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# Linear Algebra and Differential Equations MATH 2250

Utah State University

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This 5 page Class Notes was uploaded by Darren Schulist on Wednesday October 28, 2015. The Class Notes belongs to MATH 2250 at Utah State University taught by Bryan Bornholdt in Fall. Since its upload, it has received 9 views. For similar materials see /class/230413/math-2250-utah-state-university in Mathematics (M) at Utah State University.

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Date Created: 10/28/15

MATH 2250 Linear Algebra and Differential Equations Chapter 1 15 Linear First Order Differential Equations Our second method of solving a differential equation is using an integrating factor A linear firstorder differential equation has the form Pxy Qx dx Notice that the equation is linear in the dependent variable 1 virtually anything is allowed in terms of the independent variable x The key is to recognize that the only terms involving y are the derivative term y39 and y times some function of x This is a special form of differential equation having a distinct solution method Due to the subtleties of specific examples we will always use the method of solution instead of attempting to use a formula to solve these types of equations This method examines the left hand side and recognizes that it can be forced to be the derivative of a product assuming that Px comes from the chain rule for some function p x For every linear firstorder differential equation we must have p x eIPxdx I eJ PmdX as an integrating factori since multiplying both sides of the IPxdx lWe refer to p x differential equation by p x e g eIPxdx dx gives PxeIPxdx y eIPxdx 2IPxa xy This allows us to rewrite the left side of the differential equation as a derivative x PO 93K my BI Qr Therefore integrating both sides with respect to x gives PM y J memgam Solving for y results in y e lPU J Zlmxmga It is tempting to offer a formula representing y we could but there are subtleties regarding the constant of integration corresponding to given initial values It is advised that you learn to rely on the method itself rather than a formulated solution MATH 2250 Linear Algebra and Differential Equations Chapter 1 Consider the following examples Example First verify that y xequot3 Cequot3 satis es the differential equation dy 7x We have y39 6quot 3x26quot3 3xCe39x3 so substitution into the differential equation gives 3 3x2y e39x 3 3 3 3 3 y393x2y e39x 3x36quot 3x2Cequot 3x2 xe39x Ce39x 3 e71C therefore y xequot3 Cequot3 is a solution to the differential equation Now let s solve the differential equation d z 3 l 3x2y e39x Here Px 3x2 and Qx e39x x using an integrating factor In this case we determine the integrating factor to be px efsxlax ex Multiplying both sides of the differential equation by p yields d 6 3 l 3xze 3y exze39x3 dx allowing us to rewrite the equation as d 3 7 ex 1 dx y Integrating both sides with respect to x gives 6 3 y x C Solving for y we obtain the aforementioned solution y xequot3 Ce39xz The integral obtained on the right side was pretty simple Consider how a small change can effect the integration Example Solve the differential equation 3x2y e39le Here Px 3x2 and Qx e39le x The integrating factor in this case is again p x elk dz 6 3 Multiplying both sides by p yields d 6 3 l 3xze 3y 6 5 dx allowing us to rewrite the equation as e 3y l MATH 2250 Linear Algebra and Differential Equations Chapter 1 Integrating the right side requires a clever substitution 116 3 du 1 3 2udu dx 266k 77 This transforms the right side to I2ue du 2ue 2e C 2e u lC integrating by parts Therefore we have exzy2e 6 1C Solving fory gives y 2equot35 l Ce However using an integrating factor does not always result in an exponential solution Example Solve the differential equation d x2 1l3xy 6x dx We first rewrite the equation to fit the standard form dy 3x 6x 7 2 y 2 dx x 1 x 1 3 6 Here Px 27x and Qx x l x 1 The integrating factor in this case is 3 2 iln a 1 2 x2 1 27 W e I 1 e Multiplying both sides by p yields 3 l 3 x2 12 3xx2 l2y 6xx2 12 x allowing us to rewrite the equation as 3 3 ix2 12 y 6xx2 12 dx Integrating yields 3 3 x2 l2y 2x2 12 C and solving for y results in yx 2 Cx2 l7 Notice in the differential equation that ify 2 we have x2 l 6x 6x so 0 for all x x values of x Thus we call y 2 an equilibrium solution See page 49 of the text for a diagram of the slope field for this differential equation MATH 2250 Linear Algebra and Differential Equations Chapter 1 Application Our primary example from this section is a mixture problem that we can create the resulting differential equation from scratch Consider a tank reservoir vat swimming pool lemonade dispenser whatever containing a solution 7 a mixture of a solute and solvent 7 such as salt dissolved in water There is in ow and out ow and we seek to determine a formula for the amount xt of salt solute in the tank at time t given the amount x0 x0 at time t 0 We note the following relationships ri rate owing into the tank c concentration of solute per unit volume coming into the tank V t volume of tank at time t r0 rate owing out of the tank 00 concentration of solute per unit volume coming out of the tank x0 Vt The differential equation thus describes the change in amount of solute with respect to time t Therefore we approximate the change in solute by Ax E r101 r000 At Vita Dividing by At and passing to the limit as At gt 0 we obtain Volume Vt d 1 rlcl rgcg r161 r0 xt r0 dt Vt This results in the linear differential equation dx r 7 70 x0 no dt Vt Example Set up and solve the differential equation for a 150 gallon tank that initially t 0 contains 60 pounds of salt dissolved in 90 gal of water Brine salt water containing 2 lb gal of salt ows into the tank at a rate of 5 galmin and the wellstirred mixture ows out of the tank at a rate of 3 galmin Find the amount of salt xt in the tank tminutes later How much salt is in the tank when it is full Differential Equation 10 ixt dt 90 2t Standard Linear FirstOrder Form 3 ix10 dt 902t In this case the integrating factor is 3 71n 9022 2 3 pt elgwd e 90 20 MATH 2250 Linear Algebra and Differential Equations Chapter 1 Going directly to the resulting integral equation we have 90 2tx 11090 20 ch 290 20 C Solving for xt gives xt 290 2t 90 2t 2 Imposing the initial condition that x0 60 pounds of salt gives 60 180L3 2 C 12090 904 The solution is 3 90 3 xt 2 902t 120 7 90 2t The tank is full when t 30 minutes so there is 3 x30 300 1202 s 2442 pounds of salt in the tank

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