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 p p • 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) = p1 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 = p 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 = = p 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 p v0 − 1 = v p v0 = v + 1 Step 3: Solve the separable DE p p 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 7.