or du = k UU, k U > 0, t 0. The number k U is called the decay rate constant. Since the ODE is both separable and linear, we find U(t) = U 0 e k U t for the general solution where U 0 is the amount present at time t = 0. Let t 1 < t 2 be two arbitrary times and consider the equation U(t 2 ) = U(t 1) 2, or U 0 e k U t 2 = U 0 2 e k U t 1. 13/15

14 : Radioactive Decay Canceling U 0 and rearranging, we find e k U (t 2 t 1 ) = 1 2 or t 2 t 1 = ln 2 k U. Thus the time interval required after t 1 for decay by 50% is the same regardless of what t 1 is. Hence, we can speak of this time interval of length t 2 t 1 as the half-life of the substance Uranium. Now let T (t) be the amount of Thorium-230 in the same sample. Then dt = k T T, k T > 0, t 0. Since the amount of thorium present, and hence its rate of change is enhanced by the rate k U U > 0 at which uranium decays to thorium, we have dt = k T T + k U U, t 0. 14/15

15 : Radioactive Decay Using the formula for U, we find dt + k T T = k U U 0 e k U t. By solving with T (0) = 0, we obtain T (t) = which helps us to conclude T (t) U(t) = k ( ) U U 0 e k U t e k T t, k T k U k ( ) U 1 e (k U k T )t. k T k U Assuming T (t) U(t) = 10% for some t representing the present and thanks to k U = ln , k T = ln , by solving for t, we find that t = (years). 15/15