Solving differential equations (Sect. 7.4) Review: Overview of differential equations.
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1 Solving differential equations (Sect. 7.4 Previous class: Overview of differential equations. Exponential growth. Separable differential equations. Review: Overview of differential equations. Definition A differential equation is an equation, where the unknown is a function, and both the function and its derivative appear in the equation. Recall: (a All solutions y to the exponential growth equation y (x = k y(x, with constant k, are given by the exponentials where y(0 = y 0. y(x = y 0 e kx, (b All solutions y to the separable equation h(y y (x = g(x, with functions h, g, are given in implicit form, where H = h and g = g. H(y = G(x + c,
2 Review: Overview of differential equations. Find all solutions y to the equation y (x = e2x y e x+y. Solution: Rewrite the differential equation, y = e2x e y e x e y = e 2x e y 1 e x 1 e y = e2x e x e y e y. y = e x e 2y = ex e 2y e 2y y = e x. Hence, the equation is separable. We integrate on both sides, e 2y(x y (x dx = e x dx. Review: Overview of differential equations. Find all solutions y to the equation y (x = e2x y e x+y. Solution: Recall: e 2y(x y (x dx = e x dx. The usual substitution u = y(x, and then du = y (x dx, e 2u du = e x dx 1 2 e2u = e x + c. We now substitute back u = y(x, e 2y(x = 2 (e x + c 2y(x = ln ( 2 (e x + c We conclude that y(x = 1 2 ln( 2 (e x + c.
3 Solving differential equations (Sect. 7.4 Population growth Assume the world population growth is described by y(t = y 0 e k(t t 0, with t measured in years. (a If in the population increased by 2%, find k. (b If the population in t 0 = 1960 was 3 billion people, find the actual population predicted by the law above. Solution: (a y(1961 = y 0 e k(1961 t 0 = y 0 e k(1960 t 0 ( y(1960, 100 e k1961 e kt 0 = 1.02 e k1960 e kt 0 e k( = e k = 1.02 k = ln( Hence y(t = y 0 e (0.02(t t 0.
4 Population growth Assume the world population growth is described by y(t = y 0 e k(t t 0, with t measured in years. (a If in the population increased by 2%, find k. (b If the population in t 0 = 1960 was 3 billion people, find the actual population predicted by the law above. Solution: Recall: y(t = y 0 e (0.02(t t 0. (b If y represents billions of people, 3 = y(t 0 = y 0 e (0.02(t 0 t 0 y 0 = 3 y(t = 3 e (0.02(t We only need to evaluate y(2012 = 3 e ( = 8.5 billions. Solving differential equations (Sect. 7.4
5 Radioactive decay Remarks: Some atoms can spontaneously break into smaller atoms. This process is called radioactive decay. It can be seen that the concentration y of a radioactive substance in time t follows the law, We know the solution is y (t = k y(t, k > 0. y(t = y 0 e kt, y(0 = y 0. The half-life of the material is the τ such that y(τ = y 0 2. y = y 0 e kτ kτ = ln( τ = ln(2 2 k. Radioactive decay The half-life of a radioactive material is τ = 5730 years. If a material sample contains 14% of the original amount, find the date the material sample was created. Solution: Let us fix the time of the original amount at t = 0, and denote the present time by t 1. Also denote y(t the material amount at time t. y(t = y 0 e kt y 0 e kt 1 = y(t 1 = 14 y 0 e kt 1 = 14 ( y 0 kt 1 = ln y(0 = y 0. t 1 = 1 k ln ( Recall τ = ln(2/k and τ = 5730 years. So 1/k = 5730/ ln(2, ( 100 We obtain t 1 = [5730/ ln(2] ln, hence t 1 = 16, 253 years. 14.
6 Solving differential equations (Sect. 7.4 Newton s Law of Cooling. Remarks: The temperature difference T = T T 0 between the temperature of an object, T, and the constant temperature of the surrounding medium where it is placed, T s, evolves in time t following the equation ( T = k( T, T (0 = T 0, k > 0. The solution is ( T (t = ( T 0 e kt, that is, (T T s (t = (T 0 T s e kt T (t = (T 0 T s e kt + T s. The constant k depends on the material and the surroundings.
7 Newton s Law of Cooling. A cup with water at 45 C is placed in the cooler held at 5 C. If after 2 minutes the water temperature is 25 C, when will the water temperature be 15 C? while Solution: We know that T (t = (T 0 T s e kt + T s, and also T 0 = 45, T s = 5, T (2 = 25. Find t 1 such that T (t 1 = 15. First we find k, T (t = (45 5 e kt + 5 T (t = 40 e kt = T (2 = 40 e 2k ln(1/2 = 2k k = 1 2 ln(2. T (t = 40 e t ln( = 40 e t 1 ln( 2 t 1 = 4.
1.5. Applications. Theorem The solution of the exponential decay equation with N(0) = N 0 is N(t) = N 0 e kt.
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