Sect2.1. Any linear equation:

Sect2.1. Any linear equation: Divide a 0 (t) on both sides a 0 (t) dt +a 1(t)y = g(t) dt + a 1(t) a 0 (t) y = g(t) a 0 (t) Choose p(t) = a 1(t) a 0 (t) Rewrite it in standard form and ḡ(t) = g(t) a 0 (t) dt +p(t)y = ḡ(t) Find integrating factor: µ(t) = e p(t)dt.

Multiply µ(t) = e p(t)dt on both sides µ(t) dt +µ(t)p(t)y = µ(t)g(t) µ(t) dt +(µ(t)) y = µ(t)g(t) {µ(t)y} = µ(t)g(t) Integrate directly on both sides µ(t)y = µ(t)g(t)dt + c ( ) 1 y(t) = e at g(t)dt+c µ(t)

Sect2.2. Separable Equations: dx = f(x)g(x) Example 1: Show that the equation is separable and find an equation for its integral curves dx = x2 1 y 2 Choose f(x) = x 2 and g(x) = 1 1 y 2. It is separable.

dx = x 2 1 y 2 (LHS : only y) (1 y 2 ) dx = x2 (RHS : only x) (1 y 2 ) dx dx = (1 y 2 ) = x 2 dx x 2 dx (1 y 2 ) = x 2 dx (integral curves) y 1 3 y3 = 1 3 x3 +c

4 Plot of y (1/3)y 3 =x 3 +c 3 c<0 2 1 y 0 1 2 3 c>0 4 3 2 1 0 1 2 3 x

Example 2: Solve the initial value problem and determine the interval in which the solution exists. dx = 3x2 +4x+2 2(y 1) y(0) = 1 Choose f(x) = 3x 2 +4x+2 and g(x) = 1 2(y 1). It is separable.

dx = 3x2 +4x+2 2(y 1) (LHS : only y) 2(y 1) dx = 3x2 +4x+2 (RHS : only x) 2(y 1) dx dx = 2(y 1) = (3x 2 +4x+2)dx (3x 2 +4x+2)dx 2(y 1) = (3x 2 +4x+2)dx (integral curves) y 2 2y = x 3 +2x 2 +2x+c

General solutions y 2 2y = x 3 +2x 2 +2x+c. Initial condition y(0) = 1 means that y = 1 when x = 0. ( 1) 2 2 ( 1) = c c = 3 Solve for y = 1± x 3 +2x 2 +2x+4 Only the one with - sign satisfies y(0) = 1. So y = 1 x 3 +2x 2 +2x+4. To determine the interval of solution: Root function is involved. Need to make sure the quantity under the radical to be positive. x+2 0 x 3 +2x 2 +2x+4 0 (x 2 +2)(x+2) 0

20 15 y=1+sqrt(x. 3 +2*x. 2 +2*x+c+1) 10 5 0 5 10 y=1 sqrt(x. 3 +2*x. 2 +2*x+c+1) 15 4 3 2 1 0 1 2 3 4

Example 3: Solve equation. Draw graphs of several integral curves. Also find a solution passing through point (0, 1). It is separable. Rewrite it as dx = 4x x3 4+y 3 (4+y 3 ) = (4x x 3 )dx Integration 4y + y4 4 = 2x2 1 4 x4 +c

3 2 y 4 +16 y+x 4 8 x 2 100 = 0 c= 100 1 0 c= 1 c=10 y c=20 1 c=30 2 3 4 4 3 2 1 0 1 2 3 4 x

Other form of separable equations Example: xdx+ye x = 0 ye x = xdx dx = x ye x dx = xex y M(x)+N(x) dx Rewrite it as = 0 or M(x)dx+N(y) = 0 N(y) = M(x)dx N(y) = M(x)dx+c

2.3 Modeling with the first order equations Example 1: At time t = 0, a tank contains Q 0 lb of salt dissolved in 100 gal of water. Assume that water containing 1 4lb of salt/gal is entering the tank at a rate of r gal/min and the well stirred mixture is draining from the tank at the same rate. Set up the initial value problem that describes this flow process. Find the amount of salt Q(t) in the tank at any time. And also find the limiting amount Q L that is present after a very long time. If r = 3 and Q 0 = 2Q L, find the time T after which the salt level is less than 1.02Q L.

Example 1: Let Q(t) denote the total amount of salt inside the tank at any time. dq = rate in rate out dt Rate in: 1 4 lb/gal r gal/min Rate out: Concentration of salt: r gal/min. Concentration of salt inside tank:. Total amount of salt Volume of the tank = Q(t) 100 (lb/gal) dq dt = r 4 rq(t) 100 Q(t) = 25+ce rt/100 Q(0) = Q 0 c = Q 0 25

Example 1: Q(t) = 25+(Q 0 25)e rt/100 50 Plot of solutions with r=3 and different Q0 Q 45 40 35 30 25 20 15 10 5 Q0=25 Q0=20 Q0=15 Q0=10 Q0=5 Q0=0 0 0 20 40 60 80 100 t

Example 1: Q(t) = 25+(Q 0 25)e rt/100 Q(t) = 25(1 e rt/100 )+Q 0 e rt/100 Q(t) 25. Find the amount of salt Q(t) in the tank at any time. And also find the limiting amount Q L that is present after a very long time. Q L = 25 If r = 3 and Q 0 = 2Q L, find the time T after which the salt level is less than 1.02Q L. Q(t) = 25+(Q 0 25)e rt/100 = 25+25e rt/100 Q(T) < 1.02Q L = 25.5

50 Plot of solutions with r=3 and different Q0 45 40 Q 35 30 25 1.02*QL=25.5 T 40 80 120 160 200 t

30 Plot of solutions with r=3 and different Q0 29 28 Q 27 26 1.02*QL=25.5 25 24 40 80 120 160 200 t

Example 1: Q(T) = 25+25e 0.03T = 25.5 25e 0.03T = 0.5 T = ln(50) 0.03 130.4(min)

Example 3: A pond initially contains 10 million gal of fresh water (V = 10 7 gal). Polluted water with undesirable chemical flows into the pond at the rate of r=5 million gal/yr. And mixture in the pond flows out at the same rate. The concentration γ(t) of chemical in the incoming water changes periodically with time γ(t) = 2 + sin(2t) (g/gal). Rate in : r γ(t) Rate out : r Q(t) V Construct a mathematical model of this process and determine the amount of chemical Q(t) in the pond at any time.

Modeling Example 3: Let Q(t) denote the total amount of chemical (gram) in pond. Time t is in years. dq dt = rate in rate out Rate in: r γ(t) = 5million (2+sin(2t)) Rate out: r Q(t) V = 5 10 6 Q(t) 10 7 = Q(t) 2. dq dt = 5 106 (2+sin(2t)) Q(t) 2 Q(0) = 0 (fresh water)

Example 3 : solution dq dt + Q(t) 2 = 5 10 6 (2+sin(2t)) Q(t) = 5 10 6 ( 20 40 17 cos(2t)+ 10 17 ) 100 sin(2t) 17 e t/2. Variation in the incoming concentration cause the oscillation in the solution.

12 x 107 10 8 6 4 2 0 0 5 10 15 20

Example 2: Suppose that a sum of money is deposited in a bank or money fund that pays interest at an annual rate r. The value S(t) of the investment at any time t depends on the frequency with which interest is compounded as well as the interest rate. Financial institutions have various policies concerning compounding: some compound monthly, some weekly, some even daily. If we assume that compounding take place continuously, then we can set up a single initial value problem that describe the growth of the investment.

Sect2.5. Autonomous Equation and Population Dynamics Autonomous Equation: An equation is called autonomous if f does not explicitly depend on independent variable dt = f(y) (separable) Example: To determine they are autonomous or not. dt = y2 +2 du dt = sin(u t) dx = y2 +x 2 dp dt = 1 2 p 450

Sect2.5. Autonomous Equation and Population Dynamics Autonomous Equation: An equation is called autonomous if f does not explicitly depend on independent variable dt = f(y) (separable) Example: To determine they are autonomous or not. dt = y2 +2 (yes) dx = y2 +x 2 (no) du dt = sin(u t) (no) dp dt = 1 2p 450 (yes)

Exponential Growth Let y = φ(t) be the population of a species at t. The rate of change of y is proportional to the current value of y. Parameter r is called dt = ry the rate of growth if r > 0 the rate of decline if r < 0 We assume that r > 0 so population is growing. = ry y(0) = y 0 dt y(t) = y 0 e rt The population will grow exponentially.

Limitation of Exponential Growth Under ideal conditions, exponential growth ever observed to be reasonably accurate for many populations. At least for limited period of time. However it is clear that such ideal conditions will be broken eventually due to the limitation on space, food supply, or other resources. That will reduce the growth rate and bring an end to uninhibited exponential growth.

Logistic Growth The growth rate actually depends on the population. We replace the constant r by h(y) = r ay to get. Logistic equatiion h(y) r when y is small. h(y) decreases as y grows. dt = (r ay)y h(y) < 0 when y is sufficiently large. It can be rewritten as dt = r (1 y K ) y with K = r a r is called Intrinsic growth rate. It is the growth rate in the absence of any limiting factors.

Equilibrium solution Let dt = 0. r( 1 y ) y = 0 K Solve for y and obtain, y = 0 or y = K They are also called critical point.

Plot of f(y) = r ( 1 y ) y K f(y) ( K 2, rk 4 ) (0, 0) K 2 K y

t Plot of solutions with different y0 y0=10 y0=20 y0=30 y0=40 y0=50 y0=60 K y K/2 y0=10 y0=20 y0=30 y0=40 y0=50 y0=60 K y K/2 y0=10 y0=20 y0=30 y0=40 y0=50 y0=60 K y K/2 y0=10 y0=20 y0=30 y0=40 y0=50 y0=60 K y K/2 y0=10 y0=20 y0=30 y0=40 y0=50 y0=60 K y K/2 y0=10 y0=20 y0=30 y0=40 y0=50 y0=60 K y K/2 y0=10 y0=20 y0=30 y0=40 y0=50 y0=60 K y K/2 y0=10 y0=20 y0=30 y0=40 y0=50 y0=60 K y K/2 y0=10 y0=20 y0=30 y0=40 y0=50 y0=60 K y K/2 y0=10 y0=20 y0=30 y0=40 y0=50 y0=60 K y K/2 y0=10 y0=20 y0=30 y0=40 y0=50 y0=60 K y K/2

General solution ( (1 y/k)y = rdt ) 1 y 1/K 1 y/k = rdt ln y ln 1 y/k = rt+c ln y 1 y/k = rt+c y 1 y/k = ±ec e rt y 1 y/k = Cert

Plug in y(0) = y 0, we have C = y 0 1 y 0 /K. y(t) = y 0 K y 0 +(K y 0 )e rt y(t) K as t K is called carrying capacity. It is an asymptotically stable solution. Equilibrium 0 is an unstable equilibrium solution.

Logistic Growth with a threshold dt = r(1 y/t)(1 y/k)y where r > 0 and 0 < T < K.

Sect 2.7 Euler s method The short straight line is an approximation to the solution passing through the point.

Approximation dt = f(t,y), y = φ(t) y(0) = φ(0) = y 0 At (t 0,y 0 ), for t 1 close to t 0, use tangent line y(t 1 ) = y 0 +f(t 0,y 0 )(t t 0 ) to approximate φ(t 1 ) y(t 1 ) = y 0 +f(t 0,y 0 )(t 1 t 0 ) At t 1, for t 1 close to t 0 φ(t 2 ) y(t 2 ) = y 1 +f(t 1,y 1 )(t 2 t 1 ) φ(t n ) y(t n ) = y n 1 +f(t n 1,y n 1 )(t n t n 1 )

Example Consider the initial value problem dt = 3 2t 1 2 y y(0) = 1 Use Euler s method to with stepsize h = 0.2 to find approximate values of the solution of the equation at t = 0.2, 0.4, 0.6, 0.8, 1. At (t 0 y 0 ) = (0,1),