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# 367 Note 3 for MATH 251 at PSU

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Date Created: 02/06/15
Differential Equations LECTURE 3 Linear First Order Equations Recall that an nth linear ordinary differential equation has the following form ModWwaaowWWwmmoymwowwmw for some continuous functions ant gt without restriction In other words a differential equation is linear if it looks like a polynomial with differentiation instead of exponentiation This means that linear rst order equations have a very simple form a1ty t a2tyt W This isn t a very suitable form to work with however What we want to do is to divide through by a1t to obtain the following standard form mmmwgt an 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 coef cient of 3 being 1 and pt having the correct sign Notice that if pt 0 or qt 0 Equation 31 is not only linear but separable as well However if pt 7 0 and qt 7E 0 separation of variables will not work We ll need to be a bit more clever to solve this equation 1 Solution Method Integrating Factors The key to solving just about any rst order equation is to put it in a form where we can integrate both sides There s no direct way to do that in this case however What we ll do is hazard a guess as to something that might help plug in and then gure out speci cs This will be a recurring theme in this class Let s suppose we multiply Equation 31 by some function Mt Mt Mtptyt M06105 32 The left hand side of Equation 32 will hopefully look familiar it resembles the equation for the product rule from calculus yomwwowwmowm am If we could choose our function Mt such that this were the case would be convenient because then the left hand side of Equation 32 would just be the derivative of some product with respect to 25 Since the right hand side is nothing more than some function of t we could then integrate both sides and solve for Differential Equations Lecture 3 Linear First Order Equations So what does Mt need to be For the left hand side of Equation 32 to be equivalent to the left hand side of Equation 33 we need Mt Mimi This is a separable equation and it s not hard to see that the solution is ta SW 34 EXERCISE Check the above assertion With this choice of Mt we can rewrite Equation 32 as and solve for yt mmommow fMW U ma 39 REMARK You may and should notice that there is no constant of integration showing up on the left hand sidedespite us having integrated it What we ve implicitly done is to combine it with whatever constant of integration would come up when we compute f Mtqt dt as you ll see in the examples to come We ve also done this with the constant that ought to appear when we nd Mt lt s imperative not to forget the constant at the nal stage however or your answer will be very very wrong W This method is known as the method of integrating factors and Mt ef W dt as the integrating factor of Equation 31 My recommendation is not to memorize the last line in the solution but rather to compute Mt multiply through and go from there which is what I will do in the examples below 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 rst order linear differential equation are 1 Put the equation in the correct form 1 pty W 2 Calculate the integrating factor Mt amt 3 Multiply both sides of the equation by Mt 4 Integrate both sides being careful not to forget the constant of integration 5 Solve for yt using the initial condition if applicable to calculate the constant of inte gration 2 Examples EXAMPLE 31 iii314 t2 241 1 Differential Equations Lecture 3 Linear First Order Equations Once again7 we start by putting this into standard form7 which requires division by t 3 it 9 t9 Now we compute the integrating factor Mt efgdt 531nt 5111M tax We continue by multiplying through by Mt and using the product rule tSyY 75 t3y t4dt 71755 5 t2 c t 77 7 ylt 5 t3 Using our initial condition to nd 0 gives 1 1 i 4 7 c c 7 5 5 and our particular solution is t2 4 t 7 7 ylt 5 5753 EXAMPLE 32 coszy sinmy 2cos3z sinm 71 y 7 We rst need to put this equation in the correct form7 so diVide through by cosz y tanzy 2 cos2z sinm 7 secz Comparing this to Equation 317 we see that pz tanm7 so our integrating factor is Mx 6f tanz dm eln secm secz7 Differential Equations Lecture 3 Linear First Order Equations using the fact that eh m So we multiply our differential equation through by secm and solve for seczy secz tanzy 2 cosz sinm 7 sec2 seczy 2 cosz sinm 7 sec2m secmyz 2cosz sinm 7 sec2m dz 7 cos2z cosz 7 sinm ccosm Now we use our initial condition to compute 0 7 cos cos 7 sin ccos 7 c a c 71 Thus our particular solution is 7 cos2z cosz 7 sinm 7 cosz EXAMPLE 33 23 7 3y 4t y0 71 First divide by 2 to put this in the correct form 3y 772t y 2 Next compute Mt at Mt efi dt 5 7 Multiply through by Mt and write the left hand side as a product 57 10 2255 Integrate both sides and solve for 452 8g 77 277 2 35 95 6 t7 4t 8 y 7 3 9 C5 Finally we compute the constant of integration 1 8 i 1 7 7 C C 9 9 Differential Equations Lecture 3 Linear First Order Equations Thus our particular solution is 4 yt 737E 7 6 7 75 D EXAMPLE 34 253 7 2y t4 sint t3 7 325 y7r 2 First7 divide by t to put this in the correct form 2 3 7 2y t3 sint t2 7 324 Next7 compute Mt 1175 ef7dt e721nt 7572 Multiply through and write the left hand side as a product t zyy tsint 17 322 Integrate both sides and solve for t zy tsint 17 322 dt 7tcost sint t 7 t3 c yt 723 cost t2 sint t3 7 t5 0752 Plug in the initial condition to compute c 2 771393 cos7r 71392 sin7r 7139s 7 7r5 071392 27r527r3c7r2 7 2 71395 271393 7 W2 Our particular solution is then 2 5 yt t3 cost t2 sint t3 7 t5 252 7139 D

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