AMS361 Recitation 03/04 Notes 3

1 Mathematical Models

The following mathematical models are based on the natural growth and decay model.

dP • Population Growth: dt = kP , where P is the number of individuals in a population.

dA • Compound Interest: dt = rA, where A is the number of dollars in a saving account.

dN • Radioactive Decay: dt = −kN, where N is the number of atoms of a certain radioac- tive isotope.

dA • Drug Elimination: dt = −λA, where A is the amount of a certain drug in the bloodstream.

There are two more models: Newton’s Law of Cooling and Torricelli’s Law.

dT • Newton’s Law of Cooling: dt = k(A − T ), where T is the temperature of a body, A is the constant temperature of a medium.

dy √ √ • Torricelli’s Law: A(y) dt = −k y = −a 2gy, where y is the height of the water surface, A(y) is the area of the water surface and a is the area of the bottom hole.

1.1 Problem 1.4.35 Question: Carbon extracted from an ancient skull contained only one-sixth as much 14C as carbon extracted from present-day bone. How old is the skull? Solution: The differential equation is dN = −kN ⇒ N(t) = N e−kt, where k = 0.0001216 dt 0 Then, set the number to be one-sixth of the initial number. 1 N(t) = N e−kt = N 0 6 0 ln 6 ⇒ t = ≈ 14735 years k

1 1.2 Problem 1.4.64 Question: A 12-h water clock is to be designed with the dimensions shown in the figure., shaped like the surface obtained by revolving the curve y = f(x) around the y-axis. What should be this curve, and what should be the radius of the circular bottom bole, in order that the water level will fall at the constant rate of 4 inches per hour (in/h)? Solution: The differential equation is

y

1ft

4ft

y=fHxL or x=gHyL

x water flow

Figure 1: The clepsydra in Problem 1.4.64

dy A(y) = −ap2gy dt dy ⇒ π(g(y))2 = −πr2p2gy1/2 dt dy y1/2 ⇒ = −r2p2g dt (g(y))2 where r is the radius of the bottom hole and g(y) is the radius of the water surface. 1 dy 1 Since we require the water level falls at the constant rate of 4 in/h ( 3 ft/h), then dx = − 3 . y1/2 1 −r2p2g = − (g(y))2 3

y1/2 1 Because r, g are constants, so (g(y))2 = c , where c is an constant. y1/2 1 = ⇒ g(y) = c1/2y1/4 (g(y))2 c

2 Solve for c with condition g(4) = 1 ⇒ c = 1 . So the function of the curve is g(y) = √1 y1/4, 2 2 or equivalently, y = 4x4. Then, substitute g(y) back into the equation. y1/2 1 − r2p2g = − (g(y))2 3 1 ⇒ − 2r2p2g = − 3 s 1 ⇒r = √ 6 2g Here, we need to unify the scale. g = 9.8 m/s2 = 32 ft/s2 = 32 × (3600)2 ft/h2. s 1 1 r = = √ ft 6p2 × 32 × (3600)2 240 3 ≈ 0.00240563 ft ≈ 0.0288675 in

2 Linear First-Order DEs

All first-order differential equations can be written in the following form: A(x)y0 + B(x)y = C(x). • If A(x) = 0, it is not a DE. Now suppose A(x) 6= 0.

• If B(x) = 0, it is a separable DE.

• If C(x) = 0, it is also a separable DE. Generally, we have y0 + P (x)y = Q(x). Steps to solve the linear first-order DEs: • Step 1: Convert DE to the form y0 + P (x)y = Q(x);

R • Step 2: Compute integrating factor ρ(x) = e P (x)dx;

• Step 3: Multiply both sides of the converted DE with the integrating factor;

• Step 4: Write the LHS as a whole derivative R R d R e P (x)dxy0 + e P (x)dxP (x)y = [e P (x)dxy] dx • Step 5: Finally, the solution is given by R Z R y(x) = e− P (x)dx[ Q(x)e P (x)dx].

3 2.1 Problem 1.5.3 Question: Find general solution of the DE y0 + 3y = 2xe−3x R Solution: P (x) = 3 ⇒ ρ(x) = e P (x)dx = e3x. Multiply the DE by ρ(x). d (e3xy) = 2xe−3xe3x dx e3xy = x2 + C y(x) = (x2 + C)e−3x

2.2 Problem 1.5.9 Question: Solve the following DE xy0 − y = x, y(1) = 7 0 1 1 Solution: First, convert the DE to the standard form y − x y = 1. Then, P (x) = − x ⇒ R ρ(x) = e P (x)dx = e− ln x. Multiply the DE by ρ(x). d 1 1 ( y) = dx x x 1 y = ln x + C x y(x) = x ln x + Cx Use the initial condition to find the constant C. y(1) = x × ln 1 + C × 1 = 7 ⇒ C = 7 ⇒ y(x) = x ln x + 7x

3 Substitution Method

3.1 Polynomial Substitution: Problem 1.6.16 Question: Find the general solutions of the following DE y0 = px + y + 1 Solution: Step 1: Introducing a new variable v = x + y + 1 Step 2: Transform the Dew into a DE of v v0 = 1 + y0 y0 = v0 − 1 √ v0 − 1 = v √ v0 = v + 1 Step 3: Solve the separable DE √ √ 2 v − 2ln( v + 1) = x + C Then, the general solution for the original DE is implicitly given by 2px + y + 1 − 2ln(px + y + 1 + 1) = x + C

4 3.2 Homogeneous Equations: Problem 1.6.10 Question: Find the general solution of the DE

xyy0 = x2 + 3y2

Solution: Divided by xy x y y0 = + 3 y x y Introduce a new variable v = x , then y = vx ⇒ y0 = v0x + v

Substitute back to the original DE 1 v0x + v = + 3v v 1 + 2v2 = Cx4 x2 + 2y2 = Cx6

3.3 Bernoulli DEs: Problem 1.6.19 Question: Find the general solution of the following DE:

x2y0 + 2xy = 5y3

Solution: step1: Convert to the standard form of Bernoulli DEs 2 5 y0 + y = y3 x x2 step2: Identify the order of the non-linear term n = 3 step3: Introduce a new variable v = y1−n = y−2, then 1 v0 = −2y−3y0 ⇒ y0 = − v0y3 2

Substitute back to the DE 1 2 5 − v0y3 + y = y3 2 x x2 4 10 v0 − v = − x x2 Then, it can be solved by integrating factor method. R ρ(x) = e P (x)dx = x−4

Substitute to the new DE and solve it. d 10 (x−4v) = x−4(− ) dx x2

5 x−4v = 2x−5 + C v = 2x−1 + Cx4 y−2 = 2x−1 + Cx4 1 y2 = 2x−1 + Cx4

4 Exact Differential Equations

General Formula:

M(x, y)dx + N(x, y)dy = 0

If My = Nx, then the DE is called exact. Furthermore, the general solution is given by F (x, y) = C. Do the total differentiation, we have

Fxdx + Fydy = 0.

• SETP1: Compare this equation with the general formula. If we want two match, then

Fx = M(x, y), (1)

Fy = N(x, y). (2)

• SETP2: Look at the equation (1), it can be solved by direct integration. Z Z Fxdx = M(x, y)dx (3) Z F (x, y) = M(x, y)dx + g(y)

Note: Or, you can look at the equation (2), it can also be solved by direct integration. Z Z Fydy = N(x, y)dy (4) Z F (x, y) = N(x, y)dx + g(x)

• STEP3: Then, plug (3) into (2) to obtain g(y). Note: Or, plug (4) into (1) to obtain g(x).

• SETP4: Finally, the solution is Z M(x, y)dx + g(y) = C.

Note: Or the solution is Z N(x, y)dy + g(x) = C.

6 4.1 Problem 1.6.31 Question: Verify that the given differential equation is exact and then solve it.

(2x + 3y)dx + (3x + 2y)dy = 0

Solution: M = 2x + 3y, N = 3x + 2y ⇒ My = 3 = Nx, so it is exact.

2 Fx = M = 2x + 3y ⇒ F = x + 3xy + g(y) 0 0 2 N = Fy = 3x + g (y) ⇒ g (y) = 2y ⇒ g(y) = y F (x, y) = x2 + 3xy + y2 = C

5 Reducible Second-Order Equations

General Form of a second-order differential equation

F (x, y, y0, y00) = 0

0 00 0 00 dp • Dependent variable y is missing: F (x, y , y ) = 0. Do substitution p = y , y = dx .

0 00 0 00 dp • Independent variable x is missing: F (y, y , y ) = 0. Do substitution p = y , y = p dy .

5.1 Problem 1.6.44 Question: Find a general solution of the reducible second-order differential equation.

yy00 + (y0)2 = 0

0 00 dp Solution: Do substitution p = y . Then, y = p dy . dp c yp + p2 = 0 ⇒ p = 1 dy y c 1 y0 = 1 ⇒ y2 = c x + c y 2 1 2

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