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Mathematical Minimum (for Aspiring Physicists)



The dependent variable y and independent variable t in Eq.(1) can be separated to each side of the equation:

dy/y = -kdt ---------- (2)

Then it can be solved by integration:

Figure 01 Integration [view large image]

where is a symbol indicating integration. For example, the first term in Eq.(3) represents the sum of the small strips under the curve 1/y with dy 0 (Figure 01). In term of the natural log "ln" (the 2nd term in Eq.(3), see derivation in Derivation of some Differential Formulas), these integrals can be expressed in the analytical form:

ln(y) - ln(A) = -kt,      or      y = A e-kt

where e = (1 + 1/n)n 2.71828... as n is the exponential base used to simplify the form of many formulas (it plays a special role in calculus similar to the "" in geometry), "ln" is the logarithms in this base, and A is the integration constant (which vanishes and the original differential equation is recovered if we reverse the process by taking the differentiation again) to be determined by the initial condition.
For example, if at t = 0, y = yo, then A = yo. Alternatively, the integration constant can be replaced by the definite integral such that f(y) dy = F(y) = F(b) - F(a), where F(y) is the result of the integration.

Radiactive Decay

In either way, the final solution for Eq.(2) is :

y = yo e-kt --------------------------------- (4)

The green curve in Figure 02 shows the radioactive decay of Iodine 131 with a half life of 8.1 days. Half life is the interval when half of the original is gone. This graph corresponds to a value of 1/k = 8.1/loge(2) = 11.57 days, and yo = 1 gm. in Eq.(4).

Figure 02 Radiactive Decay [view large image]

Alternatively, Eq.(3) can be solved numerically by a computer. If we denote y to be the width of the strips in Figure 01, then the area of the strip y/y = -kt. The relationship between t and y can be approximated by the summation (together with the initial condition at t = 0, y = yo):

ny/y = -k nt = -k [(0 - t1) + (t1 - t2) + + (tn-1 - tn)] = k tn ---------- (5)

where n denotes the number of strips, and the vaule of y can be taken at the middle of each one (as y1, y2, , yn). The result (for a given y) becomes progressively more accurate as the width of the strip is reduced at the expense of computing time. Numerical integration is the only option to solve the differential equation when there is no analytical solution. Eq.(5) is a very simple example, numerial solution can be very complicated involving iterations and finding the root of an equation.

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