58_pdfsam_math 54 differential equation solutions odd

58_pdfsam_math 54 differential equation solutions odd -...

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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 ...
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