This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 2
is a general solution to the given diﬀerential equation. Unless C = 0 and so y = x, the function y = x + C/x2 is not deﬁned when x = 0. Therefore, among all solutions, the only function deﬁned at x = 0 is φ(x) = x, and the initial value problem with y (0) = y0 has a solution (and unique) if and only if y0 = φ(x)
x=0 = 0. (b) Standard form of the equation xy − 2y = 3x is 2 dy − y = 3. dx x This gives P (x) = −2/x, µ(x) = exp d x−2 y = 3x−2 dx ⇒ x−2 y = (−2/x)dx = x−2 , and 3x−2 dx = −3x−1 + C ⇒ y = −3x + Cx2 . Therefore, any solution is a polynomial and so is deﬁned for all real numbers. Moreover, any solution satisﬁes the initial condition y (0) = 0 because −3x + Cx2
x=0 = −3(0) + C (0)2 = 0 and, therefore, is a solution to the initial value problem. (This also implies that the initial value problem with y (0) = y0 = 0 has no solution.) 35. (a) This part of the problem is similar to Problem 33 in Section 2.2. So, we proceed in the same way. Let A(t) denote the mass of salt in the tank at t minutes after the process begins. Then we have rate of input = 5 L/min × 0.2 kg/L = 1 kg/min , A(t) A(t) rate of exit = 5 L/min × kg/L = kg/min , 500 100 dA A 100 − A =1− = . dt 100 100 Separating this diﬀerential equation yields dA/(100 − A) = dt/100. Integrating, we obtain − ln |100 − A| = 54 t + C1 100 ⇒ |100 − A| = e−t/100−C1 = e−C1 e−t/100 ...
View Full Document