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Solving differential equations (Sect. 7.4)
Today: Applications.
I (^) Review: Overview of differential equations. I (^) Population growth. I (^) Radioactive decay. I (^) Newton’s Law of Cooling.
Previous class:
I (^) Overview of differential equations. I (^) Exponential growth. I (^) 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 y (x) = y 0 ekx^ , where y (0) = y 0. (b) All solutions y to the separable equation h(y ) y ′(x) = g (x), with functions h, g , are given in implicit form, H(y ) = G (x) + c, where H′^ = h and g ′^ = g.
Review: Overview of differential equations.
Example
Find all solutions y to the equation y ′(x) =
e^2 x−y ex+y^
Solution: Rewrite the differential equation,
y ′^ =
e^2 x^ e−y ex^ ey^
= e^2 x^ e−y^
ex
ey^
= e^2 x^ e−x^ e−y^ e−y^.
y ′^ = ex^ e−^2 y^ =
ex e^2 y^
⇒ e^2 y^ y ′^ = ex^.
Hence, the equation is separable. We integrate on both sides, ∫ e^2 y^ (x)^ y ′(x) dx =
ex^ dx.
Review: Overview of differential equations.
Example
Find all solutions y to the equation y ′(x) =
e^2 x−y ex+y^
Solution: Recall:
e^2 y^ (x)^ y ′(x) dx =
ex^ dx.
The usual substitution u = y (x), and then du = y ′(x) dx, ∫ e^2 u^ du =
ex^ dx ⇒
e^2 u^ = ex^ + c.
We now substitute back u = y (x),
e^2 y^ (x)^ = 2 (ex^ + c) ⇒ 2 y (x) = ln
2 (ex^ + c)
We conclude that y (x) =
ln
2 (ex^ + c)
. C
Population growth
Example
Assume the world population growth is described by y (t) = y 0 ek(t−t^0 ), with t measured in years. (a) If in 1960 − 1961 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−1960).
We only need to evaluate y (2012) = 3 e(0.02)52^ = 8.5 billions. C
Solving differential equations (Sect. 7.4)
Today: Applications.
I (^) Review: Overview of differential equations. I (^) Population growth. I (^) Radioactive decay. I (^) Newton’s Law of Cooling.
Radioactive decay
Remarks:
I (^) Some atoms can spontaneously break into smaller atoms. I (^) This process is called radioactive decay. I (^) It can be seen that the concentration y of a radioactive substance in time t follows the law, y ′(t) = −k y (t), k > 0.
I (^) We know the solution is y (t) = y 0 e−kt^ , y (0) = y 0.
I (^) The half-life of the material is the τ such that y (τ ) = y^0 2
y 0 2
= y 0 e−kτ^ ⇒ − kτ = ln
⇒ τ =
ln(2) k
Radioactive decay
Example
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 ) =
y (0) =
y 0.
y 0 e−kt^1 =
y 0 ⇒ − kt 1 = ln
⇒ t 1 =
k
ln
Recall τ = ln(2)/k and τ = 5730 years. So 1/k = 5730/ ln(2),
We obtain t 1 = [5730/ ln(2)] ln
, hence t 1 = 16, 253 years.C
Newton’s Law of Cooling.
Example
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 − Ts ) e−kt^ + Ts , and also T 0 = 45, Ts = 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^ + 5.
20 = T (2) = 40 e−^2 k^ ⇒ ln(1/2) = − 2 k ⇒ k =
ln(2).
T (t) = 40 e−t^ ln(
√
- (^) + 5 ⇒ 10 = 40 e−t 1 ln( √
- (^) ⇒ t 1 = 4. C
