First Order Differential Equations Chapter 1

First Order Differential Equations Chapter 1 Doreen De Leon Department of Mathematics, California State University, Fresno 1 Differential Equations and Mathematical Models Section 1.1 Definitions: An equation relating an unknown function and one or more of its derivatives is a differential equation (DE). An ordinary differential equation (ODE) is a differential equation whose unknown is a function of one variable. You may have first seen some simple ODEs in Calculus I and Calculus II: 1.1 Mathematical Models = x2. Idea: We want to translate a real world problem into mathematical terms, e.g., as a differential equation. Example: Malthus law for population growth states that the time rate of change of a population P (t) with constant birth and death rates is proportional to population size: dp (t) = kp (t), dt where k is the constant of proportionality. The next stage in modeling is to analyze/solve the resulting problem. This could be either quantitative solution (via exact or numerical solutions) or qualitative analysis. Then, interpret the results in the context of the real-world situation. 1

1.2 Terminology The order of a differential equation is the order of the highest derivative that appears in the equation. Example: d2 y dt 2 = g is second order. A function y = y(x) that has a derivative and satisfies the differential equation is a solution of the differential equation. Example: = x2 y = x 2 y = x3 + c, c arbitrary general solution 3 The general solution defines a family of solution curves, one for each value of c. Choosing a particular value of c, say c = 1, gives a particular solution. Usually, the value of c is determined by prescribed conditions. 1.3 Applications 1.3.1 Drug Elimination In many cases, the amount A(t) of a certain drug in the bloodstream, measured by the amount in excess over the natural level of the drug, will decline at a rate proportional to the current excess amount. The model is given by the differential equation da dt = λa, (1) where λ > 0 is the elimination constant. The solution to Equation (1) is A(t) = ce λt. Suppose that we know initially the amount of drug administered (called an initial condition). Suppose A(t = 0) = 250 mg. To get the particular solution, we just do the following: 250 = A(0) = ce λ(0) = c = 250. So, A(t) = 250e λt mg. (Verify that this solves Equation (1).) Definition: A differential equation together with an initial condition is called an initial value problem (IVP). 2

1.3.2 Geometric Example Find the curve through the point (0, 1) with slope x y. Solution: The differential equation modeling this problem is = x y = x2 + y 2 = c (Verify.) Find c: Plug in the initial conditions, x = 0, y = 1. So, the curve is x 2 + y 2 = 1, the unit circle. 1 2 + 0 2 = c = c = 1. 1.4 Separable Equations and Applications Section 1.4 Motivating Examples: Dynamics of Tumor Growth One of the simplest models of tumor cell growth is that the increase in the volume of tumor cells is proportional to the volume of the cells, leading to the model dv dt = kv. (Okay, actually even the simplest model is a bit more complicated than this, but it actually does involve this separable first order equation in part. Also, this does serve as a cool motivational problem.) Definition: The first order differential equation is separable if it can be written in the form = we can separate the variables: We solve by integrating both sides. = H(x, y) = g(x) f(y). f(y) = g(x). Examples: 3

1) Find a general solution of y = 4y. = 4y y = 4 ln y + c 1 = 4x + c 2 ln y = 4x + c 2 c 1 = 4x + c 3 e ln y = e 4x+c 3 y = e 4x e c 3 y = ce 4x. 2) Find a general solution of y = 3y x. 3) Find a general solution of = 1.4.1 Initial Value Problems Example:Solve the initial value problem y = = 3y x 1 3 y = x ln y = 3 ln x + c 1 e ln y = e 3 ln x +c 1 y 1 + x 2. y = x 3 e c 1 y = cx 3. 1 y = 1 1 + x 2 ln y = tan 1 x + c e ln y = e tan 1 x+c 1 y = ce tan 1 x. e y, y(0) = 2. 1 + x 4

First, find the general solution: = e y 1 + x e y 1 = 1 + x e y = ln 1 + x + c = y = ln(ln 1 + x + c). Then, find c: y(0) = 2 = 2 = ln(ln 1 + 0 + c) = 2 = ln c = c = e 2. So, y = ln(ln 1 + x + e 2 ). 1.4.2 Some Applications of Separable Equations Modeling Radiocarbon Dating Carbon-14 is a radioactive isotope of carbon produced in the upper atmosphere by radiation from the sun. Plants absorb carbon dioxide from the air, and living organisms eat plants. The ratio of normal carbon (Carbon-12) to Carbon-14 in the air and living things is nearly constant. When a living creature dies, the Carbon-14 begins to decrease due to radioactive decay. By comparing the amounts of Carbon-12 and Carbon-14, the amount of Carbon-14 that has decayed can be determined. Mathematical Model Let y(t) = amount of Carbon-14 present at time t after death. The model of radioactive decay: y = ky, k > 0. We need additional information to find k. For example, if we have the half-life of Carbon-14 (which is standard information for a radioactive isotope), we can determine k. Sample Case The half-life of Carbon-14 is approximately 5,730 years. Suppose remains have been found in which it is estimated that 30% of the original Carbon-14 is present. Estimate 5

the age of the remains. Step 1: Solve the IVP y = ky, y(0) = y 0. Note: We can also set y(0) = 100 (for 100%) or y(0) = 1 (for 100% in decimal form). By separation of variables, we see that the general solution is Using the initial condition, c = y 0, so Step 2: Find k y = ce kt. y = y 0 e kt. To find k, we will use the half-life of Carbon-14. Since the half-life is the amount of time for half of the sample to decay, we know that y(5730) = 1 2 y 0. = 1 2 y 0 = y 0 e k(5730) = ln 1 2 = 5730k Step 3: Determine the age of the remains = k = ln 1 2 5730 0.000121. At time t (the age), we have 30% of the original amount of Carbon-14 = y(t ) = 0.3y 0. So, plugging this into our solution for y(t) will allow us to solve for the age of the remains. 0.3y 0 = y 0 e 0.000121t ln 0.3 = 0.000121t ln 0.3 t = 9, 950 years. 0.000121 See the textbook for cooling/heating and Torricelli s law problems. 1.5 Linear First Order Equations Section 1.5 Motivating Example: Home Loans Suppose you want to take out a 30-year loan to purchase a house. If interest rates are 4.75% and you know that you can afford at most $1,250 per month, you would like to determine the most expensive house that you can purchase in order pay off the loan on time. To do 6

this, define L(t) as the amount that you owe after t years. Since you can afford $1,250 per month, your annual payments will be $1, 250 12 = $15, 000. The balance due will gain 4.75% interest, so that amount added to the balance annually is 0.0475L. Thus, the rate of change in the amount of money due is given by the difference between the amount the loan increases and the amount that you pay on the loan, or dl = 0.0475L 15000. The initial dt condition is L(30) = 0, because you want the entire loan to be paid off by the end of the 30 years. This gives the initial value problem dl dt = 0.0475L 15000, L(30) = 0. (2) This is a linear equation, as we will see momentarily. The solution to (2) is L(t) = 6000000 19 ( ) 1 e 19 57 t 400 40. The size of the loan you should take is the amount you owe at the beginning, or L(0), which is $239,839.43. Definition A first-order differential equation is linear if it can be written in the form y + p(x)y = q(x). (3) Notice that the equation must be linear in y and y to be a linear differential equation, but p(x) and q(x) can be any function. Definition: If q(x) 0, then (3) is homogeneous. If q(x) is not identically 0, then (3) is nonhomogeneous. Examples: 1) y = y + 1 is linear, nonhomogeneous. 2) y = y sin t is linear (rewrite as y 1 y = 0), homogeneous. sin t 3) t 2 (y + t) = ty + 1 is linear (rewrite as y 1 t y = t 1 t 2 ), nonhomogeneous. 4) (y ) 2 = t + y is nonlinear. 5) yy = t is nonlinear. 6) y = sin y is nonlinear. 7) y = x + e y is nonlinear. 7

Solution to (3): Define µ(x) = e p(x) (called an integrating factor). If we multiply (3) by µ(x), we get Looking at the left-hand side: d e p(x) y + p(x)e p(x) y = q(x)e p(x). ( e p(x) y ) = d ( e ) p(x) y + e p(x) = p(x)e p(x) y + e p(x) y. So, we have: d ( e ) p(x) y = q(x)e p(x). To solve, integrate both sides with respect to x: d ( e ) p(x) y = q(x)e p(x) e p(x) y = q(x)e p(x) + c [ ] y = e p(x) q(x)e p(x) + c. Examples 1) Find a general solution of (x 2 + 1)y + 3xy = 6x. First, we need to divide both sides of the equation by x 2 + 1 to get the equation in the proper form: y + 3x x 2 + 1 y = 6x x 2 + 1. Find the integrating factor: p(x) = 3x x 2 + 1 = µ(x) = e Multiply both sides by µ(x) and solve: (x 2 + 1) 3 2 y + 3x x 2 + 1 (x2 + 1) 3 2 y = 3x x 2 +1 = e 3 2 ln(x2 +1) = (x 2 + 1) 3 2. 6x x 2 + 1 (x2 + 1) 3 2 (x 2 + 1) 3 2 y + 3(x 2 + 1) 1 2 y = 6x(x 2 + 1) 1 2 d ) ((x 2 + 1) 3 2 y = 6x(x 2 + 1) 1 2 d ) ((x 2 + 1) 3 2 y = 6x(x 2 + 1) 1 2 (x 2 + 1) 3 2 y = 2(x 2 + 1) 3 2 + c y = 2 + c(x 2 + 1) 3 2. 8

2) Solve the IVP y y = 8 11 e x 3, y(0) = 1. First, find the general solution: Find the integrating factor: p(x) = 1 = µ(x) = e 1 = e x. Multiply both sides of the equation by µ(x) = e x and solve: e x y e x y = 8 11 e 4x 3 d ( e x y ) = 8 4x 11 e 3 d ( e x y ) 8 4x = 11 e 3 e x y = 6 11 e 4x 3 + c y = 6 11 e x 3 + ce x. Find c, using y(0) = 1: 1 = 6 11 + c = c = 5 11. Therefore, y = 6 11 e x 3 5 11 ex. 1.5.1 Application Mixture Problem Example: A tank initially contains 1,000 gal of water in which is dissolved 20 lb of salt. A valve is opened and water containing 0.2 lb of salt per gallon flows into the tank at a rate of 5 gal/min. Assume that the mixture is kept uniform and drains from the tank at a rate of 5 gal/min. Howe long will it be before the tank contains twice the salt? Model: rate of change in amount of salt = rate entering - rate leaving (i.e., y = rate in - rate out). The rates in and out are found by multiplying the concentration of salt by the rate fluid enters and exits, so ( ) amount of salt rate = (rate of fluid flow). volume of solution 9

For our problem, since the rate of fluid flowing into the tank is equal to the rate flowing out of the tank, the volume of solution at any time t is the same as the initial volume, 1000. So, we have rate in (into tank) = 5 gal lb 0.2 min rate out (out of tank) = 5 gal min y = 1 0.005y. y 1000 Rewriting in the form for a linear equation, we obtain First, find the general solution: Find the integrating factor: Multiply by ρ(t): Find c: So, we have that y + 0.005y = 1, y(0) = 20. gal = 1 lb min lb gal p(t) = 0.005 = µ(t) = e 0.005dt = e 0.005t. e 0.005t y + 0.005e 0.005t y = e 0.005t d ( e 0.005t y ) dt = e 0.005t dt dt e 0.005t y = 200e 0.005t + c 1 y = 200 + ce 0.005t. = 0.005y lb min y(0) = 20 = 20 = 200 + c = c = 180. y(t) = 200 180e 0.005t. The problem asks us to find out how long it will take for the amount of salt in the tank to double. In other words, we want to find time t so that y(t ) = 40. 40 = 200 180e 0.005t 160 = 180e 0.005t 160 180 = e 0.005t ( ) 8 0.005t = ln 9 ) t = ln ( 8 9 0.005 10 23.56 min.

1.6 Substitution Methods Section 1.6 of Text 1.6.1 Homogeneous Equations Motivating Example: Solar Collector Adapted from exercises in A First Course in Differential Equations with Modeling Applications, 8th ed., by Dennis G. Zill. If we assume that when the plane curve C shown below is revolved about the x-axis, it generates a surface of revolution with the property that all light rays L parallel to the x-axis striking the surface are reflected to a single point O (the origin). Using the fact that the angle of incidence is equal to the angle of reflection, you can determine a differential equation that describes the shape of the curve C. Such a curve C is important in applications ranging from construction of telescopes to satellite antennas, automobile headlights, and solar collectors. From geometry, we can see that the angle between the reflected line and the x-axis, denoted φ, and the angle between L and C at the point of incidence, denoted θ are related by φ = 2θ. Using this, we obtain the differential equation = x + x2 + y 2. y This equation is nonlinear and homogeneous, where homogeneous is defined below. The solution of this problem requires a bit more work than the examples which we will do, because it results in a complicated integral. A general solution is y 2 = c(c 2x), which is a parabola that is symmetric with respect to the x-axis. Try to see if you can obtain this solution. Note: This problem may also be solved by a different type of substitution, namely, u = x 2 + y 2. We will discuss such substitutions a bit later. 11

Definition: A homogeneous first order differential equation is a first order differential equation that can be written as ( y ) = F. (4) x To solve, make the substitution Substituting into (4) gives: v = y x = y = vx = = xdv + v. a separable equation. x dv + v = F (v) = xdv = F (v) v, Example: Solve the initial value problem xyy = 2y 2 + 4x 2, y(2) = 4. Divide by xy: y = 2 y x + 4x y. This is a homogeneous differential equation, so let Plugging into our DE gives v = y x = y = vx = y = xv + v. xv + v = 2v + 4 v xv = v + 4 v = v2 + 4 v = x dv = v2 + 4 v v v 2 + 4 dv = 1 x 12

1 2 ln(v2 + 4) = ln x + c 1 ln(v 2 + 4) = 2 ln x + c 2 e ln(v2 +4) = e 2 ln x +c 2 (v 2 + 4) = cx 2 ( y ) 2 = + 4 = cx 2 x = y 2 + 4x 2 = cx 4. Find c: y(2) = 4 = 4 2 + 4(2) 2 = c(2) 4 = c = 2. So, y 2 + 4x 2 = 2x 4. 1.6.2 Other Substitutions Example: Find a general solution of Let v = x + y + 3. 1.6.3 Bernoulli Equations = v = 1 + y y = (x + y + 3) 2. = v = 1 + v 2 (since y = (x + y + 3) 2 = v 2 ). dv 1 + v = 2 tan 1 v = x + c v = tan(x + c) = x + y + 3 = tan(x + c) Motivating Example: A Falling Chain y = tan(x + c) x 3. Adapted from an exercise in A First Course in Differential Equations with Modeling Applications, 8th ed., by Dennis G. Zill. A portion of a uniform chain of length 8 feet is loosely coiled around a peg at the edge of a high horizontal platform, and the remaining portion of the chain hangs at rest over the edge of the platform. Suppose the length of the overhanging chain is 3 feet, that the chain weighs 2 lb/ft, and that the positive direction is downward. Starting at t = 0 seconds, the weight of the overhanging portion causes the chain on the table to uncoil smoothly and to fall to the floor. If x(t) denotes the length of the chain overhanging the table at time t > 0, 13

then v = is its velocity. When all resistive forces are ignored, it can be shown that a dt mathematical model relating v to x is given by xv dv + v2 = 32x. This differential equation is a Bernoulli equation (to be discussed immediately following). Also, interestingly enough, one can determine an integrating factor, which, when multiplied by the equation, turns it into an exact equation (the final type of equation to be discussed). If you solve this equation either way, you can determine that (1) v satisfies and x v(x) = 8 3 9 x, 2 (2) the chain leaves the table with velocity 12.7 ft/s (evaluate v(8), since the chain is 8 ft. long). Definition: A Bernoulli equation is a first order differential equation of the form Note: If n 0, 1 then (5) is nonlinear. To reduce to linear form, do a change of variables: Let v = y 1 n Example: Find a general solution of y + p(x)y = q(x)y n. (5) = v = (1 n)y n y (From (5), y = q(x)y n p(x)y, so) v = (1 n)y n (q(x)y n p(x)y) = v = (1 n)q(x) (1 n)p(x)y 1 n = v = (1 n)q(x) (1 n)p(x)v = v + (1 n)p(x)v = (1 n)q(x)(a linear DE). xy + y = 3x 3 y 3. Divide both sides by x: y + 1 x y = 3x2 y 3. 14

This is Bernoulli with n = 3. Let v = y 1 n = y 2 = v = 2y 3 y = 2y 3 ( 3x 2 y 3 1 x y ) = v = 6x 2 + 2 x y 2 = v = 6x 2 + 2 x v = v 2 x v = 6x2. We now have a linear DE, so we must find the integrating factor. Multiplying by µ(x) gives: p(x) = 2 x = µ(x) = e 2 x = e 2 ln x = x 2. x 2 v 2x 3 v = 6 d ( x 2 v ) = 6 x 2 v = 6x + c v = 6x 3 + cx 2 y 2 = 6x 3 + cx 2. 1.7 Exact Differential Equations Section 1.6 Often, the general solution of a differential equation is defined implicitly by an equation of the form F (x, y(x)) = c, c = constant. (6) Given (6), we can reconstruct the differential equation by differentiating with respect to x: F x + F y = 0. In other words, the differential equation is of the form: It is convenient to write (7) as M(x, y) + N(x, y) = 0. (7) M(x, y) + N(x, y) = 0 (called the differential form). (8) 15

(7), or equivalently (8), defines an exact differential equation with solution F (x, y) = c if M(x, y) = F F and N(x, y) =. x y Question: When is (7) (or (8)) exact? Answer: When there is a function F (x, y) such that M(x, y) = F x and N(x, y) = F y. Obviously, this is not particularly helpful. However, if M(x, y) and N(x, y) are defined and have continuous first partial derivatives in an open region in the xy-plane, then M y = 2 F x y and N x = 2 F y x. By assumption of the continuity of the first partial derivatives, we obtain the following condition that must be satisfied for the differential equation to be exact: M y = N x. If (7) (or, equivalently (8)) is exact, we can solve for F (x, y): 1. Integrate M(x, y) with respect to x (viewing y as constant) to obtain: F (x, y) = M(x, y) + g(y). Then, using the fact that N(x, y) = F, solve for g(y). y 2. Integrate N(x, y) with respect to y (viewing x as constant) to obtain: F (x, y) = N(x, y) + h(x). Then, using the fact that M(x, y) = F, solve for h(x). x Examples: 1) Find a general solution of (2x 2 y 2 ) + (2y 2 2xy) = 0. So, M(x, y) = 2x 2 y 2, N(x, y) = 2y 2 2xy. Now, check if the equation is exact: M y = 2y, N x = 2y = M y = exact. 16

Next, find F (x, y). F (x, y) = = M(x, y) (2x 2 y 2 ) = 2 3 x2 xy 2 + g(y). Now, since N(x, y) = F, find the partial derivative of F (x, y) with respect to y and y set it equal to N(x, y) to find g(y). N(x, y) = F y = 2y2 2xy = 2xy + g (y) = g (y) = 2y 2 = g(y) = 2 3 y3 (+c 1 ). So, F (x, y) = 2 3 x3 xy 2 + 2 3 y3. Note: We do not need to include the +c 1 in the above calculation, because our solution is of the form F (x, y) = c, and so the constants will be combined in the solution. Solution: F (x, y) = c = 2 3 x3 xy 2 + 2 3 y3 = c. 2) Solve the initial value problem y 3 + 3xy 2 So, M(x, y) = y 3, N(x, y) = 3xy 2. Now, check if the equation is exact: = 0, y(1) = 2. Next, find F (x, y). M y = 3y2, N x = 3y2 = M y F (x, y) = = N(x, y) 3xy 2 = exact. = xy 3 + h(x). 17

Now, since M(x, y) = F, find the partial derivative of F (x, y) with respect to x and x set it equal to M(x, y) to find h(x). M(x, y) = F x = y3 = y 3 + h (x) = h (x) = 0 = h(x) = 0(+c 1 ). As before, we do not need to include the +c 1 in the calculation, so F (x, y) = xy 3, and the general solution is F (x, y) = c = xy 3 = c. Find c: y(1) = 2 = x = 1, y = 2 = 1( 2) 3 = c = c = 8. So, xy 3 = 8. 1.8 Chapter 1 Review Problems Find a general solution of 1. sin yy 2x = 0. 2. e y y cos x = (sin x xe y ) y. Solve the initial value problem 3. y + 2xy = xy 1, y(0) = 1. 18