Question

3. Assume x/x .10 and ż/z .02, and suppose that x(0) 2 and z(0) 1.Calculate thenumerical values of y(t) fort1, t 2, and t 10 for the following cases: (a) y - xz (b) y z/x (c) y = xßZ1-β, where β = 1 /3

Answer 1

We have $\dpi{80} \frac{\mathrm{d} x/\mathrm{d} t}{x} = 0.1$ or $\dpi{80} \frac{\mathrm{d} x}{x} = 0.1 \mathrm{d} t$ , and taking integral on both sides, we have $\dpi{80} \int \frac{1}{x}\mathrm{d} x = 0.1 \int \mathrm{d} t$ or $\dpi{80} ln(x) = 0.1 t + c_1$ or $\dpi{80} x = e^{0.1 t + c_1}$ or $\dpi{80} x = e^{c_1}e^{0.1 t}$ or $\dpi{80} x = C_1e^{0.1 t}$ . Now, for t=0, we have x=2 or $\dpi{80} 2 = C_1e^{0.1 *0}$ or $\dpi{80} C_1 = 2$ , and hence, $\dpi{80} x = 2e^{0.1 t}$ .

Simlarly, we have $\dpi{80} \frac{\mathrm{d} z/\mathrm{d} t}{z} = 0.02$ or $\dpi{80} \frac{\mathrm{d} z}{z} = 0.02 \mathrm{d} t$ , and taking integral on both sides, we have $\dpi{80} \int \frac{1}{z}\mathrm{d} z = 0.02 \int \mathrm{d} t$ or $\dpi{80} ln(z) = 0.02 t + c_2$ or $\dpi{80} z = e^{0.02 t + c_2}$ or $\dpi{80} z = e^{c_2}e^{0.02 t}$ or $\dpi{80} z = C_2e^{0.02 t}$ . Now, for t=0, we have z=1 or $\dpi{80} 1= C_2e^{0.02 *0}$ or $\dpi{80} C_2 = 1$ , and hence, $\dpi{80} z = e^{0.02 t}$ .

(a) For $\dpi{80} y = xz$ , we have $\dpi{80} y = (2e^{0.1 t})(e^{0.02 t})$ or $\dpi{80} y = 2e^{0.1 t + 0.02 t}$ or $\dpi{80} y = 2e^{0.12 t}$ . Hence,

• $\dpi{80} y(0) = 2e^{0.12 *0}$ or $\dpi{80} y(0) = 2$ .
• $\dpi{80} y(1) = 2e^{0.12 *1}$ or $\dpi{80} y(1) = 2e^{0.12} = 2.255$ .
• $\dpi{80} y(2) = 2e^{0.12 *2}$ or $\dpi{80} y(2) = 2e^{0.24} = 2.5425$ .
• $\dpi{80} y(10) = 2e^{0.12 *10}$ or $\dpi{80} y(10) = 2e^{1.2} = 6.6402$ .

(b) For $\dpi{80} y = x/z$ , we have $\dpi{80} y = (2e^{0.1 t})/(e^{0.02 t})$ or $\dpi{80} y = 2e^{0.1 t - 0.02 t}$ or $\dpi{80} y = 2e^{0.08 t}$ . Hence,

• $\dpi{80} y(0) = 2e^{0.08 *0}$ or $\dpi{80} y(0) = 2$ .
• $\dpi{80} y(1) = 2e^{0.08 *1}$ or $\dpi{80} y(1) = 2e^{0.08 } = 2.1666$ .
• $\dpi{80} y(2) = 2e^{0.08 *2}$ or $\dpi{80} y(2) = 2e^{0.16} = 2.347$ .
• $\dpi{80} y(10) = 2e^{0.08 *10}$ or $\dpi{80} y(10) = 2e^{0.8} = 4.4511$ .

(c) For $\dpi{80} y = x^\beta z^{1-\beta}$ or $\dpi{80} y = x^{1/3} z^{1-1/3}$ or $\dpi{80} y = x^{1/3} z^{2/3}$ , we have $\dpi{80} y = (2e^{0.1 t})^{1/3}(e^{0.02 t})^{2/3}$ or $\dpi{80} y = (2e^{0.1 t/3})(e^{0.04 t/3})$ or $\dpi{80} y = 2e^{0.1 t/3 + 0.04 t/3}$ or $\dpi{80} y = 2e^{0.14 t/3}$ . Hence,

• $\dpi{80} y(0) = 2e^{0.14 *0/3}$ or $\dpi{80} y(0) = 2$ .
• $\dpi{80} y(1) = 2e^{0.14 *1/3}$ or $\dpi{80} y(1) = 2e^{0.14/3} = 2.0955$ .
• $\dpi{80} y(2) = 2e^{0.14 *2/3}$ or $\dpi{80} y(2) = 2e^{0.28/3} = 2.1957$ .
• $\dpi{80} y(10) = 2e^{0.14 *10/3}$ or $\dpi{80} y(10) = 2e^{1.4/3} = 3.1893$ .