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PSU MATH 251 - Linear First Order Equations

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Lecture 3. Linear First Order Equations1. Solution Method: Integrating Factors2. ExamplesDifferential EquationsLECTURE 3Linear First Order EquationsRecall that an nth linear ordinary differential equation has the following form:an(t)y(n)(t) + an−1(t)y(n−1)(t) + . . . + a1(t)y0(t) + a0(t)y(t) = g(t)for some continuous functions an(t), g(t) without restriction. In other words, a differential equationis linear if it looks like a polynomial, with differentiation instead of exponentiation.This means that linear first order equations have a very simple form.a1(t)y0(t) + a2(t)y(t) = g(t)This isn’t a very suitable form to work with, however. What we want to do is to divide throughby a1(t) to obtain the following standard form:dydt+ p(t)y(t) = q(t) (3.1)Remark. It is absolutely essential that the equation be in this form before we try to solve it.The method is completely dependant on the coefficient of y0being 1 and p(t) having the correctsign.Notice that if p(t) = 0 or q(t) = 0, Equation 3.1 is not only linear, but separable as well.However, if p(t) 6= 0 and q(t) 6= 0, separation of variables will not work. We’ll need to be a bitmore clever to solve this equation.1. Solution Method: Integrating FactorsThe key to solving just about any first order equation is to put it in a form where we canintegrate both sides. There’s no direct way to do that in this case, however. What we’ll do ishazard a guess as to something that might help, plug in, and then figure out specifics. This will bea recurring theme in this class.Let’s suppose we multiply Equation 3.1 by some function µ(t).µ(t)dydt+ µ(t)p(t)y(t) = µ(t)q(t) (3.2)The left hand side of Equation 3.2 will hopefully look familiar; it resembles the equation for theproduct rule from calculus.y(t)µ(t) + µ0(t)y(t) =ddt[µ(t)y(t)] (3.3)If we could choose our function µ(t) such that this were the case, would be convenient, becausethen the left hand side of Equation 3.2 would just be the derivative of some product with respectto t. Since the right hand side is nothing more than some function of t, we could then integrateboth sides and solve for y(t).1Differential Equations Lecture 3: Linear First Order EquationsSo what does µ(t) need to be? For the left hand side of Equation 3.2 to be equivalent to theleft hand side of Equation 3.3, we needµ0(t) = µ(t)p(t).This is a separable equation, and it’s not hard to see that the solution isµ(t) = eRp(t) dt. (3.4)Exercise. Check the above assertion.With this choice of µ(t), we can rewrite Equation 3.2 asddt[µ(t)y(t)] = µ(t)q(t)and solve for y(t):µ(t)y(t) =Zµ(t)q(t) dty(t) =Rµ(t)q(t) dtµ(t).Remark. You may (and should!) notice that there is no constant of integration showing upon the left hand sidedespite us having integrated it. What we’ve implicitly done is to combine itwith whatever constant of integration would come up when we computeRµ(t)q(t) dt, as you’ll seein the examples to come. We’ve also done this with the constant that ought to appear when wefind µ(t). It’s imperative not to forget the constant at the final stage, however, or your answer willbe very, very wrong.This method is known as the method of integrating factors and µ(t) = eRp(t) dtas the integratingfactor of Equation 3.1. My recommendation is not to memorize the last line in the solution, butrather to compute µ(t), multiply through, and go from there, which is what I will do in the examplesbelow. Also, notice that we always get an explicit solution using this method.Before we do some examples, let’s summarize the previous discussion. The steps to solve a firstorder linear differential equation are:(1) Put the equation in the correct formy0+ p(t)y = q(t).(2) Calculate the integrating factor,µ(t) = eRp(t) dt.(3) Multiply both sides of the equation by µ(t).(4) Integrate both sides, being careful not to forget the constant of integration.(5) Solve for y(t), using the initial condition (if applicable) to calculate the constant of inte-gration.2. ExamplesExample 3.1.ty0+ 3y = t2y(1) = 12Differential Equations Lecture 3: Linear First Order EquationsOnce again, we start by putting this into standard form, which requires division by t:y0+3ty = tNow we compute the integrating factor:µ(t) = eR3tdt= e3 ln(t)= eln(t3)= t3We continue by multiplying through by µ(t) and using the product rule.(t3y)0= t4t3y =Zt4dt=15t5y(t) =t25+ct3.Using our initial condition to find c gives1 =15+ c ⇒ c =45and our particular solution isy(t) =t25+45t3. Example 3.2.cos(x)y0+ sin(x)y = 2 cos3(x) sin(x) − 1 yπ4= −√2We first need to put this equation in the correct form, so divide through by cos(x).y0+ tan(x)y = 2 cos2(x) sin(x) − sec(x)Comparing this to Equation 3.1, we see that p(x) = tan(x), so our integrating factor isµ(x) = eRtan(x) dx= eln sec(x)= sec(x),3Differential Equations Lecture 3: Linear First Order Equationsusing the fact that eln f(x)= f(x). So, we multiply our differential equation through by sec(x) andsolve for y(x).sec(x)y0+ sec(x) tan(x)y = 2 cos(x) sin(x) − sec2(x)(sec(x)y)0= 2 cos(x) sin(x) − sec2(x)sec(x)y(x) =Z2 cos(x) sin(x) − sec2(x) dx=Zsin(2x) − sec2(x) dx= −12cos(2x) − tan(x) + cy(x) = −12cos(2x) cos(x) − sin(x) + c cos(x).Now we use our initial condition to compute c.−√2 = −12cosπ2cosπ4− sinπ4+ c cosπ4= −√22+ c√22⇒ c = −1Thus our particular solution isy(x) = −12cos(2x) cos(x) − sin(x) − cos(x). Example 3.3.2y0− 3y = 4t y(0) = −1First, divide by 2 to put this in the correct form.y0−3y2= 2tNext, compute µ(t).µ(t) = eR−32dt= e−3t2Multiply through by µ(t) and write the left hand side as a product.e−3t2y0= 2te−3t2Integrate both sides and solve for y(t).e−3t2y(t) =Z2te−3t2dt= −43te−3t2−89e−3t2+ cy(t) = −43t −89+ ce3t2Finally, we compute the constant of integration.−1 = −89+ c ⇒ c = −194Differential Equations Lecture 3: Linear First Order EquationsThus our particular solution isy(t) = −43t −89−19e3t2. Example 3.4.ty0− 2y = t4sin(t) + t3− 3t5y(π) = 2First, divide by t to put this in the correct form.y0−2ty = t3sin(t) + t2− 3t4Next, compute µ(t).µ(t) = eR−2tdt= e−2 ln(t)= t−2Multiply through and write the left hand side as a product.t−2y0= t sin(t) + 1 − 3t2Integrate both sides and solve for y(t).t−2y =Zt sin(t) + 1 − 3t2dt= −t cos(t) + sin(t) + t − t3+ cy(t) = −t3cos(t) + t2sin(t) + t3− t5+ ct2Plug in the initial condition to


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