This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 2: First Order Different Equations
Yanxiang Zhao
Department of Mathematics, The Pennsylvaina State University January 23, 2011 Outline
1 First Order Linear Equations: Integrating Factors Equations we are able to solve General linear equations 2 1st Order Nonlinear Equations I: Separable ODE 3 1st Order Nonlinear Equations II: Nonseparable ODE Exact Equations Inexact equations 4 Existence and Uniqueness of the Solution of 1st ODE 5 Applications Autonomous Equations Mixture Problems Type I: d f (t) = g(t) or f (t) = g(t), g(t) is known dt d f (t) = sin t. dt f (t) = sin t dt = − cos t + C. d f (t) = eπt . dt 1 f (t) = eπt dt = eπt + C. π • Example 1: Solve • Example 2: Solve • Example 3: Solve d f (t) = g(t), g(t) is known. dt t f (t) = g(t) dt = g(s) ds.
t0 Type II: d µ(t) · f (t) = g(t) or µ(t) · f (t) = g(t), µ(t)&g(t) are known dt • Example 1: Solve t · f (t ) = 2 d 2 t f (t) = cos t. dt cos t dt = sin t + C⇒ f (t) = 1 (sin t + C). t2 • Example 2: Solve d µ ( t ) · f ( t ) = g( t ) . dt 1 µ(t) · f (t) = g(t) dt⇒ f (t) = g(t) dt µ( t ) Type III: µ(t) d d f (t) + f (t) µ(t) = g(t), µ(t)&g(t) are known dt dt Notice that µ( t ) d d d f ( t ) + f ( t ) µ( t ) = µ(t) · f (t) ⇒ Type II = Type III dt dt dt • Example 1: Solve t4 f (t) + 4t3 f (t) = et . t4 f (t) + 4t3 f (t) = et ⇒ t 4 f ( t ) = et ⇒ t 4 f ( t ) = et + C 1 ⇒ f (t) = 4 (et + C). t Type IV: d f (t) = g · f (t), g is a constant. dt d f (t ) = f (t ). dt df (t) = f (t ) dt • Example 1: Solve df (t) df ( t ) = f (t ) ⇒ = dt ⇒ dt f (t ) ⇒ ln f (t) = t + C ⇒ f (t) = et+C = C · et .
• Example 2: Solve d f (t ) = g · f (t ). dt df (t) = f (t ) g dt df (t) df (t) = g · f (t ) ⇒ = g dt ⇒ dt f (t ) ⇒ ln f (t) = gt + C ⇒ f (t) = egt+C = C · egt . Type V: d f (t) = g(t) · f (t), g(t) is a given function. dt df (t) df (t) df (t) = g( t ) · f ( t ) ⇒ = g(t) dt ⇒ = g(t) dt dt f (t ) f (t ) ⇒ ln f (t) = g(t) dt ⇒ f (t) = e g(t) dt . General Form: df + p(t)f = q(t), p&q are given functions. dt • p(t) = 0 ⇒ Type I. If neither p(t) nor q(t) is zero, how can we solve the equation? Compare d d µ( t ) f ( t ) + µ ( t ) f ( t ) = g( t ) dt dt d f (t ) + p( t ) f ( t ) = q( t ) dt d µ(t) f (t) + µ(t)p(t) f (t) = µ(t)q(t) dt • q(t) = 0 ⇒ Type V. Method of Integrating Factors To change the general form into solvable Type III, we set
d µ(t) = p(t)µ(t)(Type V) ⇒ µ(t) = e p(t) dt dt Then general form can be solved, df df + p(t)f = q(t) ⇒ µ(t) + µ(t)p(t)f = µ(t)q(t) dt dt df d ⇒ µ ( t ) + µ ( t ) f = µ ( t ) q( t ) dt dt ⇒ µ ( t ) f ( t ) = µ ( t ) q( t ) t 1 1 ⇒ f (t ) = µ(t)q(t) dt = µ(s)q(s) ds µ( t ) µ ( t ) t0 Method of Integrating Factors: Examples dy 1 + y = t 3 ( t > 0) . dt t dy • + y tan x = sin(2x) dx
• Separable ODE A separable ODE is the one which can be rewritten as M (x)dx = −N (y)dy,
• or M (x)dx + N (y)dy = 0. dy x2 = , y + y2 sin x = 0, xy = (1 − y2 )1/2 are separable. dx 1 − y2 dy • y = 1 + xy, = sin(xy) are nonseparable. dx Solving Separable ODE: Separation of Variables M (x)dx + N (y)dy = 0 ⇒ N ( y) dy = 0 M (x)dx = −N (y)dy ⇒ M (x)dx = − N (y)dy. dy x2 =. dx y ydy = x2 dx ⇒ y2 /2−x3 /3 = C. M (x)dx + • Example 1: Solve dy x2 = ⇒ ydy = x2 dx ⇒ dx y Solving Separable ODE: Separation of Variables • Example 2: Solve dy x − e−x = . dx y + ey dy x − e−x = ⇒ (y + ey )dy = (x − e−x )dx dx y + ey y ⇒ y + e dy = x − e−x dx ⇒ y2 /2 + ey = x2 /2 + e−x + C.
• Example 3: Initial value problem y = 2y2 + xy2 , y(0) = 1. dy = (2 + x)dx ⇒ −1/y = 2x + x2 /2 + C. y2 determine C : −1/1 = 2 × 0 + 0 × 12 + C ⇒ C = −1. y = 2y2 + xy2 ⇒ What is Exact Equation? Nonseparable 1st Order Nonlinear Equations: M (x, y) + N (x, y)y = 0. In particular, if there exists a function φ = φ(x, y) such that M (x, y) = φx , N (x, y) = φy , then one has φx + φy y = 0 ⇔ [φ(x, y)] = 0 ⇔ φ(x, y) = C. In this case, (1) is called an EXACT equation, and φ is called a KERNEL function. (2) (1) Exact Equations: Examples • Example 1: Solve (2x + 3) + (2y − 2)y = 0. Consider φ = x2 + y2 + 3x − 2y, it satisﬁes φx = 2x + 3 and φy = 2y − 2. Therefore, original equation is exact, and the solution is φ = C. (ex sin y − 2y sin x)dx + (ex cos y + 2 cos x)dy = 0. Consider φ = ex cos y + 2 cos x, it satisﬁes φx = M (x, y), φy = N (x, y). Therefore, original equation is exact, and the solution is φ = C. • Example 2: Solve Kernel Function φ • Question: How do we know that if a given function of form M (x, y) + N (x, y)y = 0 is EXACT? If it is exact, how do we ﬁnd a KERNEL φ? (3) • Theorem: Equation (3) is EXACT if and only if the conservation equality holds. My = Nx Solve Exact Equations Follow the 3step rule to solve the exact equation: M (x, y) + N (x, y)y = 0 (4) • 1st step: Check the conservation equality My = Nx to tell if (4) is exact. • 2nd step: If (4) is NOT exact, other methods; If (4) is exact, ﬁnd a kernel function φ(x, y) by φx = M (x, y), φy = N (x, y). • 3rd step: set φ(x, y) = C. Done. Solve Exact Equations: Examples • Example 1: (y cos x + 2xey ) + (sin x + x2 ey − 1)y = 0. • 1st step: My = cos x + 2xey , Nx = cos x + 2xey ⇒ My = Nx ⇒ Exact
• 2nd step: Find a kernel function φ(x, y): φy = sin x + x2 ey − 1 ⇒ φ = sin x · y + x2 ey − y φ = y sin x + x2 ey − y
• 3rd step: y sin x + x2 ey − y = C is the solution. φx = y cos x + 2xey ⇒ φ = y sin x + x2 ey Solve Exact Equations: Examples • Example 2: (ex sin y − 2y sin x)dx + (ex cos y + 2 cos x)dy = 0
• 1st step: My = ex cos y − 2 sin x, Nx = ex cos y − 2 sin x ⇒ My = • 2nd step: Find a kernel function φ(x, y): Nx ⇒ Exact. φx = ex sin y − 2y sin x ⇒ φ = ex sin y + 2y cos x φy = ex cos y + 2 cos x ⇒ φ = ex sin y + 2y cos x φ = ex sin y + 2y cos x • 3rd step: ex sin y + 2y cos x = C is the solution. What is Inexact Equation? Given a nonseparable 1st Order Nonlinear Equation: M (x, y) + N (x, y)y = 0. (5) If the conservation equality fails, namely My = Nx , equation (5) is called an INEXACT equation. It is difﬁcult to solve such an inexact equation.
• A special case: Introduce an integrating factor µ = µ(x), such that µ(x)M (x, y) + µ(x)N (x, y)y = 0 is possibly exact. Integrating factor
• If integrating factor µ = µ(x) makes µ(x)M (x, y) + µ(x)N (x, y)y = 0 exact, then the conservation equality reads [µ(x)M (x, y)]y = [µ(x)N (x, y)]x ⇒ µMy = µNx + µ N Nx − My ⇒ µ + µ = 0. N
• Theorem: For M + Ny = 0, if My − Nx depends on x only, then N there exists a integrating factor µ(x) satisfying µ = My − Nx µ N such that µM + µNy = 0 is exact. Solving Inexact equations Follow the 4step rule to solve the inexact equation: M (x, y) + N (x, y)dy = 0. (6) • 1st step: Check the conservation equality My = Nx , if it holds, exact; if it fails, inexact. • 2nd step: If it is inexact, check if • 3rd step: Solve My − Nx depends on x only. If N it is, go to 3rd step; if it is not, nonsolvable. My − Nx µ N and ﬁnd the integrating factor µ. µ = • 4th step: Study the equation µM + µNy = 0 which is exact. Solving Inexact equations: Examples • Example: (3xy + y2 ) + (x2 + xy)y = 0 • 1st step: Inexact. • 2nd step: My − Nx 3x + 2y − (2x + y) 1 = = N x2 + xy x
• 3rd step: depends on x only. µ = 1 µ ⇒ µ = x. x • 4th step: Study the equation (3x2 y + xy2 ) + (x3 + x2 y)y = 0 which is exact. Linear ODE • Theorem: Consider equation y (t) + p(t)y(t) = q(t), y ( t0 ) = y 0 . (7) If p, q are continuous in (a, b) with t0 ∈ (a, b), then there exists a unique solution of (7).
• Example: Consider the problem sin(t)y + y = sin t , ln(t − 1) y(π /3) = 17. Find the largest interval I where the solution to this problem is certain to exist. A) 1 < t < 2; B) 1 < t < π ; C) 2 < t < π ; D) π < t < 2π . Nonlinear ODE • Theorem: Consider equation y (t) = f (t, y), y ( t0 ) = y 0 . (8) If f , fy are continuous in (t1 , t2 ) × (y1 , y2 ), (t0 , y0 ) ∈ (t1 , t2 ) × (y1 , y2 ), then there exists a unique solution of (8) in the neighborhood of t0 . √ • Example: For the equation y (t) = −1 + y − 1, consider two initial conditions I : y(1) = −1, II : y(1) = 1. Which one of the initial conditions guarantees existence and uniqueness of the solution? A) Both; B) Neither; C) I only; D) II only. What is Autonomous Equation? • For a general 1st order ODE of form y = f (t, y), if f only depends on y, namely, y = f (y), we call it an autonomous equation.
• A constant function y = y1 , where y1 is the root of f (y) = 0, is a EQUILIBRIUM SOLUTION of y = f (y). • Question: Is the constant solution y = y1 stable? Stability of Equilibrium Solutions • For a constant solution y = y1 of a autonomous equation y = f (y), • If f (y) changes the sign from + to − as y crosses y1 from left to • If f (y) changes the sign from − to + as y crosses y1 from left to • If f (y) does not change the sign as y crosses y1 from left to right, right, then y = y1 is stable. right, then y = y1 is unstable. then y = y1 is semistable. Examples • Example: Consider the ﬁrst order autonomous equation y = y2 ( 9 − y2 ) .
• Find all of its equilibrium solutions. • For each equilibrium solution, classify its stability, justify your • If y(10) = −3, what is y(0)? • If y(1) = −π , what is limt→∞ y(t)? • If y(−π ) = 1, what is limt→∞ y(t) answer. Examples • Example: Consider the ﬁrst order autonomous equation y = (y2 − 9)(y + 3)y.
• Identify the equilibrium solutions and classify their stability. • Find the limit value of the solution that satisﬁes the initial condition y(−4) = 4. Concavity of The Solutions • Notice that y (t) = y (t) = f (y) = f (y)y = f (y)f (y) • If f , f have the same sign, the solution y = y(t) is concave up. • If f , f have the opposite sign, the solution y = y(t) is concave down. • Example: Sketch the solutions of y = −(1 − y)y. • Example: Sketch the solutions of y = −(1 − y)(1 − y/2)y. Mixture Problem Q0 lb: initial amount of salt; V0 gal: initial volume of solution; r1 gal/min: incoming solution rate; s1 lb/gal: incoming salt concentration; r2 lb/gal: outgoing solution rate. • At time t = 0 a tank contains Q0 lb of salt dissolved in V0 gal of water. Assume that water containing s1 lb of salt/gal is entering the tank at a rate of r1 gal/min and that the wellstirred mixture is draining from the tank at a rate of r2 gal/min. Set up the initial value problem that describes the rate of change of the amount of salt Q(t). Find the amount of salt Q(t) in the tank at any time. Mixture Problem
• Answer: the rate of change of the amount of salt satisﬁes: dQ = rate in − rate out dt
• rate in = incoming solution rate × incoming salt concentration • rate out = outgoing solution rate × outgoing salt concentration rate in = r1 × s1 Finally, the initial value problem reads Q dQ = r · s − r · 1 1 2 dt V0 + (r2 − r1 )t Q ( 0) = Q
0 rate out = r2 × s2 total amount of salt in tank at time t rate out = r2 × total volume at time t Q( t ) rate out = r2 × V0 + (r1 − r2 )t Mixture Problem: Examples • Example 1: Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dye solution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water ﬂowing in at a rate of 2 L/min, the well stirred solution ﬂowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value. mixture containing a concentration of γ g/L salt enters the tank at a rate of 2 L/min, and the wellstirred mixture leaves the tank at the same rate. Find an expression in terms of γ for the amount of salt in the tank at any time t. Also ﬁnd the limiting amout of salt in the tank as t → ∞. • Example 2: A tank initially contains 120 L of pure water. A Mixture Problem: Examples
• Example 3: A tank originally contains 100 gal of fresh water. Then water containing 1/2 lb of salt/gal is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 min. 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt/gal is entering at a rate of 3 gal/min, and the mixture is allowed to ﬂow out of the tank at a rate of 2 gal/min. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overﬂow. Find the concentration of salt in the tank when it is on the point of overﬂowing. Compare this concentration with the theoretical limiting concentration if the tank had inﬁnite capacity. • Example 4: A tank with a capacity of 500 gal originally contains Mixture Problem: Examples • Example 5: A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of 1 (1 + sin t/2) oz/gal 4 ﬂows into the tank at a rate of 2 gal/min, and the mixture in the tank ﬂows out at the same rate. Find the amount of salt in the tank at any time. The longtime behavior of the solution is an oscillation about a certain constant level. What is this level? What is the amplitude of the oscillation? ...
View
Full
Document
This note was uploaded on 01/31/2011 for the course MATH 250 taught by Professor Gyrya,vitaliy during the Spring '07 term at Penn State.
 Spring '07
 GYRYA,VITALIY
 Math, Linear Equations, Equations, Factors

Click to edit the document details