First Order Linear Differential Equations
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1 LECTURE 8 First Orer Linear Differential Equations We now turn our attention to the problem of constructing analytic solutions of ifferential equations; that is to say,solutions that can be epresse in terms of elementary functions (or formulae. We consier first the case of first orer linear ifferential equations. 1. Linear vs Non-Linear Differential Equations An orinary or partial ifferential equation is sai to be linear if it is linear in the unknowns f, f etc.. Thus,a general,linear,orinary,n th orer,ifferential equation woul be one of the form a n ( n f ( +a n 1( n 1 f n n 1 ( + +a 1( f ( +f(=g(. It is important to note that the functions a n (,...,a1(,g( nee not be linear functions of. following two eamples shoul convey the general iea., 2 f 2, The Eample f isa2 n orer,linear,partial,ifferential equation. Eample f + z y 2 3 f 3 + f 2 = ezy + f 2 =1 is a non-linear,orinary,ifferential equation of orer 3. The equation is non-linear arises because of the presence of the term f 2 which is a quaratic function of the unknown function f. 2. Solving First Orer Linear ODEs A linear first orer orinary ifferential equation is a ifferential equation of the form (8.1 a(y + b(y = c(. Here y represents the unknown function,y its erivative with respect to the variable,an a(, b(c an c( are certain prescribe functions of (the precise functional form of a(,b(,an c( will fie in any specific eample. So long as a( 0,this equation is equivalent to a ifferential equation of the form (8.2 y + p(y = g( where p( = b( a(, g( = c( a(. 23
2 2. SOLVING FIRST ORDER LINEAR ODES 24 We shall refer to a ifferential equation (8.2 as the stanar form of ifferential equation (8.1. (In general,we shall say that an orinary linear ifferential equation is in stanar form when the coefficient of the highest erivative is 1. Our goal now is to evelop a formula for the general solution of (8.2. To accomplish this goal,we shall first construct solutions for several special cases. Then with the knowlege gaine from these simpler eamples, we will evelop a general formula for the solution of any ifferential equation of the form (8.2. Case (i p( =0,g( = arbtrary function. In this case,we have y = g(, an so we are looking for a function whose erivative is g(. Calculus (integral anti-erivative,we have y( = y = where C is an arbitrary constant of integration. Eample 8.3. y g( = g( + C, = 3 cos(4 Applying the Funamental Theorem of y( = 3cos(4 + C = 3 sin(4 +C 4 Case (ii: g( =0,p( = arbitrary function. In this case we are trying to solve a ifferential equation of the form (8.3 y + p(y =0. To construct a solution,we first re-write (8.3 as an equation involving ifferentials y + p(y = 0 y = y p( Integrating both sies of the latter equation (the left han sie with respect to y an the right han sie with respect to yiels or,eponentiating both sies ln(y = p( + C [ y = ep ] p( + C
3 [ ] 2. SOLVING FIRST ORDER LINEAR ODES 25 y = ep p( + C [ ] = e C ep p( ] p( = A ep [ In the last step we have simply replace the constant e C,which is arbitrary since C is arbitrary,by another arbitrary constant A. There is nothing tricky here; the point is that in the general solution the numerical factor in front of the eponential function is arbitrary an so rather than writing this factor as e C we use the simpler form A. Thus,the general solution of y + p( =0 is y = A ep [ p( ]. We note that in both Cases (i an (ii,we constructe a solution by carrying out a single integration,an in oing so an arbitrary parameter (ue to a constant of integration was introuce. This is typical of first orer ifferential equations. Inee,a general solution to an n th -orer ifferential equation will involve n arbitrary parameters. We shall see latter that in physical applications these arbitrary constants correspon to initial conitions. Case (iii: g( 0, p(= a,a constant. In this case we have y + ay = g(. To solve this equation we employ a trick. (This will not be the last trick you see in this course. Let s multiply both sies of this equation by e a : e a y + ae a y = e a g(. Noticing that the right han sie is y (via the prouct rule for ifferentiation we have,equivalently, (ea e a y = We now take anti-erivatives of both sies to get (e a y=e a g(. e a g( + C or Eample 8.4. y( = 1 e a e a g( + Ce a. y 2y = 2 e 2 This equation is of type (iii with So we multiply both sies by e 2 to get p = 2 g( = 2 e 2. ( e 2 y = 2 e (y 2y =e 2( 2 e 2 = 2
4 3. THE GENERAL CASE 26 Integrating both sies with respect to,an employing the Funamental Theorem of Calculus on the left yiels e 2 y = C or y = e 2 + Ce 2. Let us now confirm that this is a solution y = 2 e e 2 +2Ce 2 2y = e 2 2Ce 2 so y 2y = 2 e 2 3. The General Case We are now prepare to hanle the case of a general first orer linear ifferential equation; i.e.,ifferential equations of the form (8.4 y + p(y = g( with p( an g( are arbitrary functions of. Note: This case inclues all the preceing cases. We shall construct a solution of this equation in a manner similar to case when p( is a constant; that is, we will try fin a function µ( satisfying (8.5 µ((y +p(y= (µ(y Multiplying (8.4 by µ(,we coul then obtain which when integrate yiels or (µ(y =µ(g( µ(y = µ( g( + C y = 1 (8.6 µ( g( + C µ( µ( It thus remains to fin a suitable function µ(; i.e.,we nee to fin a function µ( sothat This will certainly be true if (8.7 (µ(y +µ(yp( =µ(y µ( =p(µ(.
5 3. THE GENERAL CASE 27 Thus,we have to solve another first orer,linear,ifferential equation of type (iii. As before we re-write (8.7 in terms of ifferentials to get an then integrate both sies; yieling µ µ = p(, Eponentiating both sies of this relation yiels So a suitable function µ( is ln(µ = p( + A. µ =ep( p( + A µ = ep( p( + A = A ep( p( µ(=a ep( p( Inserting this epression for µ( into our formula (14 for y yiels ( y( = A ep p( ( 1 [ A ep p( g( + C ] It is easily see that the constant A in the enominator is irrelevant to the final answer. This is because it can be cancele out by the A within the integral over,an it can be absorbe into the arbitrary constant C in the secon term. Thus,the general solution to a first orer linear equation y + p(y = g( is given by (8.8 y( = 1 µ( g( + C µ( µ( µ( =ep ( p( Eample 8.5. (8.9 y +2y= sin( Putting this equation in stanar form requires we set Now p( g( p( = 2 = 2 = sin( = 2 ln( =ln ( 2,
6 3. THE GENERAL CASE 28 so Therefore, [ ] µ( = ep [ p( ( =ep ln 2] = 2 1 y( = µ( g( + C µ( µ( = 1 2 ( 2 sin( + C 2 = 1 2 sin( + C 2 Now can be integrate by parts. Set sin( u =, v = sin( Then u =, v = v = cos( an the integration by parts formula, uv = uv vu, tells us that sin( = cos( + cos( Therefore,we have as a general solution of (8.9, = cos( + sin(. y( = 1 2 ( cos( + sin( + C 2 = 1 2 sin( 1 cos( + C 2.
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