Section 9.2 Reading Assignment: First-Order Linear Equations

Answer Only Exercise 1, 2, and 3 by using a screenshot provided Calculus Pearson textbook. Make sure you read these three questions very carefully and see on what it is asking for and what is really about.

References: Thomas' Calculus: Early Transcendentals | Calculus | Calculus | Mathematics | Store | Pearson+

Feedback: Exercise #1: The algebra here is correct, but this doesn't get at the heart of the idea. What does the integrating factor do to make this problem solvable? The reason is crucial to this argument.

Section 9.2 Reading Assignment: First-Order Linear Equations Instructions. Read through this assignment and complete the three exercises below by reading the appropriate passages of the textbook. In this section, we consider another class of differential equations that we can solve: the firstorder linear differential equations. The solution method and its justification are different from separable equations, so it's important to distinguish between these two types of equations. Exercise 1. Read the preface to the subsection \"Solving Linear Equations" (p. 549 550). Explain how the integrating factor is used to solve a linear differential equation in standard form. It's worth noticing the formula forthe integrating factor in the boxed information on p. 550, but it's important to understand why this formula works rather than just what it is. If you are having trouble with this first exercise, pay special attention to the blue text on the side of this reading passage. The videos may also help with this. Understanding why this method works will help a great deal with performing this computation. Exercise 2. Read Example 2 (p. 550]. Despite the formula for the integrating factor, not every integrating factor is an exponential. Describe how the integrating factor is simplified into 3713 in this example. This bit of algebra is worth knowing for these integrating factors, even if you just memorize the formula. This algebra is often needed in order to solve these equations, since the equation can often be unsolvable without this simplification. Exercise 3. Read Example 2 (p. 550) again. This time we're concerned about the final answer of this example. What happens to the constant of integration [+C) in this example? Explain where it ends up in the expression in the end. We mentioned this before in Section 7.2 too, but what happens to the constant of integration in general is important. Example 3 provides another example of this, where the value of C is found too. It's very important to notice that when solving differential equations, you don't just stick a + C at the end. The constant becomes algebraically involved into the final expression. The last subsection \"RL Circuits" in. 552] provides a good example of how relevant differential equations can be to other fields, but we won't be covering it here. Chapter 9 First-Order Differential Equations 9.2 First-Order Linear Equations 549 Solution dy dy Divide by x. dy Standard form with P(x) = -3/x d.x and (Xx) = x Notice that P(x) is -3/x, not +3/x. The standard form is y' + P(xjy = Q(), so the minus sign is part of the formula for P(x). Solving Linear Equations We solve the equation dy 7 + Poly = Qu) by multiplying both sides by a positive function w(x) that transforms the left-hand side into the derivative of the product v(x) . y. We will show how to find us in a moment, but first we want to show how, once found, it provides the solution we seek. Here is why multiplying by v(x) works: dy Original equation is 7 + Puxly = Q() in standard form. dy U( x ) 1x Multiply by positive vix). m(x) is chosen to make dy Integrate with respect to .X. U(x) v(x)Q(x) dx Divide by u(r). (2)Equation (2) expresses the solution of Equation (1) in terms of the functions v(x) and ((x). We call w(x) an integrating factor for Equation (1) because its presence makes the equation integrable. Why doesn't the formula for P(x) appear in the solution as well? It does, but indi- rectly, in the construction of the positive function v(x). We have (vy)=UT+PUy dy Condition imposed on v dy du dy dx + Pvy Derivative Product Rule dx du dy - Puy The terms u cancel. This last equation will hold if AN = PU dx du = P dx Variables separated, > > 0 * = Pax Integrate both sides.Chapter 9 First-Order Differential Equations 550 Chapter 9 First-Order Differential Equations Inv = / pax Since v > 0. we do not need absolute value signs in In v. elmv = el Par Exponentinte both sides to solve for u. (3) Thus a formula for the general solution to Equation (1) is given by Equation (2), where v(x) is given by Equation (3). However, rather than memorizing the formula, just remem- ber how to find the integrating factor once you have the standard form so P(x) is correctly identified. Any antiderivative of P works for Equation (3). To solve the linear equation y' + P(x)y = O(x), multiply both sides by the inte- grating factor v(x) = off) and integrate both sides. When you integrate the product on the left-hand side in this procedure, you always obtain the product ulxjy of the integrating factor and solution function y because of the way v is defined. EXAMPLE 2 Solve the equation dy = x + 3y, I >0. HISTORICAL BIOGRAPHY Solution First we put the equation in standard form (Example 1): Adrien Marie Legendre (1752-1833) = x. www . goo . g1/1zai86so P(x) = -3/x is identified. The integrating factor is Constant of integration is 0, so v is as simple as possible. 120 Next we multiply both sides of the standard form by v(x) and integrate: dx 1 dy x dx dx = Left-hand side is - (v - y). dix Integrate both sides. By = - -+ C. Solving this last equation for y gives the general solution:Chapter 9 First-Order Differential Equations 552 Chapter 9 First-Order Differential Equations RL Circuits /Switch The diagram in Figure 9.9 represents an electrical circuit whose total resistance is a con- stant R ohms and whose self-inductance, shown as a coil, is & henries, also a constant. There is a switch whose terminals at a and b can be closed to connect a constant electrical source of V volts. Ohm's Law, V = R/, has to be augmented for such a circuit. The correct equation L accounting for both resistance and inductance is FIGURE 9.9 The RL circuit in L- + Ri = V. (5) Example 4. where i is the current in amperes and / is the time in seconds. By solving this equation. we can predict how the current will flow after the switch is closed. EXAMPLE 4 The switch in the RZ. circuit in Figure 9.9 is closed at time r = 0. How will the current flow as a function of time? Solution Equation (5) is a first-order linear differential equation for / as a function of t. Its standard form is di + Ri = L' (6) and the corresponding solution, given that i = 0 when / = 0, is V V R (7)(We leave the calculation of the solution for you to do in Exercise 28.) Since R and L are positive, -(R/L) is negative and e (8/4 0 as / - co. Thus, lim i = lim R R = - 4.0 = R R 15 At any given time, the current is theoretically less than V/R. but as time passes, the cur- rent approaches the steady-state value V/R. According to the equation FIGURE 9.10 The growth of the current L di + RI = V. in the RL circuit in Example 4. / is the current's steady-state value. The number / = V/R is the current that will flow in the circuit if either & = 0 (no inductance) or 1 = 1./R is the time constant of the di / de = 0 (steady current, i = constant) (Figure 9.10). circuit. The current gets to within 5% of Equation (7) expresses the solution of Equation (6) as the sum of two terms: a steady- its steady-state value in 3 time constants state solution V/ R and a transient solution -(V/R) @-(/ that tends to zero as / -+ 00. (Exercise 27)