Calculus 1120, Spring 2012 Dan Barbasch October 16 Dan Barbasch - - PowerPoint PPT Presentation

Calculus 1120, Spring 2012

Dan Barbasch October 16

Dan Barbasch Calculus 1120, Spring 2012 October 16 1 / 1

First Order Equations

y′ = f (x, y) A first order differential equation Initial Value Problem: Find a solution to a first order equation satisfying the extra condition y(x0) = y0.

Dan Barbasch Calculus 1120, Spring 2012 October 16 2 / 1

First Order Equations

Motivation/Examples:

1 Exponential Growth/Decay, population growth/radioactive

decay/interest

2 Logistic Equation 3 Circuits 4 Newton’s Law of Cooling 5 Motion with resistance proportional to velocity 6 Mixture Problems Dan Barbasch Calculus 1120, Spring 2012 October 16 2 / 1

First Order Equations

Solving Equations: Finding solutions directly, separable/linear equations Graphical Methods, slope fields Numerical Methods, Euler’s Method

Dan Barbasch Calculus 1120, Spring 2012 October 16 2 / 1

Population Growth/Radioactive Decay

Material and Problems in section 7.2. A quantity y(t) increases/decreases at a rate proportional to the amount present y′ = ky General Solution is obtained by separating variables: dy dt = ky ⇔ dy y = kdt ⇔ dy y =

• k dt ⇔ ln |y| = kt + c

⇔ ⇔ |y| = ekt+c ⇔ y = ±ecekt ⇔ y = Cekt

Dan Barbasch Calculus 1120, Spring 2012 October 16 3 / 1

Population Growth/Radioactive Decay

Initial Value Problem: If you start with the initial amount y0 = 5, and time is measured in hours, what is the amount y after 3hrs. Answer: y(0) = 5 implies 5 = Cek·0 = C. So y(t) = 5ekt, and y(3) = 5e3k. If you start with y(0) = 5, and the amount doubles in 3hrs, what is the constant k? Answer: y = 5ekt, and 10 = 5e3k, then ln 2 = ln

• e3k

= 3k. So k = ln 2

3 , y(t) = 5et ln 2

3 . Dan Barbasch Calculus 1120, Spring 2012 October 16 3 / 1

Population Growth/Radioactive Decay

Solve dy dx = xy + 3x − 2y − 6.

Dan Barbasch Calculus 1120, Spring 2012 October 16 3 / 1

Population Growth/Radioactive Decay

Newton’s Law of Cooling: dH dt = −k(H − HS), where t is time H(t) is the temperature at time t, HS is the temperature of the ambient space k > 0 is a constant depending on the substance cooling, and the ambient space Material is in Section 9.4; separable equation as in section 7.2.

Dan Barbasch Calculus 1120, Spring 2012 October 16 3 / 1

Population Growth/Radioactive Decay

dH H − HS = −kdt ⇔

• dH

H − HS = −

• k dt ⇔

⇔ ln |H − HS| = −kt + c ⇔ eln |H−HS| = e−kt+c = ece−kt ⇔ ⇔ |H − HS| = ece−kt ⇔ H = HS + Ce−kt The initial condition H(0) = H0 gives the final answer H(t) = HS + (H0 − HS)e−kt. Long Term Behaviour: As t → ∞, H → HS. How do we recognize a separable equation? f (x, y) = a(x)b(y), the function f (x, y) factors.

Dan Barbasch Calculus 1120, Spring 2012 October 16 3 / 1

Population Growth/Radioactive Decay

Dan Barbasch Calculus 1120, Spring 2012 October 16 3 / 1

Examples of Separable Equations

Motion Motion where the force is proportional to the velocity: ma = −kv, same as mdv dt = −kv, k > 0. dv dt = − k mv ⇔ dv v = − k mdt ⇔ dv v = − k m

• dt ⇔ ln |v| = − k

mt + c. so v(t) = ±e− k

m t+c = ±ece− k m t, v(t) = Ce− k m t. Plugging in v(0) = v0,

v(t) = v0e− k

m t.

s(t) =

• v(t) =
• v0e− k

m t dt ⇔ s(t) = −v0

m k e− k

m t + C ⇔

⇔ s(t) = −v0 m k e− k

m t + v0

m k + s0. Example in Section 9.3

Dan Barbasch Calculus 1120, Spring 2012 October 16 4 / 1

Examples of Separable Equations

Circuits LdI dt + RI = V ⇔ I ′ = V L − R L I separable I(t) is the intensity of the current at time t. R is the resistance V is the voltage L is the self-inductance

Dan Barbasch Calculus 1120, Spring 2012 October 16 4 / 1

Examples of Separable Equations

Problem 28 in Section 9.2 I = V R + (I0 − V R )e− R

L t

As t → ∞, I → V

R . This is called the long term behaviour of I.

For t large, V ≈ IR, Ohm’s Law.

Dan Barbasch Calculus 1120, Spring 2012 October 16 4 / 1

Logistic Equation

This is a population growth model where the rate of change in the population P(t) at time t is not kP but r(M − P)P with r > 0. dP dt = r(M − P)P = rMP − rP2. M is called the carrying capacity. In the exponential growth model, k can be interpreted as the birth rate. As the population gets closer to the carrying capacity, the birth rate goes down. Material in Section 9.4, separable equation as in Section 7.2. Same type as homework problem 60 in Section 8.4.

Dan Barbasch Calculus 1120, Spring 2012 October 16 5 / 1

Logistic Equation

dP dt = r(M − P)P ⇔ dP P(M − P) = rdt ⇔ ⇔ LHS :=

• dP

P(M − P) =

• r dt = RHS.

Dan Barbasch Calculus 1120, Spring 2012 October 16 5 / 1

Logistic Equation

The RHS is easy to integrate, rt + c. For the LHS, this is why we learned about partial fractions. 1 P(M − P) = 1 M 1 P + 1 M − P

• .

Integrating, 1 M (ln P − ln(M − P)) = rt + c ⇔ ln P M − P = Mrt + cM ⇔ ⇔ P M − P = erMt+cM = ecMeMrt So P M − P = CerMt.

Dan Barbasch Calculus 1120, Spring 2012 October 16 5 / 1

Logistic Equation

Plugging in P(0) = P0 and solving for P (a lot of arithmetic), P(t) = P0M P0 + (M − P0)e−rMt = P0erMt 1 1 + P0

M (erMt − 1)

. The long term behaviour is P(t) → M as t → ∞. When t is close to 0, the solution looks like P0erMt, the exponential model with birth rate rM.

Dan Barbasch Calculus 1120, Spring 2012 October 16 5 / 1

Linear Equations

y′ + P(x)y = Q(x) f (x, y) = a(x)y + b(x). y′ = x + y

Dan Barbasch Calculus 1120, Spring 2012 October 16 6 / 1

Linear Equations

y′ − y = x. We can solve this equation explicity. Use the formula d(uv) = udv + vdu. Multiply both sides by a factor (called integrating factor) so that the LHS becomes the derivative of a product. The factor we need is e−x : e−xy′ − e−xy = xe−x ⇔

• e−xy

′ = xe−x ⇔

• e−xy

′ = xe−x. Integrate e−xy =

• xe−x dx = −xe−x − e−x + C ⇔ y = −x + Cex − 1.

Solving for C from y(x0) = y0, y = −x − 1 + (y0 + x0 + 1)ex−x0. If y0 + x0 + 1 = 0, the solution is a line, y = −x − 1.

Dan Barbasch Calculus 1120, Spring 2012 October 16 6 / 1

Linear Equations

Integrating Factors An equation is called linear if f (x, y) = Q(x) − P(x)y. The circuit equation is linear. To solve, Rewrite as y′ + P(x)y = Q(x). Look for a function m(x) such that m′(x) = P(x). You can do this in principle by integrating, m(x) =

• P(x) dx.

Multiply by em(x) : y′em(x)+P(x)em(x)y = em(x)Q(x) ⇔ (yem(x))′ = Q(x)em(x) ⇔ ⇔ yem(x) =

• Q(x)em(x) dx ⇔ y = e−m(x)
• Q(x)em(x) dx.

In the example m(x)′ = −1 ⇒ m(x) = −x.

Dan Barbasch Calculus 1120, Spring 2012 October 16 6 / 1

Linear Equations

Example: (Problem 16 in section 9.2) xy′ + 2y = x3, x > 0, y(2) = 1. xy′ + 2y = x3 ⇔ y′ + 2 x y = x2 m′ = 2 x ⇔ m = 2 ln x = ln

• x2

. em = x2 : y′x2 + 2xy = x4 ⇔ (yx2)′ = x4 ⇔ yx2 =

• x4 dx = x5

5 + C ⇔ ⇔ y = x3 5 + C x2 . 1 = y(2) = 8 5 + C 4 ⇒ C = −12 5

Dan Barbasch Calculus 1120, Spring 2012 October 16 6 / 1

Mixing Problem (an example of a linear equation)

A 300gal tank is full of brine, containing 50lb of salt. Solution is being drained at 10gal/min, and solution is being poured in at 5gal/min containing 1lb/gal. The tank is being mixed continuously so that the solution is uniform. How much salt is there when the tank is half full?

Dan Barbasch Calculus 1120, Spring 2012 October 16 7 / 1

Mixing Problem (an example of a linear equation)

Answer: First, express the text mathematically Let x(t) be the amount of salt at time t. x(0) = 50. V (t) amount of solution at time t, V (t) = 300+5t − 10t = 300− 5t. Second, write down the equation satisfied by x(t). x′(t) = (amount in) − (amount out) Amount in equals (1lb/gal) · (5gal/min) = 5. Amount out equals (salt per gal) · (gal per unit time) = x(t)

V (t) · 10.

The equation is x′(t) = 5 − x(t) 300 − 5t · 10 x′ + 2 60 − t x = 5 linear equation.

Dan Barbasch Calculus 1120, Spring 2012 October 16 7 / 1

Mixing Problem (an example of a linear equation)

Dan Barbasch Calculus 1120, Spring 2012 October 16 7 / 1