Introductory Differential Equations using Sage - William Stein

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24 CHAPTER 1. FIRST ORDER DIFFERENTIAL EQUATIONS Example 1.4.1. Are the following ODEs separable? If so, solve them. (a) (t 2 + y 2 )y ′ = −2ty, (b) y ′ = −x/y, y(0) = −1, (c) T ′ = k · (T − T room ), where k < 0 and T room are constants, (d) ax ′ + bx = c, where a ≠ 0, b ≠ 0, and c are constants (e) ax ′ + bx = c, where a ≠ 0, b, are constants and c = c(t) is not a constant. (f) y ′ = (y − 1)(y + 1), y(0) = 2. (g) y ′ = y 2 + 1, y(0) = 1. Solutions: (a) not separable, (b) y dy = −xdx, so y 2 /2 = −x 2 /2+c, so x 2 +y 2 = 2c. This is the general solution (note it does not give y explicitly as a function of x, you will have to solve for y algebraically to get that). The initial conditions say when x = 0, y = 1, so 2c = 0 2 + 1 2 = 1, which gives c = 1/2. Therefore, x 2 + y 2 = 1, which is a circle. That is not a function so cannot be the solution we want. The solution is either y = √ 1 − x 2 or y = − √ 1 − x 2 , but which one? Since y(0) = −1 (note the minus sign) it must be y = − √ 1 − x 2 . (c) dT T −T room = kdt, so ln |T − T room | = kt + c (some constant c), so T − T room = Ce kt (some constant C), so T = T(t) = T room + Ce kt . (d) dx dt = (c − bx)/a = − b a (x − c dx b ), so x− c b = − b a dt, so ln |x − c b | = − b at + C, where C is a constant of integration. This is the implicit general solution of the DE. The explicit general solution is x = c b + Be− b a t , where B is a constant. The explicit solution is easy find using Sage: Sage sage: a = var(’a’) sage: b = var(’b’) sage: c = var(’c’) sage: t = var(’t’) sage: x = function(’x’, t) sage: de = lambda y: a*diff(y,t) + b*y - c sage: desolve(de(x),[x,t]) (c*eˆ(b*t/a)/b + c)*eˆ(-b*t/a)
1.4. FIRST ORDER ODES - SEPARABLE AND LINEAR CASES 25 (e) If c = c(t) is not constant then ax ′ + bx = c is not separable. (f) dy (y−1)(y+1) = dt so 1 2 (ln(y −1)−ln(y+1)) = t+C, where C is a constant of integration. This is the “general (implicit) solution” of the DE. Note: the constant functions y(t) = 1 and y(t) = −1 are also solutions to this DE. These solutions cannot be obtained (in an obvious way) from the general solution. The integral is easy to do using Sage: Sage sage: y = var(’y’) sage: integral(1/((y-1)*(y+1)),y) log(y - 1)/2 - (log(y + 1)/2) Now, let’s try to get Sage to solve for y in terms of t in 1 2 (ln(y −1)−ln(y+1)) = t+C: Sage sage: C = var(’C’) sage: solve([log(y - 1)/2 - (log(y + 1)/2) == t+C],y) [log(y + 1) == -2*C + log(y - 1) - 2*t] This is not working. Let’s try inputting the problem in a different form: Sage sage: C = var(’C’) sage: solve([log((y - 1)/(y + 1)) == 2*t+2*C],y) [y == (-eˆ(2*C + 2*t) - 1)/(eˆ(2*C + 2*t) - 1)] This is what we want. Now let’s assume the initial condition y(0) = 2 and solve for C and plot the function. Sage sage: solny=lambda t:(-eˆ(2*C+2*t)-1)/(eˆ(2*C+2*t)-1) sage: solve([solny(0) == 2],C) [C == log(-1/sqrt(3)), C == -log(3)/2] sage: C = -log(3)/2 sage: solny(t) (-eˆ(2*t)/3 - 1)/(eˆ(2*t)/3 - 1) sage: P = plot(solny(t), 0, 1/2) sage: show(P)
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202 BIBLIOGRAPHY [D-spr] Wikipedia
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204 BIBLIOGRAPHY [NS-intro] General
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Introductory Differential Equations using Sage - William Stein

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