Section B. Ordinary Differential Equations & its Applications Maths II

Transcription

1 Section B Ordinar Differential Equations & its Applications Maths II Basic Concepts and Ideas: A differential equation (D.E.) is an equation involving an unknown function (or dependent variable) of one or more independent variables, t,..., and its derivatives. Differential equations are classified into two categories "ordinar and partial" depending on the number of independent variables appearing in the equation. Ordinar Differential Equation (ODE): An ordinar differential equation is a DE in which the dependent variable depends onl on one independent variable sa. For eample, ' cos, '' 4 0, ''' ' e '' ( ) are ordinar differential equations. d d Notation: The first derivative is denoted b ', second derivative b ", ( n) b n d etc. n Partial Differential Equation (PDE) A partial differential equation is a DE in which the dependent variable, sa u, depends on two or more independent variables,,..., (so that the derivatives of u are partial derivatives). For eample, u u 0, is a partial differential equation. Order of a DE is the order of the highest derivative appearing in the equation. In the eamples above, () is a first-order equation, () is of the second order, and (3) is of the third order. Degree of a DE is the degree of the highest order derivative (when the derivatives are cleared of radicals and fractions). Linear: A differential equation is said to be linear if the unknown function, together with all of its derivatives, appear in the differential equation with positive integral power not greater than one and not as products either. In other words, a differential equation is linear in the dependent variable if it is linear in the unknowns and its derivatives when considered as an algebraic equation. A nonlinear differential equation is a differential equation which is not linear.

2 For eample, 0 is a linear differential equation and 0 is a nonlinear equation. The general form of an n th order linear ODE in is n n d d d 0 n n n a ( ) a ( )... a ( ) an( ) b( ) where a 0 () 0, a (),..., a n () and b() are given functions of. Note: A linear D.E. is of first degree but a first degree D.E. need not be linear. Solution of a Differential Equation A function = g() (4) is called a solution of a given first-order differential equation f (, ) on some interval, sa, a < < b if g() is defined and differentiable throughout that interval and is such that the equation becomes an identit when and ' are replaced b g and g', respectivel, that is, g ( ) f (, g( )). For eample, the function = g() = e is a solution of the first-order differential equation = for all, because b differentiation we obtain g' e and b inserting g and g' we see that the equation reduces to the identit e e. Sometimes a solution of a differential equation will appear as an implicit function, that is, implicitl given in the form G(, ) = 0, and is then called an implicit solution, in contrast to the eplicit solution (4). For eample,, 0, is an implicit solution of the differential equation = - on the interval - < <. A differential equation ma have man solutions.

3 Eample : Each of the functions sin, sin 3, sin 4 5 is a solution of the differential equation = cos and we know from calculus that ever solution of the equation is of the form = sin + c (5) where c is a constant. If we regard c as arbitrar, then (5) represents the totalit of all solutions of the differential equation. (See figure) Solutions of ' = cos Eample: One can verif that each of the functions 6 e, e, e 5

4 is a solution of the differential equation for all. Ever solution of this equation is of the form equation ce, where c is a constant and it represents the totalit of all solutions of the differential. The above eamples illustrate that a differential equation ma have infinitel man solutions, which can be represented b a single formula involving an arbitrar constant c. Such a solution which contains an arbitrar constant is called a general solution of the corresponding differential equation of the first order. If we assign a definite value to that constant, then the solution so obtained is called a particular solution. Remarks: () In some cases there ma be solutions which cannot be obtained b assigning a definite value to the arbitrar constant in the general solution; such a solution is called a singular solution of the equation. For eample, the equation ' ' 0 has the general solution c c We can easil verif (b substitution in the equation) that 4 is a singular solution, since we cannot obtain it b assigning a definite value to c in the general solution. Singular solutions rarel occur in engineering problems. () The conditions under which a given differential equation has solutions are fairl general. But we should note that there are simple equations which do not have solutions at all, and others which do not have a general solution. For eample, does not have a solution for real. (Wh?) The equation 0 does not have a general solution, because its onl solution is 0. Applications (Modeling)

5 Differential equations are of great importance in engineering. Man phsical laws and relations appear mathematicall in the form of differential equations. The following table shows some of the phsical problems that can be modeled and solved as applications of differential equations. Falling stone " = g = const. Out flowing water h' = - k h Current I in an RL- circuit L' + RI = E Parachutist Pursuit problem mv' = mg - bv Cable of a suspension bridge

6 k Vibrating mass on a spring Current I in an RLC circuit Pendulum m k 0 L I RI ( / C) I E l g sin 0 Note that whenever a phsical law involves a rate of change of a function, such as velocit, acceleration, etc., it will lead to a differential equation. Let us consider a simple phsical eample that illustrates the tpical steps of modeling, that is, the steps which lead from the phsical situation (phsical sstem) to a mathematical formulation (mathematical model) and solution, and to the phsical interpretation of the result. Eample: Radio activit, eponential deca Eperiments show that a radioactive substance decomposes at a rate proportional to the amount present. Starting with a given amount of substance, sa, grams, at a certain time, sa, t = 0, what can be said about the amount available at a later time? Solution: Step I (Mathematical description of the phsical process b a differential equation). We denote b (t) the amount of substance still present at time t. Then according to the phsical law

7 d governing the process of radiation, we have k. Here k is a definite phsical constant whose numerical value is known for various radioactive substances sec - (For eample, k.4 0 for radium). Since the amount of substance is positive and decreases with time, d is negative, and so is k. Hence the above equation is the mathematical model of the dt phsical process. Step II (Solving the differential equation) From the knowledge of calculus we can sa that () t where c is an constant, is the general solution of the equation. ce kt Step III (Determination of a particular solution) The amount of substance (t) still present at time t will depend on the initial amount of substance given. This amount is grams at t = 0. Hence we have to specif the value of c so that = when t = 0. This condition is called an initial condition, since it refers to the initial state of the phsical sstem. B inserting this condition (0) = in the general solution, we obtain (0) = c e o = or c =. If we use this value of c in the general solution, then we get the particular solution ( t) e kt The above solution gives the amount of substance still present at an time t 0. The phsical constant k is negative, and (t) decreases, as is shown in the following figure. Radioactivit (eponential deca) Step IV (Checking) We have d dt 0 ke kt k and (0) e Therefore the differential equation and the initial condition are satisfied b the solution function.

8 Methods of Solving First-order First-degree Differential Equation I. Differential equations with variables separable (or Separable Equations) A first-order first-degree differential equation is of the form If the function f turns out to be of the form d f (, ).. () f(, ) = p() q(). () (i.e., the product of a function of with a function of ), then the differential equation () is said to have the variables, separable. In this case the differential equation () can be written in the alternative form as ( q ) d p( ), provided q () 0. Integrating both sides with respect to, we get ( q ) d p( ), or, d q( ) p( ) c The above gives the general solution of the differential equation () with c as a parameter. Eample: Solve the differential equation 9 ' 4 0. Solution: B separating variables we have 9d 4.

9 B integrating on both sides we obtain the general solution The solution represents a famil of ellipses. Initial Value Problem (IVP) In man engineering applications we are not interested in the general solution of a given differential equation but onl in the particular solution () satisfing a given initial condition ( i.e., the condition that at some point 0 the solution () has a prescribed value 0. A first-order differential equation together with an initial condition is called an initial value problem: 0 ) f (, ), ( (IVP) 0 ) 0 0 Eample: Solve the initial value problem ( ) ' 0, (0). Solution: Step I Separating variables, we find B integration we obtain or Taking the tangent on both sides, we have tan (arc tan + arc tan ) = tan c.

10 Now the addition formula for the tangent is F or a = arc tan and b = arc tan this becomes Step II We determine c from the initial condition. Setting = 0 and =, we have = tan c, so that Step III Check the result. Newton's law of cooling: Eperiments show that the time rate of change of the temperature T of an object is proportional to the difference between T and the temperature of the surrounding medium. Eample: A copper ball is heated to a temperature of 00 C. Then at time t = 0 it is placed in water which is maintained at a temperature of 30 C. At the end of 3 minutes the temperature of the ball is reduced to 70 C. Find the time at which the temperature of the ball is reduced to 3 C. Solution: Eperiments show that heat flows so rapidl in copper that at an time the temperature is practicall the same at all points of the ball. I. The mathematical formulation of Newton's law of cooling is ( 30) dt kt dt

11 where we denoted the constant of proportionalit b -k in order that k > 0. kt II. The general solution is obtained b separating variables; we find T ( t) c e 30. III. The given initial condition is T(0) = 00. The particular solution satisfing this condition is kt T ( t) 70e 30. IV. k can be determined from the given information T(3) = 70. We obtain T(3) 70e 3k or k ln Using this value of k we see that the temperature T(t) of the ball is T ( t) 0.865t 70e 30. It follows that the value T = 3 C is reached when 0.865t ln 70 70e or 0.865t ln 70, t , that is, after approimatel 3 minutes. V. Check the result. II. Homogeneous differential equations A function f of two variables is said to be a homogeneous function of degree k if k k f (, ) g or h The first-order first-degree differential equation f (, ) is said to be homogeneous, if f is a homogeneous function of degree zero, i.e. f (, ) g( / ). The substitution = v (or = v ) reduces a homogeneous differential equation to a separable equation in v and (or in v and ) that can be solved b the previous method. III. Eact Differential Equations

12 The (total or eact) differential of a function f(. ) is denoted b df and is given b Consider the differential equation Suppose there eists a function f (, ) such that M(, ) + N(, ) d = 0 () f M (, ) 3 And f N(, ) 4 If equations (3) and (4) are true, then M + N d is said to be an eact differential and the differential Equation () is called an eact differential equation. Using (3) and (4), the given D.E. () becomes 0 f f M Nd d df i.e., df = 0. Integrating, we get f (, ) = c, where c is an arbitrar constant, as the general solution of (). Necessar Condition for Eactness: Differentiating (3) and (4) partiall with respect to and respectivel, we get M f f f f N

13 Thus the necessar condition for D.E. () to be an eact D.E. is M N Method of finding f Step : Integrating (3) partiall with respect to, we get f (, ) M (, ) g( ) (5) where g() is the constant of integration which depends onl on. Step II: To find g() Differentiating (5) partiall with respect to and using (4), we get dg d N M Integrating with respect to, we have g( ) N M d c (Note: Similar result can be obtained from the above procedure starting with (4) also.) Step III: Substituting g() from (6) in (5) The required general solution of the eact D.E. () is f(, ) = c. Eample: Solve

14 Solution: Here so that Hence the given equation is eact. Integrating with respect to, Differentiating with respect to Integration gives g() = + c. Substituting g() in f Thus the desired solution is Eample: Solve Solution:

15 So the equation is eact. Integration with respect to, ields Differentiating with respect to and equating the result to N, we have i.e., g() = constant = c. Required solution is Eample: Solve (cos cos ) d (sin sin ) 0 Solution: M = - sin - sin, N = cos - cos Integrating with respect to, we get f(, ) = - sin + cos + h() Differentiating with respect to and equating it to N

16 dh d 0, i.e. h = constant. The required solution is cos - sin = c Eample: Solve Solution: Here Differentiating So that M N and the D.E. is not eact. IV. Integrating Factor Consider. M(, ) + N(, ) d = 0 () which is not eact. Suppose there eists a function F(, ) such that F(, ) [M + N d] = 0 () is eact, then F (, ) is called an integrating factor (I F) of (). A differential equation ma have more than one integrating factor. For eample, the differential equation d 0 has the functions, and as its integrating factors, because

17 d d, d d log e, d d arc tan. Methods to find Integrating Factors: Case : Integrating Factors b Inspection (Grouping of Terms) In some cases, integrating factors can be obtained b recognizing and regrouping the terms of the given equation such that the form the part of certain common eact differentials. Listed below are some of the eact differentials that are useful in finding I F: S. No Group of terms Integrating factor Eact Differential d d d ln( ) d n, n d d n ( ) n n 3 d d d ln 4 d n, n d n d n n 5 d d d

18 6 d d d 7 d d d 8 d d d ln 9 d d d tan Eample: Solve the differential equations Solution: Regrouping the terms, we have Integrating,

19 Rearranging, where c = c. Eample: Solution: Dividing b throughout, we get Regrouping Eample: Solution: Regrouping, Dividing throughout b 3,

20 Integrating, we get the general solution Case : ( M N) g( ) N (Integrating Factor is a function of alone) Suppose F(, ) is an I F, then F (M + N d) = 0 is an eact differential equation, therefore Solving we get the formula for F(, ) as If F is a function of alone, then () reduces to Since L H S of () is a function of alone, sa g()

21 Then integrating with respect to, we get where Case 3: (N M )/M = h() (Integrating Factor is a function of alone) Suppose that F is a function of alone, then () reduces to Integration gives an integrating factor as where h() = (N M )/M. Eample: Solve Solution: Here and so the given equation is not eact. Since is a function of alone, we get an I F as

22 Multipling the given D.E. b I F e, we get which must be an eact equation. Solving the above eact equation we get the general solution as Eample: Solve the following IVP: ( - ) - d = 0, (0) =. Solution: M = -, N = -, M = -, N = 0, so the equation is not eact. But is a function of. Therefore Multipling both sides of the differential equation b I F, we get which is an eact equation that can be solved b the standard method. Alternative Method Rewriting the above equation as

23 Therefore the general solution is Put = 0, =, so that c = -3. is the particular solution (or the solution of the given IVP) Eample: Solve the following Solution: Here and the equation is not eact. Since = function of alone, we get an I F

24 Multipling the given differential equation b (/ ), we have the following eact equation that ma be solved b the standard method Alternativel, it can also be solved b rearranging the terms as follows: Regrouping the terms, we have Integration gives the general solution as Case 4: Homogeneous D.E. with M + N 0 If M (, ) and N (, ) are homogeneous functions of the same degree then ( M + N) - is an integrating factor of M + N d = 0, provided ( M+ N) 0. In case M + N = 0 then or or are integrating factors. Eample: Solve Solution: Here are both homogeneous functions of degree 3. Since

25 unless =, D E has an Multipling the D E b I F, we get Rewriting Regrouping Integrating we get the general solution as or where c = e c.

26 Case 5: D.E. of the form g() + h() d = 0 If the differential equation M + N d = 0 is of the form g() + h() d = 0 where g() and h() are functions of the product and g() h() then M N { g( ) h( )} is an integrating factor provided ( M N) 0. Note: If ( M N) = 0 then its solution. M N and the given D.E. reduces to d + = 0 with = c as Eample: Solve ( ) ( ) d 0 Solution: Clearl the above equation is not eact. Here M = g() and N = h() so the given D.E. is of the form g() + h() d = 0 which has an integrating factor given b M N ( ) ( ) = Multipling the D.E. with I.F., we get ( Rearranging ) ( ) + d 0

27 Regrouping the terms Integrating Case 6: D.E. of the form a b (m + n d) + c d (p + q d) = 0 The D.E. is of the form a b (m + n d) + c d (p + q d) = 0 where a, b, m, n, c, d, p, q are all constants and (m p n q) 0 has an integrating factor of the form h k,where the unknown constants h, k are determined from the two equations and Eample: Solve ( + 4 ) + ( ) d = 0

28 Solution: Here the equation is not eact. Let us tr to find an integrating factor of the form Consider the D.E. h k 4 4d 3 3 d 0 Rearranging the terms (4 3d) ( 4d) 0 Comparing this with a b (m + n d) + c d (p + q d) = 0, we have a =, b = 0, m = 4, n = 3, c = 0, d =, p, q = 4 Also mp - nq = 8 - = The unknown constants in the integrating factor are determined from the following: i.e.. 4k - 3h = 5 i.e. k - = 0 Solving for h, k, we get h =, k =. Thus the required integrating factor is. Multipling the given D.E. b the integrating factor, we get ( + 4 ) + ( )d = 0

29 or, d d = 0 Regrouping the terms (st and 3rd) and (nd and 4th), we get d( 4 ) + d( 4 3 ) = 0 Integrating, c is the general solution. V. Linear First-Order Differential Equations A first-order differential equation is said to be linear in if it can be written in the form () ' + f() = r(), where f and r ma be an given functions of. If r() = 0 for ever in the domain of the function r(), the equation is said to be homogeneous; otherwise, it is said to be non-homogeneous. Formula for the general solution: It can be shown that an integrating factor (I.F.) of () is the following formula: e f ( ) and its general solution of is given b ( I. F.) c ( I. F.) r( ), where c is an arbitrar constant Or, e f ( ) c ( e f ( ) r( ) Eample: Solve = 0 Solution: We write this equation in the form (), that is, Hence, f = /, r = - 4/ and, therefore, I. F. e f ( ) e e ln

30 So, the general solution is given b Alternativel, the given equation can be written as ( ) + 4 = 0, and the solution is obtained b integration. Note: In simple cases (such as the above eample and the eample given below) one ma solve the equation without using the formula for the general solution. Eample: Solve the linear equation Solution: We can write the equation in the form d() = sin. Integrating, we get Let us now illustrate the procedure of solving initial value problems. Eample: Solve the initial value problem Solution: Here f = tan, r = sin = sin cos, and I. F. e ln sec sec, and the general solution is

31 According to the initial condition, = when = 0, that is and the solution of our initial value problem is = c - or c = 3, Eample: Solve the initial value problem Solution: The equation ma be written as B integration we obtain From this and the initial condition we have () = ½ - + c = 0 or c = ½. Thus Electric Circuits Linear first-order differential equations have various applications in phsics and engineering. The transition from the phsical sstem to a corresponding mathematical model is alwas the first step in engineering mathematics, and this important step requires eperience and training which can onl be gained b considering tpical eamples from various fields. To illustrate this, let us consider some standard eamples in connection with electric circuits. The simplest electric circuit is a series circuit in which we have a source of electric energ (Electromotive force) such as a generator or a batter, and a resistor, which uses energ, for eample an electric light bulb (see the figure). If we close the switch, a current I will flow

32 through the resistor, and this will cause a voltage drop, that is, the electric potential at the two ends of the resistor will be different; this potential difference or voltage drop can be measured b a voltmeter. Eperiments show that the following law holds. The voltage drop E R across a resistor is proportional to the instantaneous current I, sa, where the constant of proportionalit R is called the resistance of the resistor. The current I is measured in amperes, the resistance R in ohms, and the voltage E R in volts. The other two important elements in more complicated circuits are inductors and capacitors. An inductor opposes a change in current, having an inertia effect in electricit similar to that of mass in mechanics. Eperiments ield the following law. The voltage drop E L across an inductor is proportional to the instantaneous time rate of change of the current I sa, where the constant of proportionalit L is called the inductance of the inductor and is measured in henrs; the time t is measured in seconds. A capacitor is an element which stores energ. Eperiments ield the following law. The voltage drop E C across a capacitor is proportional to the instantaneous electric charge Q on the capacitor, sa,

33 where C is called the capacitance and is measured in farads; the charge Q is measured in coulombs. Since this ma be written The current I(t) in a circuit ma be determined b solving the equation (or equations) resulting from the application of the following phsical law: Kirchhoff's voltage law (KVL) The algebraic sum of all the instantaneous voltage drops around an closed loop is zero, or the voltage impressed on a closed loop is equal to the sum of the voltage drops in the rest of the loop. Kirchoff's current law (KCL) At an point of a circuit, the sum of the inflowing currents is equal to the sum of the out flowing currents. Eample (RL-circuit): For the "RL-circuit" we obtain from Kirchhoff's voltage law and (), () RL-circuit Case A (Constant electromotive force). If E(t) = E 0 = constant, then the general solution of the above equation (4) is

34 The last term approaches zero as t, so after a sufficientl long time, I will practicall be constant (= E 0 /R). The particular solution corresponding to the initial condition I(0) = 0 is where L L / R is called the inductive time constant of the circuit. Current due to a constant e.m.f. Case B (Sinusoidal electromotive force). If E(t) = E 0 sin ωt, then the general solution of (4) in this current section is Integration b parts ields This ma be written The eponential term will approach zero as t approaches infinit. This means that after a sufficientl long time the current I(t) eecutes practicall harmonic oscillations. Figure shows the phase angle as a function of L / R. If L = 0, then = 0, and the oscillations of I(t) are in phase with those of E(t).

35 Current due to a sinusoidal electromotive force with / 4 An electrical (or dnamical) sstem is said to be in the stead state when the variables describing its behavior are periodic functions of the time or constant, and it is said to be in the transient state (or unstead state) when it is not in the stead state. The corresponding variables are called stead-state functions and transient functions, respectivel. In the above eample, Case A, the function E o / R is the stead-state function or stead-state solution of (4), and in Case B the stead-state solution is represented b the last term in (6). Before the circuit (practicall) reaches the stead state it is in the transient state. It is clear that such an interim or transient period occurs because inductors and capacitors store energ, and the corresponding inductor currents and capacitor voltages cannot be changed instantl. Similar situations arise in various phsical sstems. For eample, if a radio receiver having heater-tpe vacuum tubes is turned on, a transient interval of time must elapse during which the tubes change from cold to hot. Practicall, transient state of our present sstem will last onl for a short time. Eample: RC-circuit B appling Kirchhoff's voltage law and (), (3) to the RC-circuit we obtain the equation To get rid of the integral we differentiate the equation with respect to t, finding RC-circuit Solving the above linear DE (8), we have the general solution as Case A (Constant electromotive force). If E is constant, then

36 and (9) assumes the simple form where C RC is called the capacitive time constant of the circuit. Current in an RC-circuit due to a constant electromotive force Case B (Sinusoidal electromotive force). If E( t) E0 sin t, then B inserting this in (9) and integrating b parts we find where The first term decreases steadil as t increases, and the last term represents the stead-state current, which is sinusoidal. VI. Reduction of Nonlinear equations to Linear Form Certain nonlinear first-order differential equations ma be reduced to linear form b a suitable change

37 of the dependent variable. (a) Bernoulli equation The differential equation ' + f() = r() n, n an real number, is called the Bernoulli equation. For n = 0 and n = the equation is linear in, and otherwise it is non-linear. Rewriting the equation as n f ( ) n n g( ) and setting [ ( )] u( ), one can show that the equation reduces to the following linear form (in the dependent variable u) that can be solved b the method discussed above: u ( n) f ( ) u ( n) g( ) Eercise: Solve the following Bernoulli equations (i) (ii) ( ) (iii) Orthogonal Trajectories If for each fied real value of c the equation () F(,, c) = 0 represents a curve in the -plane and if for variable c it represents infinitel man curves, then the totalit of these curves is called a one-parameter famil of curves, and c is called the parameter of the famil. Eamples: The equation () F(,, c) = + + c = 0 represents a famil of parallel straight lines; each line corresponds to precisel one value of c. The equation represents a famil of concentric circles of radius c with center at the origin. The general solution of a first-order differential equation involves a parameter c and thus represents a famil of curves. Therefore one ma represent man one-parameter families of curves b such differential equations. Eample: B differentiating () we see that

38 is the differential equation of that famil of straight lines. Similarl, the differential equation of the famil (3) is If the equation obtained b differentiating () still contains the parameter c, then we have to eliminate c from this equation b using (). Eample: The differential equation of the famil of parabolas is obtained b differentiating (4), and b eliminating c from (5). From (4) we have c = /, and b substituting this into (5) we find the desired result Alternativel, we solve (4) for c, finding c = /, and differentiate this equation with respect to, obtaining

39 In man engineering applications, a famil of curves is given, and it is required to find another famil whose curves intersect each of the given curves at right angles (The angle of intersection of two curves is defined to be the angle between the tangents of the curves at the point of intersection). Then the curves of the two families are said to be mutuall orthogonal, and the curves of the famil to be obtained are called the orthogonal trajectories of the given curves (and conversel). Curves and their orthogonal trajectories Eamples: The meridians on the earth's surface are the orthogonal trajectories of the parallels. In electrostatics the equipotential lines and the lines of electric force are orthogonal trajectories of each other. Equipotential lines and lines of electric force Given a famil F(,, c) = 0 which can be represented b a differential equation f (, ) we ma find the corresponding orthogonal trajectories as follows. A curve of the given famil which passes through a point ( o, o ) has the slope f( o, o ) at this point. The slope of the orthogonal trajector through ( o, o ) at this point should be [ / f ( 0, 0 ) ]. Consequentl, the differential equation of the orthogonal trajectories is / f (, ), and the trajectories are obtained b solving this equation.

40 Eample: Find the orthogonal trajectories of the parabolas = c. Solution: The differential equation of the orthogonal trajectories is (see (6)) B separating variables and integrating we find that the orthogonal trajectories are the ellipses Eample: Find the orthogonal trajectories of the circles ( c) c. () Solution: B differentiating () with respect to we obtain ( c) 0 () Solving () for c, we have B inserting this into () and simplifing we get Circles and their orthogonal trajectories (dashed) Hence, the differential equation of the orthogonal trajectories is Solving the above homogeneous equation, we have the orthogonal trajectories as

41 Eercise: (Isogonal trajectories) Isogonal trajectories of a given famil of curves are curves which intersect the given curves at a constant angle. Show that at each point the slopes m l and m of the tangents to the corresponding curves satisf the relation Further, determine the curves which intersect the circles c at an angle of 45. Polar coordinates Suppose the polar equation of a curve C is r = f(θ). B writing the rectangular coordinates (, ) of a point P(r, θ) of C as r cos f ( )cos r sin f ( )sin and considering them as parametric equations with parameter θ, we get the formula for the slope m at P: m d d/ d / d f ( )cos f ( )( sin ) f ( )sin f ( )cos f f ( ( )sin )cos f ( f ( )cos )sin A curve ma also be epressed in polar coordinates b

42 tan = d r.. (A) dr where the angle is measured (positive) in counter clockwise direction from radius vector OP to the tangent line at an point P on C (see the figure below). If is the angle that this tangent makes with the Also, we have tan tan( ), and tan tan tan tan [( f sin f cos ) /( f cos f sin )] f sin f cos sin f cos f sin cos sin / cos After simplification, we get initial line (polar ais), then m d tan. f tan = f r dr d = d r dr Orthogonalit Two curves c and c (see the figure below) are said to intersect orthogonall at the point P if.

43 Therefore, at the point of intersection P, the value of the product r( d / dr) for one curve c must be the negative reciprocal of value of that product for the other curve c. That is, if c is represented b the equation (A) then its orthogonal dr trajector c must be given b tan. r d Method of obtaining orthogonal trajectories in polar coordinates Suppose the first famil of curves has differential equation Then and the differential equation corresponding to the orthogonal trajectories is

44 The solution of the above equation is the required orthogonal trajectories. Self-orthogonal A given famil of curves is said to be "self-orthogonal" if its famil of orthogonal trajectories is the same as the given famil. Eample: Find the orthogonal trajectories of the famil of curves given b the following: r acos Solution: Differentiating r acos with respect to r, we get Eliminating we get Therefore the differential equation corresponding to orthogonal trajector is Solving we get

45 So the orthogonal trajector is given b Linear Differential Equations of Second and Higher Order Linear Equations of Arbitrar Order A differential equation of nth order is said to be linear if it can be written in the form where the function r on the right-hand side and the coefficients f 0, f,..., f n are an given (n) functions of, and is the nth derivative of with respect to, etc. An differential equation of order n which cannot be written in the form () is said to be nonlinear. If r() 0, equation () becomes and is said to be homogeneous. If r() is not identicall zero, the equation is said to be nonhomogeneous. Using operator notations, we can also write () in the form

46 and similarl for (), where D denotes differentiation with respect to, and L = P(D). Eistence and uniqueness theorem If f0 ( ), f( ),..., fn ( ) in () are continuous functions on an open interval I, then the initial value problem consisting of the equation () and the n initial conditions has a unique solution () on I; here 0 is an fied point in I, and K, K,..., K n are given numbers. Linear dependence and linear independence The n ( ) functions ( ), ( ),..., n ( ) are said to be linearl dependent on some interval I where the are defined, if (at least) one of them can be represented as a "linear combination" of the other n - functions on I. If none of the functions can be represented in this wa, the are said to be linearl independent on I. 3 For eample, the functions,, 3 are linearl independent on an interval. The functions, 3, 3 are linearl dependent on an interval, because Theorem: n functions ), ( ),..., n ( ) are linearl dependent on an interval I where the are ( defined if and onl if we can find constants k, k,..., kn, not all zero, such that the relation holds for all on I.

47 A solution of a differential equation of the order n (linear or not) is called a general solution if it contains n arbitrar independent constants. Here independence means that the solution cannot be reduced to a form containing less than n arbitrar constants. If we assign definite values to the n constants, then the solution so obtained is called a particular solution of that equation. A set of n linearl independent solutions ( ),..., n ( ) of the linear homogeneous equation () on I is called a basis or a fundamental sstem of solutions of () on I. If ( ),..., n ( ) is such a basis, then is a general solution of () on I. Theorem (Test for linear dependence and independence of solutions): Suppose that the coefficients f ), f ( ),..., fn ( ) of () are continuous on an open interval I. 0 ( ( ), ( ),..., ( Then n solutions n ) of () on I are linearl dependent on I if and onl if their Wronskian is zero for some = o in I. (If W = 0 at = o, then W 0 on I.) Eample: The functions e 3, e, and 3 e are solutions of the equation The Wronskian is

48 which shows that the functions constitute a basis of solutions of (6) for all, and the corresponding general solution is Homogeneous Linear Equations of Arbitrar Order with Constant Coefficients To solve a homogeneous linear equations of order n with constant coefficients we put = e and its derivatives into (). On simplification, we obtain The equation () is called the characteristic equation of the differential equation (). The equation () is a polnomial equation of degree n in and so it has at least one and at most n distinct roots (or solutions). Further, from elementar algebra we know that, since the coefficients a 0, a,, a n-, are real, the characteristic equation ma have (Case I) simple real root, (Case II) multiple real root of multiplicit k (Case III) simple comple conjugate roots, or (Case IV) multiple comple conjugate roots of multiplicit k We use the following table to find the general solution of (): Case Characteristic root (or solution of ()) Solution of DE () I, a simple real root e

49 II, multiple real root of multiplicit k e, e,.., k k e III p iq, a simple pair of comple roots IV p iq, multiple pairs of comple roots of multiplicit k p e cos q, p e sin q. p e cos p k p q, 3 e cos q,, k e cos q, p e sin p q, e sin k p 4 q,, k e sin q. It can be shown that the n functions ), ( ),..., n ( ) obtained from column 3 of the above ( table are linearl independent solutions of the differential equation. Therefore the general solution of () is given b ( ) c ( ) c ( )... cn n( ), where c, c,..., cn are arbitrar constants. The most difficult aspect of solving constant-coefficient equations is finding the roots of auiliar equations f ( ) 0 of degree greater than two. One wa to solve such an equation is to guess a root and then divide f ( ) b to obtain the factorization ( ) Q ( ). We then tr to find the roots of Q ( ) 0. Eample: Solve Solution: The auiliar equation is and b inspection we find that is one of its 3 root [since () 3() 4 ]. Now if we divide b we find = ( )( 4 4 ) = ( )( ), and so the other roots are 3 (multiple real root of multiplicit ). Thus the general solution is e ce c3e. c

50 Eample: Solve 4 d 4 d 0. Solution: The auiliar equation 4 ( ) 0 has i as comple conjugate roots of multiplicit. Thus the general solution is c cos c sin c3cos c4sin. Non-homogeneous Linear Equations We shall now discuss the methods to find the general (or complete) solution of non-homogeneous linear equations. Theorem (General solution): A general solution () of the linear non-homogeneous differential equation is the sum of a general solution h () of the corresponding homogeneous equation and an arbitrar particular solution (or particular integral) p () of (): ( ) ( ) ( ) h p. Equation () is called the subsidiar (or related homogeneous) equation of () and the general solution () h of the related homogeneous equation is known as complementar function of (). With these terminolog, general solution () of () = complementar function of () + a particular integral of ().

51 Method for Solving Non-homogeneous Equations For obtaining a general solution of a given non-homogeneous linear differential equation we need complementar function and a particular integral p () of that equation. There are several methods to find particular integrals. We shall discuss the following two methods. I. Method of Variation of Parameters Method of variation of parameters is the most general method for solving non-homogeneous equations if its complementar function is known. We consider linear differential equations of the form '' f ( ) ' g( ) r( ) () Assuming that f, g, and r are continuous on an open interval I. We shall obtain a particular solution p () of () b using the method of variation of parameters as follows. We assume that the corresponding homogeneous differential equation '' f ( ) ' g( ) 0 () has a general solution of the form The method consists in replacing c and c b functions u() and v() to be determined so that the

52 resulting function ( p ) u ( ) ( ) ( ) ( ) v (3) is a particular solution of (). B differentiating (3) we obtain We shall see that we can determine u and v such that u ' v ' 0. (4) This reduces the epression for p to the form ' u ' v '. (5) p B differentiating this function we have '' u ' ' u '' v' ' v ''. (6) p B substituting (3), (5), and (6) into () and collecting terms containing u and terms containing v we obtain Since and are solutions of the homogeneous equation (), this reduces to

53 Equation (4) is Solving the above two linear algebraic equations for the unknown functions u' and v', we find u r W r W ', v', (7) where is the Wronskian of and. Clearl, W 0 since, constitute a basis of solutions. Integration of (7) gives Substituting these epressions for u and v into (3), we obtain the desired solution of (), Eample: Solve the differential equation Solution: The related homogeneous equation is 0 and its characteristic equation 0 has i as simple comple conjugate roots. Therefore the functions cos, sin constitute a basis of solutions of the homogeneous equation. Their Wronskian is

54 we thus obtain the following particular solution The corresponding general solution of the differential equation is II. Method of Inverse Operator Consider the nth order linear differential equation with constant coefficients n d n n n d d d k k kn kn X n... n () where k, k,..., kn, are given constants and X is a given function of. The above equation ma also be written as n n n ( D k D kd... kn D kn ) X, () d where D ( ) denotes differentiation with respect to, and n n d D, n n n d n D etc. Equation () is the differential operator form of (). The above takes the form [ f ( D)] X, (3) n n n where f ( D) D k D kd... kn D kn (called a linear differential operator of degree n with constant coefficients).

55 If we define f ( D) X as that function of which does not contain arbitrar constants and when operated upon b f (D) gives X, that is, f ( D) X X, then X satisfies the equation f(d) = f ( D) f ( D) X. Thus X becomes a particular integral of equation (). Obviousl, f(d) and f ( D) operators. are inverse f ( D) Let us find X for some simple differential operators f(d): f ( D) (I) f(d) = D Let D X. (i) Operating on both sides b D, Integrating both sides w.r.t., no constant being added as (i) must not contain an arbitrar constant (b definition of Thus X X. D f ( D) X ). (II) f(d) = D - a Let X... ( ii) D a Operating b D - a,

56 ( D a ). ( ). ( D a) X D a d d or X a, i. e a X which is a linear equation of first order with integrating factor e a and its solution is a a e Xe, no constant being added as (ii) doesn't contain an constant. a a Thus X e Xe. D a Rules for finding the Particular Integral of equation (3) Case : X is an function of Let f ( D) ( D m )( D m )... ( D mn ). Resolve f ( D) into its partial fractions: f ( D) A D m A D m... An D mn The particular integral of (3) = A X = A An [... ] X f ( D) D m D m D mn = A X A X... An X D m D m D m m = A e m Xe m + A e m Xe m m... A e n n Xe n b the application of (II) to each term. n

57 Case : X = a e a a a a n a n a Since De ae, D e a e,..., D e a e, and so on, therefore n n a D... kn e = ( D k ) n n ( a k a... k ) e n a a i. e. f ( D) e f ( a) e a Operating on both sides b, f ( D) Dividing b f (a), a a e e f ( D) f ( a), provided f(a) 0.. () If f(a) = 0, the above rule fails and we proceed further. This means that (D a) is a factor of f (D). Suppose Then i.e. a a e e f ( D) f ( a).. ()

58 If f (a) 0, then appling () again, we get Eample: Find the particular integral of (D +5D + 6) = e Solution: Eample: Find the P.I.. of (D + )(D ) = e - + sinh. Solution: Let us evaluate each of these terms separatel.

59 Case 3: X = sin (a + b) or cos (a + b) Since D sin (a + b) = a cos (a + b), D sin (a + b) = - a sin (a + b), D 3 sin (a + b) = - a 3 cos (a + b), D 4 sin (a + b) = a 4 sin (a + b), we can write D sin (a + b) = (- a ) sin (a + b), (D ) sin (a + b) = (- a ) sin (a + b), and in general Appling the operation l/f (D ) on both sides, Dividing b f (- a )

60 If f(- a ) = 0, the above rule fails and we proceed further. i(a + b) Since cos (a + b) + i sin (a + b) = e [Euler's theorem] and so on. Similarl, and so on. 3 Eample: Find the particular integral of ( D ) cos( ). Solution:

61 Eercise: Find the P.I. of 3 d 3 4 d sin Solution: The given equation in operator form is 3 ( D 4D) sin Case 4: X = m Here Epand [f (D)] - in ascending powers of D as far as the term in D m and operate on m term b term. Since the (m + l) th and higher derivatives of m are zero, we need not consider terms beond D m.

62 Eample: Find the P.I. of Solution: Given equation in operator form is (D + D) = Case 5: X = e a V, V being a function of If u is a function of, then and in general, Operating both sides b /f(d), Now put f(d + a) u = V, that is

63 Eample: Find P.I. of (D -D + 4) = e cos. Solution: Replacing D b D+, we get Cauch-Euler Equation An equation of the form n n d n n n d a... n an d an X where e z coefficients. a i ' s are constants and X is a function of, is called a Cauch-Euler equation. The substitution ( z log ) reduces the above equation to linear differential equation with constant B chain rule, we have d d dz dz d dz, or, d d dz Dz, where Dz d dz, and

64 d d d dz d dz d dz dz d dz d dz dz d d d or, D D [ D ( D )]. z z z z dz dz Proceeding as above we find 3 3 d 3 n n d [ Dz ( Dz )( Dz )],..., and [ Dz ( Dz )( Dz )... ( Dz n )]. n Substituting these in the Cauch-Euler equation, we get a linear differential equation with constant coefficients in the independent variable z that can be solved b the methods discussed earlier. Eample: Solve d d log. Solution: With the substitution e z or z log it follows from the chain rule that d d d d Dz, where Dz, and dz dz Dz ( Dz ). Substituting in the given differential equation and simplifing ields ( D z D z ) z. Since the above equation has constant coefficients, its auiliar equation is 0, ( ) 0. Thus we obtain the complementar function For a particular integral we appl the method of inverse operator: z z h c e cz e.

65 p ( Dz ) ( ) ( )( ) z = ( D z ) z = [ ( Dz ) ( Dz )...] z!! d d z ( z) 3 ( z) terms containing higher powers of D dz z (greater than ) operated on z. dz = z () 3(0) 0 = z. Hence equation is h p = z z h c e cz e + z and so the general solution of the original differential c clog + log. Eercise: Solve [Hint: Multipl throughout b ] Legendre s Linear Equation An equation of the form n n n d n d ( a b) a ( a b)... an ( a n n b) d an X where a i ' s are constants and X is a function of, is called a Legendre s linear equation. The above equation can be reduced to linear differential equation with constant coefficients b the substitution a b e z or z log( a b). Application of chain rule gives d d dz dz d dz a b, b or, (a d b) d b dz bdz, where Dz d dz, and

66 d d a b b d dz ( a b b) d dz a b b d dz dz ( a b b) d dz ( a b b) d dz ( a b b) d dz d dz d or, ( a b) b ( D D ) b [ D ( D )]. z z z z 3 3 d 3 Similarl, ( a b) b [ D ( D )( D )],..., 3 z z z and n n d n ( a b) b [ Dz ( Dz )( Dz )... ( Dz n )]. n These substitutions convert the given Legendre s linear equation to a linear differential equation with constant coefficients that can be solved b the methods alread discussed before. Eample: Solve the following Legendre's linear equation. (3 ) d d 3(3 ) z Solution: Put 3 + = e i.e., z = log(3 + ), so that d ( 3 ) 3Dz, (3 ) d d 3 [ Dz ( Dz )], where D z. dz Substituting these values in the given equation, it reduces to [3 D z ( Dz ) 3 3Dz 36] 3 z e 3 4 z e 3 or z 9( D z - 4) e 3, 3

67 or z ( D z 4) ( e 7 ), which is a linear equation with constant coefficients in the dependent variable z. Its auiliar equation is 4 0 and the roots are and. Thus the complementar function is For the particular integral, we appl the inverse operator method: Hence the general solution (or complete solution) of the original equation is c (3 ) c (3 ) [(3 ) log(3 08 ) ]. Simultaneous Linear Equations with Constant Coefficients Simultaneous ordinar differential equations involve two or more equations containing derivatives of two or more unknown functions of a single independent variable. If,, and z are functions of the variable t, then

68 and are two eamples of sstems of simultaneous differential equations. A solution of a sstem of differential equations is a set of differentiable functions = f(t), = g(t), z = h(t), and so on, which satisfies each equation of the sstem on some interval I. The method that we shall consider for solving sstems is based on the fundamental principle of sstematic algebraic elimination of variables. Eample: Write the sstem of differential equations in operator notation. Solution: Rewrite the given sstem as so that Method of Solution

69 Consider the simple sstem of linear first-order equations D D 3 () or equivalentl, Operating on the first equation in () b D while multipling the second b and then subtracting will eliminate from the sstem. It follows that Since the roots of the auiliar equation are 6, we obtain Whereas multipling the first equation b -3 while operating on the second b D and then adding gives the differential equation for, D - 6 = 0. It follows immediatel that Now (3) and (4) do not satisf the sstem () for ever choice of c, c, c3, and c4. Substituting (t) and (t) into the first equation of the original sstem () gives, after simplifing, Since the latter epression is to be zero for all values of t we must have

70 Hence we conclude that a solution of the sstem must be It can be verified that the same relationship (5) between the constants is obtained if (3) and (4) are substituted into the second equation of (). Eample: Solve Solution: Operating on the first equation b D - 3 and on the second b D and subtracting eliminates from the sstem. It follows that the differential equation for is Since the characteristic equation of this last differential equation is 6 ( )( 3) 0, we obtain the solution Eliminating in a similar manner ields (D + D - 6) = 0, from which we find

71 Substituting (7) and the above into the first equation of (6) gives and so Solving for c 3 and c 4, we get Accordingl, a solution of the sstem is Notes: () Since we could just as easil solve for c 3 and c 4 in terms of c and c, the solution in above eample can be written in the alternative form () It sometimes pas to keep one's ees open when solving sstems. Had we solved for first, then could be found, along with the relationship between the constants, b simpl using the last equation of (6):

72 Eample: Solve Solution: First write the sstem in differential operator notation Then, b eliminating, we obtain Since the roots of the auiliar equation ( 4) = 0 are 0, i,and 3 i, the complementar function is c c cos t c sin t. h 3 Particular integral: p ( t t) = ( t t) = D ( t t) 3 ( D 4D) D 4D 4 4D( ) 4 D ( )( ) D [ ( )...]( t 4D 4! 4 t) = 3 ( t t) dt D( t t) D ( t t) terms containing higher powers (greater than 4) of D 3 t t 4 t 8 8

73 Thus 3 h p c c cos t c3 sin t t t t, () 4 8 where c C 8, is another constant. Eliminating from the sstem (0) leads to Its complementar function is given b [(D - 4) - D(D + )] = t or (D + 4) = - t. h = c 4 cos t + c 5 sin t Particular integral can be obtained b the method of inverse operator. A particular integral is p t 4 8 and so h p = c 4 cost c5 sin t t 4 8 Now c4 and c 5 can be epressed in terms of c and c3 b substituting () and () into either equation of (9). B using the second equation we find, after combining terms, so that and Solving for c4 and c5 in terms of c and c3 gives and Finall, a solution of (9) is found to be

74 Applications of Linear Differential Equations I. Simple Pendulum A heav particle attached b a light string to a fied point and oscillating under gravit constitutes a simple pendulum. Let O be the fied point, l be the length of the string and A be the position of the bob initiall. If P be the position of the bob at an time t, such that arc AP = s and AOP =, then s = l. The equation of motion along PT is to first approimation. The characteristic equation is ( g / l) 0, and we have ( g / l ) i. Thus the general solution is