Contents: V.1 Ordinary Differential Equations - Basics

Transcription

1 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 35 CHAPTER V ORDINARY DIFFERENTIAL EQUATIONS Contents: V. Ordinar Differential Equations - Basics V. st Order Ordinar Differential Equations V.3 Linear Ordinar Differential equations V.4 Power Series Solution Method of Frobeneous V.5 Sstems of st Order Linear ODEs

2 36 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 Engineering design focuses on the use of models in developing predictions of natural phenomena. These models are developed b determining relationships between e parameters of the problem. Usuall, it is difficult to find immediatel the functional dependence between needed quantities in the model; at the same time, often, it is eas to establish relationships for the rates of change of these quantities using empirical laws. For eample, in heat transfer, directional heat flu is proportional to the temperature gradient (Fourier s Law) dt q d where the coefficient of proportionalit is called the coefficient of conductivit. Also, during radiative heat transfer in the absorbing media, the rate of change of total intensit I with distance is proportional to itself (Lambert s Law) di ds I where the coefficient of proportionalit is called the absorptivit of the media. The basic approach to deriving models is to appl conservation laws and empirical relations for control volumes. In most cases, the governing equation for a phsical model can be derived in the form of a differential equation. The governing equations with one independent variable are called ordinar differential equations. Here we will stud the methods of solution of ordinar differential equations.

3 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 37 V.. ODE BASICS Basic Concepts, Definitions, Notations and Classification Differential equation Definition A differential equation is an equation, which includes at least one derivative of an unnown function. Eample a) ( ) d d + ( ) e b) ( ) + sin u(, ) u(, ) c) d) + ( n) F,,,..., ( ) e) u (,t) u(,t) v ODE PDE Differential operator D Order of DE If a differential equation (DE) onl contains unnown functions of one variable and, consequentl, onl the ordinar derivatives of unnown functions, then this equation is said to be an ordinar differential equation (ODE); in a case where other variables are included in the differential equation, but not the derivatives with respect to these variables, the equation can again be treated as an ordinar differential equation in which other variables are considered to be parameters. Equations with partial derivatives are called partial differential equations (PDE). In Eample, equations a),b) and d) are ODE s, and equation c) is a PDE; equation e) can be considered an ordinar differential equation with the parameter t. It is often convenient to use a special notation when dealing with differential equations. This notation called differential operators, transforms functions into the associated derivatives. Consecutive application of the operator D f into its derivatives of different orders: transforms a differentiable function ( ) df ( ) ( ) f D : D f f d D ( ) d f f ( ) d D : f f ( ) n d f ( ) n n D f D :f f n d A single operator notation D can be used for application of combinations of operators. For eample, the operator D ad n + bd implies n n d f ( ) ( ) ( ) ( ) df ( ) Df ad f + bdf a + b n d d The order of DE is the order of the highest derivative in the DE. It can be reflected as an inde in the notation of the differential operator as D ad + bd + c Then a differential equation of second order with this operator can be written in the compact form D F( ) ( n)

4 38 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 Linear operator A differential operator D n is linear if its application to a linear combination of g ields a linear combination n times differentiable functions ( ) f and ( ) ( α f βg) αd f + βd g Dn + n n, α, β R The most general form of a linear operator of n th order ma be written as n n L a D + a D + + a D a n ( ) ( ) n ( ) + n ( ) where the coefficients a ( ) C( ) are continuous functions. i Linear and non-linear DE A DE is said to be linear, if the differential operator defining this equation is linear. This occurs when unnown functions and their derivatives appear as DE s of the first degree and not as products of functions combinations of other functions. A linear DE does not include terms, for eample, lie the following:, etc. If the do, the are referred to as non-linear DE s., ( ) 3,, ln ( ) A linear ODE of the n th order has the form n n L a + a + + a ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) n n ( ) + an ( ) ( ) F( ) where the coefficients a i ( ) and function ( ) F are, usuall, continuous functions. The most general form of an n th order non-linear ODE can be formall written as ( n) F (,,,..., ) which does not necessaril eplicitl include the variable and the unnown function with all its derivatives of order less than n. Homogeneous A homogeneous linear ODE includes onl terms with unnown functions: L n ( ) Non-Homogeneous Normal form A non-homogeneous linear ODE involves a free term (in general, a function of an independent variable): L n F ( ) ( ) A normal form of an n th order ODE is written eplicitl for the n th derivative: ( ) ( n) ( n ) f,,,..., Solution of DE Definition An n times differentiable function ( ) which satisfies a DE ( n) F (,,,..., ) is called a solution of the DE, i.e. substitution of the function into the DE ields an identit. ( ) Satisfies means that substitution of the solution into the equation turns it into an identit. This definition is constructive we can use it as a trial method for finding a solution (guess a form of a solution (which in modern mathematics is often called ansatz), substitute it into the equation and force the equation to be an identit).

5 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 39 Eample Consider the ODE + on I (, ) a Loo for a solution in the form e Substitution into the equation ields ae a + e a ( + ) e a a a divide b e > a + a Therefore, the solution is e. This solution is not necessaril a unique solution of this ODE. Solution Integral Trivial Solution The Solution of the ODE ma be given b an eplicit epression lie in eample called the eplicit solution; or b an implicit function (called the implicit solution integral of the differential equation) g, ( ) If the solution is given b a zero function ( ), then it is called to be a trivial solution. Note, that the ODE in eample posses also a trivial solution. The complete solution of a DE is a set of all its solutions. General Solution Particular Solution Integral Curve The general solution of an ODE is a solution which includes parameters, and variation of these parameters ields a complete solution. Thus, { ce c R}, is a complete solution of the ODE in eample. The general solution of an n th order ODE includes n independent parameters and smbolicall can be written as g (,, c,..., c n ) The particular solution is an individual solution of the ODE. It can be obtained from a general solution with particular values of parameters. For eample, e is a particular solution of the ODE in Eample with c. A solution curve is a graph of an eplicit particular solution. An integral curve is defined b an implicit particular solution. Eample 3 The differential equation has a general solution + c The integral curves are implicit graphs of the general solution for different values of the parameter c To get a particular solution which describes the specified engineering model, the initial or boundar conditions for the differential equation should be set.

6 33 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 Initial Value Problem An initial value problem (IVP) is a requirement to find a solution of n th order ODE ( n) F,,,..., for I ( ) subject to n conditions on the solution ( ) and its derivatives up to order n- specified at one point I : ( ) ( ) ( n ) ( ) n where,,..., n. Boundar Value Problem In a boundar value problem (BVP), the values of the unnown function and/or its derivatives are specified at the boundaries of the domain. For eample, find the solution of + satisfing boundar conditions: a ( ) a ( b) b where a,b on [ a, b] The solution of IVP s or BVP s consists of determining parameters in the general solution of a DE for which the particular solution satisfies specified initial or boundar conditions. Tpes of Boundar Conditions I) a boundar condition of the I st ind (Dirichlet boundar condition) specifies the value of the unnown function at the boundar L : u L f II) a boundar condition of the II nd ind (Neumann boundar condition) specifies the value of the derivative of the unnown function at the boundar L (flu): du d L f III) a boundar condition of the III rd ind (Robin boundar condition or mied boundar condition) specifies the value of the combination of the unnown function with its derivative at the boundar L (a convective tpe boundar condition) du + hu d L f

7 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 33 Uniqueness of solution The solution of an ODE is unique at the point (, ), if for all values of parameters in the general solution, there is onl one integral curve which goes through this point. Such a point where the solution is not unique or does not eist is called a singular point. The question of the eistence and uniqueness of the solution of an ODE is ver important for mathematical modeling in engineering. In some cases, it is possible to give a general answer to this question (as in the case of the first order ODE in the net section.) Eample 4: a) The general solution of the ODE in Eample is { ce,c R } There eists a unique solution at an point in the plane b) Consider the ODE The general solution of this equation is { c,c R} (,) is a singular point for this ODE

8 33 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 Differentiale Equations with Maple:. Solution curves - graph of the general solution g(,c) c*ep(-) (Eample 4a) > restart; > g:c*ep(-); > p:{seq(g,c-..)}; g : c e ( ) p : {, e ( ), e ( ), e ( ), e ( ) } > plot(p,-..,colorblac);. General solution of the Differential Equation > restart; > DE:diff((),)+(); > dsolve(de); DE : d + d ( ) ( ) ( ) _C e ( ) 3. Solution of IVP: > restart; > DE:diff((),)+(); DE : d + d ( ) ( ) > Solution:dsolve( {DE, ()3}, ()); > assign(solution); > plot((),-..); Solution : ( ) 3 e ( )

9 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, Solution of IVP > restart; > with(detools): > DE:diff((),)+(); DE : d + d ( ) ( ) > DEplot(DE,(), -..,[[()]],-..,stepsize.5,colorblue); 5. Solution of IVP: > restart; > DE:()*diff((),)-; DE : ( ) d d ( ) > Solution:dsolve( {DE, ()3}, ()); > assign(solution); > plot((),-5..); Solution : ( ) + 9

10 334 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 V. FIRST ORDER ODEs Contents:. Eact ODEs. Equation Reducible to Eact Integrating Factor 3. Separable Equations 4. Homogeneous Equations 5. Linear st Order ODEs 6. Special Equations 7. Approimate and Numerical Methods for st Order ODEs 8. Equations of Reducible Order 9. Orthogonal Trajectories. Eercises In this section we will consider the first order ODE, the general form of which is given b General form F (,, ) normal form f (, ) This equation ma be linear or non-linear, but we restrict ourselves mostl to equations which can be written in normal form (solved with respect to the derivative of the unnown function): or in the standard differential form: standard differential form M (, ) d + N (, ) d Note that the equation in standard form can be easil transformed to normal form and vice versa. If the equation initiall was given in general form, then during transformation to normal or standard form operations (lie division or root etraction) can eliminate some solutions, which are called suppressed solutions. Therefore, later we need to chec for suppressed solutions. Initial Value Problem Initial value problem (IVP) for the first order ODE is formulated as find a solution ( ) of the differential equation F (,, ), ( a,b) subject to the initial condition at ( a,b) : ( ),

11 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, The question of eistence and uniqueness of the solution of an IVP for a first order ODE can be given in the form of sufficient conditions for equations in normal form b Picard s Theorem: Picard s Theorem (eistence and uniqueness of the solution of IVP) Let the domain R be a closed rectangle centered at the point (,) : R {(, ) : a, b} and let the function ( ) f, be continuous and continuousl differentiable in terms of the function in the domain R: f, C R ( ) [ ] f (, ) C[ R] and let the function ( ) f (, ) M for ( ) R Then the initial value problem f, f, be bounded in R:,. ( ) ( ) has a unique solution ( ) in the interval I { h} :, where h min a, b M This theorem guarantees that under given conditions there eists a unique solution of the IVP, but it does not claim that the solution does not eist if conditions of the theorem are violated. Now we will consider the most important methods of solution of the first order ODE.

12 336 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 V.. EXACT ODEs Consider a first order ODE written in the standard differential form: M (, ) d + N(, ) d, (,) D If there eists a differentiable function f (, ) such that f (, ) M (, ) () () f (, ) N(, ) (3) for all (, ) D, then the left hand side of the equation is an eact differential of this function, namel f f df d + d M, d + N, and the function f (, ) satisfing conditions () and (3) is said to be a potential function for equation (). The equation in this case is called an eact differential equation, which can be written as df (, ) (4) direct integration of which ields a general solution of equation (): eact differential ( ) ( )d ( ) c f, (5) where c is a constant of integration. The solution given implicitl defines f,. integral curves of the ODE or the level curves of the function ( ) Eample The First order ODE 3 d + d is an eact equation 3 with the general solution f (,) + c. Then the integral curves of this equation are To recognize that a differential equation is an eact equation we can use a test given b the following theorem (the proof is also a method of solution): Test on eact differential Theorem Let functions M (, ) and ( ) differentiable on N, be continuousl D, then the differential form M (, ) d + N (, ) d (6) is an eact differential if and onl if M N in D (7) Proof: ) Suppose that the differential form is eact. According to the f, such that definition, it means that there eists a function ( ) f (, ) f M (, ) (, ) and N(, ). Then differentiating the first of these equations with respect to and the second one with respect to, we get

13 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, (, ) M ( ) (, ) N( ) f, f, and. Since the left hand sides of M N these equations are the same, it follows that. M N ) Suppose now that the condition holds for all (, ) D. To show that there eists a function f (, ) which produces an eact differential of the form (6), we will construct such a function. The same approach is used for finding a solution of an eact equation. We are looing for a function f (, ), the differential form (6) of which is an eact differential. Then this function should satisf conditions () and (3). Tae the first of these conditions: f (, ) M (, ) and integrate it formall over variable, treating as a constant, then f (, ) M (, ) d + ( ) (8) where the constant of integration depends on. Differentiate this equation with respect to and set it equal to condition (3): f (, ) d ( ) ( ) M, d + N (, ) d Rearrange the equation as shown d( ) N(, ) M (, ) d d Then integration over the variable ields: ( ) N(, ) M (, ) d d + c Substitute this result into equation (8) instead of ( ) f (, ) M (, ) d + N(, ) M (, ) d d + c (9) To show that this function satisfies conditions () and (3), differentiate it with respect to and and use condition (7). Therefore, differential form (6) is an f, constructed in equation (9). eact differential of the function ( ) The other form of the function ( ) f, can be obtained if we start first with condition (3) instead of condition (): f (, ) N(, ) d + M (, ) N(, ) d d + c () Note, that condition (7) was not used for construction of functions (9) or (), we applied it onl to show that form (6) is an eact differential of these functions. Then according to equation (5), a general solution of an eact equation is given b the implicit equations: f, M, d N, M, d d c ( ) ( ) + ( ) ( ) () or f (,) N (,) d + M (,) N (,) d d c ()

14 338 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 The other form of the general solution can be obtained b constructing a function with help of a definite integration involving an arbitrar point ( ), in the region D : f ( ) M ( t, ) dt + N(, t) dt, c (3) f ( ) M ( t, ) dt + N(, t) dt, c (4) Formulas () and () or (3) and (4) are equivalent the should produce the same solution set of differential equation (), but actual integration ma be more convenient for one of them. Eample Find a complete solution of the following equation Solution: ( 3 + ) d + ( + 3) d Test for eactness: M N 3 3 the equation is eact We can appl Equations (-4), but in practice, usuall, it is more f, as in convenient to use the same steps to find the function ( ) the derivation of the solution. Start with one of the conditions for the eact differential f (, ) M (, ) ( 3+ ) Integrate it over, treating as a parameter (this produces a constant of integration ( ) depending on ) f (, ) ( ) Use the second condition for the eact differential f (,) N (,) 3+ ( ) ( ) Solve this equation for ( ) ( ) neglecting the constant of integration. The function is completel determined and the solution of the ODE is given b f (, ) c General solution: + 6 +

15 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, Eample 3 Find a complete solution of the following equation ( ) ( ) + + d + + d Solution: Test for eactness: M N the equation is eact Use for the solution Equation (4) with f (, ) ( ) + ( ) M t, dt N,t dt M ( t,) dt ( ) + ( ) N,t dt N (,t ) M ( t, ) t + dt + t + dt ( ) 3 t t + t + t + 3t Then the general solution is given b the implicit equation: c 3 Setch the solution curves:

16 34 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 V... EQUATIONS REDUCIBLE TO EXACT INTEGRATING FACTOR Integrating factor In general, non-eact equations, which possess a solution, can be transformed to eact equations after multiplication b some nonzero function µ (, ), which is called an integrating factor. Theorem 3 The function (, ) equation M (, ) d + N(, ) d if and onl if µ (, ) µ is an integrating factor of the differential satisfies the partial differential equation M µ µ N M N µ But it is not alwas eas to find this integrating factor. There are several special cases for which the integrating factor can be determined: M N ) M N N ) f ( ) The test for eactness. The integrating factor µ (, ) The test for eactness fails but the given ratio is a function of onl. Then the integrating factor is µ ( ) e 3) M N + The test for eactness fails but the given ratio is a function of onl. Then the g( ) M integrating factor is ( ) µ e f ( ) d g( ) d 4) M N The test for eactness fails but the given ratio is a function of the product of and h( ). Then the integrating factor is N M µ (, ) h( ) d ( ) 5) M N The test for eactness fails but the given ratio is a function of the ratio. M + N Then the integrating factor is µ (,) d

17 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 34 Eample 4 Find a complete solution of the following equation ( + ) d d Test for eactness: M N equation is not eact Test for integrating factor: M N ( ) N µ ( ) e f ( ) d e d ( + ) d d f + + e ln f ( ) e ln int.factor b Eq. ) f ln + ( ) General solution (implicit): f + c + ln c c e Chec if is a suppressed solution: Suppressed solutions d ( ) + (es, satisfies this equation) d If the given differential equation is reduced to standard differential form M (, ) d + N(, ) d with some algebraic operations, then zeros of the epressions involved in division can be solutions of the differential equation not included in the general solution. Such lost solutions are called suppressed solutions. If such operations were applied for the transformation of the differential equation, then the equation has to be checed for suppressed solutions. To chec if a or ( ) is a suppressed solution of Equation (), reduce the differential equation to normal form with as a dependent variable d d (, ) (, ) M and substitute into it a. N To chec if b ( ( ) ) is a suppressed solution of Eq., reduce the differential equation to normal form with as a dependent variable d d (, ) (, ) N and substitute into it b. M The suppressed solutions should be added to the general solution.

18 34 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 V..3. SEPARABLE EQUATIONS Separable equation Definition A differential equation of the first order is called separable if it can be written in the following standard differential form: ( ) M ( ) d + N ( ) N ( ) d M () where M ( ) N( ) ( ) N ( ), are functions of the variable onl and M, are functions of the variable onl. Assuming that N ( ) and ( ) in equation () can be separated b division with M ( ) N( ) M ( ) N ( ) d + d N ( ) M ( ) M for all and in the range, variables : Then equation () can be formall integrated to obtain a general solution: () M N ( ) ( ) N d + M ( ) ( ) d c (3) where c R is an arbitrar constant. Note, that separated equation () is eact - it can be obtained from equation () b multiplication b the integrating factor µ. N( ) M ( ) The potential function for this equation is f (, ) M ( ) d + N ( ) ( ) ( ) N d M Which ields the same general solution f ( ) c Because of division b ( ) N ( ),. M, some solutions can be lost; therefore, equations should be checed for suppressed solutions. If, where R belongs to the domain and is a root of N ( ), then the function is obviousl a solution of differential equation (). Similarl, if is a real root of M ( ), then the function is also a solution. The both should be added to the general solution (3). Eample 5 Find a general solution of the following ODE: + + ln, > ( ) ( + ) ln d + d ln d + d + ( ln ) + ln( + ) c There are no suppressed solutions.

19 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, Eample 6 Find a solution of the following ODE: ( ) 4 cot Solution: Separate variables: d tan d 4 Integrate: ( ) ln cos + ln 4 lnc General solution: ( 4 ) cos c Chec for suppressed solutions: π + nπ are suppressed solutions. d ± are solutions of ( ) 4 + ( tan ) d if independent and dependent variables are reversed. Then the famil of solution curves is represented b 3π π π 3π

20 344 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 V..4. HOMOGENEOUS EQUATIONS In this section, we will stud a tpe of equations which can be reduced to a separable equation or to an eact equation. Homogeneous function Function ( ) M, is homogeneous function of degree r, if r (, λ) λ M (, ) It means that after replacing b the parameter M λ for an λ > λ and b r λ can be factored from the epression. λ in the function M (, ), Eamples 7: a) Homogeneous function of degree zero. Let M (, ), then + M Therefore, ( ) λ λ λ + λ λ for λ > + ( λ, λ) M (, ) M (, ) M, is homogeneous of degree zero. If we divide the numerator and the denominator b, then u M (,) + + u and we see that the function M (, ) depends on a single variable u. It appears to be a fact for zero degree homogeneous functions: The function (, ) M is homogeneous of degree zero if and onl if it depends on a single variable u : M, f u ( ) ( ) b) A more general fact: homogeneous functions of degree r can be written as r M (,) f or r M (,) g > To show it, choose parameters of the form λ < M, c) Consider ( ) M 3. Test on homogeneit ields ( λ, λ) ( λ) ( λ) λ λ λ M (, ) Therefore, the given function is homogeneous of degree 3.

21 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, Homogeneous equation Reduction to separable A differential equation written in standard differential form (, ) d + N(, ) d M is called a homogeneous differential equation if the functions ( ) N (, ) are homogeneous of the same degree r. If the equation written in standard differential form M, and (, ) d + N(, ) d M is homogeneous, then it can be reduced to a separable differential equation b one of the change of variable: u d du + ud or if u then d du + ud from the first ever paper on calculus NOVA METHODUS PRO MAXIMIS ET MINIMIS b Gottfried Wilhelm Leibniz A new method for finding maima and minima, and liewise for tangents, and with a single ind of calculation for these, which is hindered neither b fractions nor irrational quantities. Acta Eruditorium, 684, pp v d vd + dv Both approaches are equivalent, just because in standard differential form the variables are equivalent. But actual integration of the equation ma be more convenient with one of them. Justification: First appl the substitution to the differential equation (,u) d + N(,u) du + N(,u) ud M u and divide it formall b N (,u)d M (,u) du + + u N(,u) d If the differential equation is homogeneous then the functions ( ) (, ) M, and N are homogeneous of the same degree r and, according to Eample 7b, can be written as r r M (,) f f( u) r r N(,) f f( u) Substitute them into the previous equation, then f f ( u) ( u) du + + u d Now the variables can be separated d + f f du + u ( u) ( u) Formall this equation can be integrated to a general solution ln + f f du c + u ( u) ( u) where c is a constant of integration. The solution of the original equation can be obtained b bac substitution u.

22 346 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 Eample 8 Solve the differential equation ( ) + d + d Solution: M and N are homogeneous functions of degree. Use change of variable: u d du + ud ( u + ) d + u( du + ud ) 3 ( u + + u ) d + u du 3 ( u ) d + u du + separable d udu + u + d d + u ln 4 4 ( u + ) + ( u + ) ln c + ln general solution ( u ) c + bac substitution ( + ) c Chec for suppressed solutions: is also a solution > f:{seq(^*(^+^)i/8,i..)}: > implicitplot(f,-..,-5..5);

23 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, Change to polar coordinates Reduction of the homogeneous differential equation to a separable equation b transition to polar coordinates: This method is convenient when the solution is represented b complicated transcendental functions which are more suitable for representation in polar coordinates (ellipses, spirals, etc). Conversion formulas from Cartesian to polar coordinates: r cosθ + r d dr + dθ cosθdr r sinθdθ r θ r sinθ tanθ d dr + d θ sin θ dr + r cos θ d θ r θ Eample 9 Solve the differential equation ( ) d ( ) d Solution: It is a homogeneous equation of order. Reduce it to a separable equation b transition to polar coordinates: ( ) d + ( ) d ( θ θ)( θ θ θ) ( θ θ)( θ θ θ) r sin r cos cos dr r sin d + r cos r sin sin dr + r cos d ( ) ( ) ( cos θ sin θ ) + dθ sinθcosθ dr + r cos θ sin θ dθ separable equation dr r sinθ cosθ separate variables ( sinθ cosθ ) dr d + r sinθ cosθ integrate ln r ln sinθcosθ lnc +, c general solution General solution: r c, sinθ cosθ c equation of ellipse in polar coordinates > f:{seq(i/sqrt(-sin(t)*cos(t)),i..4)}: > polarplot(f,t..*pi,-5..5);

24 348 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 V..5. Linear st order ODE The general form of the st order linear differential equation is given b: ( ) + a ( ) f ( ) L a D () We can rewrite this equation in the standard form, if we divide it b a ( ) a ( ) f ( ) + provided ( ) a ( ) a ( ) a for all Then for simplicit, coefficients ma be renamed, and the equation becomes a Standard form + P( ) Q( ) where ( ) ( ) P and Q( ) a ( ) ( ) ( ) f () a Initial value problem For the first order ODE, an initial value problem (IVP) is formulated in the following wa: Solve the equation + P( ) Q( ) subject to the condition ( ), D In other words, we need to find a particular solution of differential equation () which goes through the given point (,). General solution We will tr to find a solution of the linear equation with the help of the method which we have alread studied (integrating factor) and to do that, we transform equation () into standard differential form [ P ( ) Q( ) ] d + d (3) from which we can identif the coefficients of the standard differential form as M (, ) P( ) Q( ) and N (, ) Chec this equation for eactness: M N φ P( ), if P( ) then the equation is not eact. From the test for an integrating factor φ P( ) N (function of onl), it follows that the integrating factor is determined b the equation P( ) d ( ) e Multiplication of our equation b the integrating factor ( ) µ (4) eact equation ( )[ P( ) Q( ) ] d + µ ( ) d µ transforms it to an µ (5) Following the nown procedure, we can find a function (, ) differential form (5) is an eact differential: f µ ( ) [ P( ) Q( ) ] f µ ( ) P ( ) Q( ) d + ( ) f µ + µ ( ) P( ) Q( ) d ( ) ( ) f for which

25 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, ( ) P( ) d + ( ) µ ( ) µ ( ) µ ( ) ( ) P( )d µ ( ) µ ( ) µ ( ) P( )d ( ) [ P( ) Q( ) ] d ( ) f µ + f f µ ( ) [ P( ) Q( ) ] d + µ ( ) ( ) P( )d µ µ ( ) ( ) Q( )d µ µ ( ) µ ( ) Q( ) d c Solving this equation with respect to, we end up with the following general solution: general solution c ( ) + µ ( ) µ ( ) Q( )d µ (6) We see that the solution of a first order linear differential equation is given eplicitl and ma be obtained with this formula provided that integration can be performed. The same result ma be obtained, if we show first that the differential equation multiplied b the integrating factor ma be written in the form d d ( µ ) µ Q then after direct integration (from inspection, up with the same general solution. µ is a function of onl) we end In a case of an equation with constant coefficients, the integrating factor ma be evaluated eplicitl µ P( ) d ad a e e e ( ) And the solution becomes a a a ce + e e Q( )d (7) Solution of IVP Using initial condition ( ), we can determine the constant of integration directl from the general solution. In another more formal approach, we can chec b inspection that ( ) µ ( ) + µ ( ) ( ) Q( ) µ µ d (8) is a solution satisfing the initial condition. For an equation with constant coefficients, the solution of the IVP is given b ( ) a a + e e Q( ) a e d (9)

26 35 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 Eample (First order linear ODE with variable coefficients) Find a general solution of equation ( cot ) sin + and setch the solution curves. Solution: The integrating factor for this equation is µ ( ) e cot d ln sin e sin then a general solution is c sin sin + sin( ) sin( )d c sin + sin + sin sin c sin sin c sin + 3 sin ( ) sin( ) cos( )d ( ) d sin( ) In Maple, create a sequence of particular solutions b varing the constant c, and then plot the graph of solution curves: > ():*sin()^/3+c/sin(); sin( ) 3 ( ) : + c sin( ) > f:{seq(subs(ci/4,()),i-..)}: > plot(f,-*pi..*pi,-5..5); π π

27 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 35 Eample An Initial Value Problem for an equation with constant coefficients Solve the equation + sin subject to the initial condition: ( ) Solution: Appling equation (9), we obtain the solution of the IVP: e e e e e ( ) a a + e e Q( ) a d ( ) + e e sin( ) + e + e + e e sin ( ) d d + e sin e sin cos + cos Use Maple to setch the graph of the solution: > : ep()+ep(-)/+(sin()-cos())/; : e + + e ( ) sin( ) cos( ) > plot(,-..,colorblac);

28 35 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 V..6. SPECIAL EQUATIONS Some first order non-linear ODE s which do not fall into one of the abovementioned tpes can be solved with the help of special substitution. These equations arise as a mathematical model of specific phsical phenomena, and the carr the names of mathematicians who first investigated these problems.. Bernoulli Equation Definition The differential equation which can be written in the form ( ) Q( ) n + P () where n is a real number is called a Bernoulli equation. Jacques (Jacob) Bernoulli ( ) n lim + e n n If n or n, then the equation is linear and it can be solved b a corresponding method, otherwise the Bernoulli equation is a non-linear differential equation. B the change of dependent variable n z n z () The non-linear Bernoulli equation ( n ) can be reduced to a linear first order ODE. Indeed, the derivative of the function can be epressed as n d d dz n n n z z z z (3) d d n d n Substitution of () and (3) into equation () ields n n n z z + P n n n ( ) z Q( ) z Dividing this equation b n n z and multipling b n, we end up with ( n) P( ) z ( n) Q( ) z + (4) Equation (4) is a linear ODE, the general solution of which can be found with a nown method (see section 4). Then solution of the Bernoulli equation is determined b bac substitution n z (5) It is eas to see that the Bernoulli equation possesses also a trivial solution when n is positive, n >. Eample (Bernoulli equation) Solution: Find a general solution of the equation + Use a change of variable linear equation z + z n 3 3 which ields a z z z 3 The integrating factor for this equation is µ e, and then the general solution is

29 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, ce z ce + e e d Bac substitution results in the general solution of the initial equation 3 3 z ce + 3 Solve it for, then in eplicit form, the general solution is determined b the following equation 3 3 ce + 3 One more solution of the given equation is a trivial solution. Use Maple to setch the solution curves: > ():(c*ep(-/3)+-3)^3; ( ) : ( c e ( / 3 ) + 3) > f:{seq(subs(ci/4,()),i-6..6)}: > plot(f,-5..6,-5..,colorblac); 3. Riccati equation Definition A differential equation which can be written in the form ( ) + Q( ) R( ) P + (6) is called a Riccati equation. If one particular solution of the Riccati equation is nown, then it can be reduced to a first order linear ODE: Jacopo Riccati ( ) Theorem 4 Suppose that ( ), Q( ), R( ) C[ D] functions on D. Then if the function ( ) P, D are continuous u, D is a solution of the Riccati equation (6) in D, then the substitution ( ) u( ) + (7) z( ) for all D for which z ( ) transforms the Riccati equation (6) into the st order ODE: [ P( ) u( ) + Q( ) ] z + P( ) z + (8) Remars: - It does not matter how simple the particular solution u( ) - For an equation with constant coefficients, this particular solution can be found as a constant (stead state solution). is.

30 354 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 B the other substitution, the Riccati equation can be reduced to a linear ODE of the nd order: Theorem 5 Suppose that ( ), Q( ), R( ) C[ D] P, D are continuous functions on D. Then the substitution w ( ) ( ) (9) P( ) w( ) for all D for which P ( ) and w ( ) transforms the Riccati equation (4) into a second order ODE: P ( ) w + Q( ) w + R( ) P( ) w () P( ) Eample 3 (Riccati equation with a nown particular solution) Find a general solution of the equation 3 Solution: Given that the equation has two obvious particular solutions: and 3 Choose the first one of them for substitution (7): z Identif coefficients of the Riccati equation: P Q R 3 Then the corresponding linear equation (8) is z 4z The general solution of this first order linear ODE is z ce + e e ( ) d ce + 4 Then the solution of the given Riccati equation becomes z 4 ce + 4 Use Maple to setch the solution curves: > p:{seq(/(i*ep(4*)/+/4)-,i-..)}: > plot(p,-..,-4..8,colorblac,disconttrue); 3

31 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, Clairaut equation Definition A differential equation which can be written in the form ( ) + f () is called a Clairaut equation. Aleis Clairaut ( ) The general solution of a Clairaut equation is given b: ( c) c + f () This can be confirmed b a direct substitution into the Clairaut equation. The Clairaut equation additionall ma include a particular solution given in parametric form: f ( t) f t tf t (3) ( ) ( ) Eample 4 Solve ( ) This equation belongs to the Clairaut tpe. Therefore, the general solution of the equation is given b the one-parameter famil c c Chec if the parametric solution (3) is also a solution of this equation: t t + t t Which can be reduced to an eplicit equation b the solution of the first equation for t and substitution into the second equation: 4 This solution defines a (limiting curve) for the famil of curves from the general solution: > p:{seq(c*-c^,c-..)}: > g:plot(p,-..,-..,colorred): > g:plot(^/4,-..,..,colorblue): > displa({g,g}); 4

32 356 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 4. Lagrange equation Definition A differential equation which can be written in the form ( ) + f ( ) g (4) is called a Lagrange equation. Note, that the Clairaut equation is a particular case of a Lagrange equation when g. ( ) Joseph-Louis Lagrange ( ) Appl the substitution v ( v) f ( v) g + (5) dv Differentiate the equation v g( v) + g ( v) + f ( v) d d dv dv v g( v) Solve this equation for d d g ( v) + f ( v) d g ( v) + f ( v) Invert the variables: dv v g( v) This equation is a linear equation for ( v) as a function of an independent variable v d g ( v) f ( v) dv v g( v) v g( v) The general solution can be obtained b integration to determine ( v c) F,, c (6) To determine a general solution of the Lagrange equation (4), use equation (5) to eliminate v (if possible) from equation (6) to get (,, c) dv ϕ (7) Otherwise, the variable v can be used as a parameter in the parametric solution organized from equation (6) and equation (5) which is replaced from equation (6): F v, c c ( ) ( v c) g( v) f ( v) F, + v Z (8) Eample 5 (Lagrange equation ) d d Find a general solution of d d General solution: ( 7 3c) ( + 4 ) ( 7 3c)( + 4 ) Plot the solution curves with Maple: > p:{seq((7**-3*i/)^*(*+4*^)-**(7**- 3*i/)*(*+4*^),i-4..4)}: > implicitplot(p,-.., -3..3,numpoints,colorblac);

33 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, Equations solvable for ( ) f, (9) This tpe of equations is a further generalization of Clairaut and Lagrange equations. Appl the substitution v ( v) f, differentiate with respect to dv ϕ, v, or d dv v ϕ, v, d dv d this equation ma be solvable for or to get a general solution d dv F Then if from the two equations (, v, c) f ( v) (, v, c), F () v can be eliminated, then it ields an eplicit general solution ( c), and if v cannot be eliminated, then the sstem of equations () can be considered as a parametric solution of equation (9) with parameter ν for fied values of the constant of integration c.

34 358 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 V..7. APPROXIMATE AND NUMERICAL METHODS FOR st ORDER ODEs. Direction field Consider a first order ODE written in normal form: f (, ) () Suppose that this equation satisfies conditions of Picard s Theorem in some domain D. Then for an point (, ) D ~ D there eists onl one solution curve which goes through this point; and equation () defines the slope of a tangent line to the solution curve at this point: So equation () gives us a wa to determine the direction of tangent lines to solution curves even without solving the equation. We can use it for visualization of the solution curves of the differential equation. Create a grid in D as a set of points (, ). At each point of the grid setch a small segment with a slope given b equation (). The obtained picture is called a direction field (or slope field) of the ODE. It gives us a general view on the qualitative behavior of solution curves of the ODE. In Maple, the direction field of an ODE is generated b the command DEplot in the pacage DEtools: > de:diff((),)*()-s*()^; d de : d ( ) ( ) ( ) > DEplot(de,(),..5,..4); actual solution curves can be added b specifing the initial conditions: > DEplot(de,(),..5,{[,.],[,.5],[,3.5]},..4);

35 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, Isoclines Isoclines of equation () are curves at each point of which, the slope of the solution curves is constant c f (, ), c () So, isoclines are curves defined b the implicit equation (). In the previous eample of the logistic equation, the function does not depend on, and isoclines are the straight lines parallel to the -ais: The approimate method of solution based on application of the direction field is called the method of isoclines. It consists in the construction of a direction field using isoclines and then drawing approimate solution curves following the direction segments.. Euler s method The direction field concept helps us to understand the idea of the Euler method, in which we use equation () to determine the slope of tangent lines to the solution curve step b step and construct an approimate solution curve of the IVP: f ( ) ( ), The solution is calculated at discrete points,,,,. For the grid with step size h, the nodes are determined b + h,,, At the point the solution is given b the initial condition ( ) Then we calculate the slope of the tangent line to the solution curve at the point, f, +. If we consider it to ( ) and draw a tangent line ( )( ) be an approimate solution for the interval [, ], the approimation is given b f (, )( ) + + h f (, ), then at the net point Now the approimate value is nown, we can calculate the slope of the, and draw a tangent tangent to the solution at the point ( ) f (, )( ) + from which the net approimation can be determined + h f (, ) Continuing this process, we get for point, that h f (, ) +

36 36 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7, we proceed following the direction field of the differential equation to get an approimate,. Starting from the point specified b the initial condition (, ) solution curve which is a piece-wise linear curve connecting points ( ) Euler s Method The algorithm for Euler s Method can be summarized as follows: Starting with the initial condition ( ) function ( ),,,, calculate the values of unnown following the eplicit algorithm:,,,...,,..., h h f (, ) (3) +,, The accurac of Euler s Method depends on the character of variation of the solution curve and the size of steps h. It can be shown that when step size h goes to zero, Euler s approimation approaches the eact solution. But it can easil deviate from the eact solution for coarse steps. If we want an accurate solution, then step-size should be ver small. It maes the Euler method a time consuming one. Some improvement can be made, in increasing the efficienc of approimation. In the modified Euler s Method the average slope of the tangent line between steps is taen into account: h Modified Euler s Method ~ h f (, ) + h ~ [ ( ) + f (, )] + f, (4),, Further improvement can be obtained b taing into account the slope of the tangent line to the solution at the intermidiate points. Depending on the number of intemidiate steps these methods are called Runge-Kutta methods of different orders. The most popular is the Fourth Order Runge-Kutta Method. Its algorithm for regular step-size h, is traditionall written in the following form: h n n n 4 th Order Runge-Kutta Method hf (, ) n n n h hf n +, n + h 3 hf n +, n + (5) hf + h, + ( ) 4 n n 3 n n n,, ( + + ) 4

37 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, Picard s Method of Successive Approimations f ( ) ( ), (, ) + f d,, (6) 4. Newton s Method (Talor series solution) We assume that the solution of the IVP for the first order differential equation in normal form f (, ) ( ) can be obtained in the form of a Talor s series ( ) ( ) ( ) + ( )( ) + ( ) +! For this epansion we need to determine the values of the unnown function and its derivatives at the point : from initial condition ( ) And b substitution and into the equation and determining the derivative ( ) f (, ( )) f (, ) to obtain values of the higher derivatives, differentiate consecuativel the equation as an implicit function and substitute d and d ( ) f (, ) d d ( ) f (, ) Solution with a finite number of terms (truncated series) is an approimate solution. In the obtained approimate solution, sometimes a Talor series epansion of a nown function can be identified, then the corresponding Talor series is an eact solution. Newton s method can be applied also and for higher order differential equations.

38 36 Chapter V ODE V. ODE Basics Eample 6 V. First Order Ordinar Differential Equations September 4, 7 Use Newton s Method to solve the following IVP + ( ) ( ) Solution: The value of the function and first derivative are alread nown. From the differential equation: ( ) ( ) Differentiate the ODE and substitute and : ( ) ( ) iv iv ( ) ( ) j j j ( ) ( ) j + j j + ( ) ( ) j j Then the Talor s series can be constructed as ( ) + ( ) + 3!! j j + ( ) j ( ) + ( ) ( j )! ( j + )! ( ) + ( ) j ( ) j Where the Talor series epansion of trigonometric functions can be recognized: ( ) cos ( ) + sin ( )

39 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, V..8. EQUATIONS OF REDUCIBLE ORDER. The unnown function does not appear in an equation eplicitl The general form of these equations which are solved for the second derivative is: f, () ( ) This equation can be reduced to a st order ODE b the change of dependent variable v () then v (3) and substitution into the equation ields a st order ODE for the new function v. Eample 7 Solve the following nd order differential equation + The dependant variable v is missing in this equation. Then substitutions (-3) ield v + v which is a first order linear ODE with constant coefficients. The general solution can be obtained b variation of parameter (with an integrating factor µ ( ) e ): v ce ce + e + e e d ( e e ) ce + Then substitution into equation () ields the first order ODE for the unnown function ce + which can be solved b direct integration c e + + c Solution curves can be setched with the help of Maple: > f:{seq(seq(i*ep(-)+^-+j,i-..),j-..)}: > plot(f,-..,-..,colorblac);

40 364 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 This approach can be applied for reduction of order of more general equations. Thus, an ODE of order n ( n) ( n ) ( ) F,,,, (4) ( ) in which the unnown function and its first derivatives are missing, b the change of variable ( ) v ( + ) v (5) ( n ) ( n) v is reduced to an ODE of order n : ( n ) ( n ) F,v,,,v (6) ( ). The independent variable does not appear in the equation eplicitl (autonomous equation) The normal form of these equations is: (, ) f (7) Such equations in which the independent variable does not appear eplicitl, are called autonomous equations. These equations can be transformed to st order ODE s b the change of the dependent variable to v (8) and then in the resulting equation consider to be the independent variable and v to be the dependent variable. These transformations of the given ODE wor as follows: ) epress derivatives of in terms of a new function v : v d d dv d dv d d d d d ) substitution into equation (7) ields a st order ODE ( ) v v v v (,v) v v f (9) 3) find (if possible) a general solution of equation (9) and write it in the form where it is solved for the function v (the general solution should include one parameter c ): v F(,c ) () 4) using bac-substitution (8), set up the equation for the unnown function F(,c ) which formall can be solved b separation of variables d d F,c ( ) d + c () F(,c ) Equation () is an implicit form of the general solution of equation (7). It also with as the independent variable: can be written as an eplicit function ( ) d + c () F (,c )

41 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, Eample 8 Solve the following nd order ODE Solution: Substitution (8) ields v v v v v ( ) from which we have two equations: v v The first equation immediatel leads to the solution c c R The second equation is a st order ODE with the general solution v + c Bac-substitution gives the equation for + c which is a separable equation d d + c Depending on the sign of the constant c, integration ields the following solutions: a) for c c tan c + c c > b) for c c c ln + c c + c < c,c R c,c R c) for c + c c,c R recall also the solution d) c c R

42 366 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 Eample 3 (outer-space radiator, [Siegel&Howell, ThermalRadiation Heat Transfer, 3 rd Ed., p.46] with different solution) q cond,in du ( aw ) d qrad 4 ( W ) εσ u du ( aw ) d qcond,out + Ecessive heat from space ships can be released onl b radiating it from the surfaces eposed to outer-space which is assumed to be at zero absolute temperature. The schematic of one section of a radiator is shown in the Figure. Fluid heated inside of the ship to the temperature u enters pipes connected b fins of thicness a and width L. Fins are from material with thermal conductivit and total surface emissivit ε. Determine the stead state temperature distribution in the fin. Assumptions for the phsical and mathematical model describing heat transfer in the fins: u u ; temperature varies onl in the -direction ( ) the ends of the fin attached to the pipes are at temperature u ; the fin surface is not eposed to direct sun radiation; because of the smmetr, there is no heat flu at the middle of the plate: du d L Energ balance for the control volume ( a W ): du d 4 ( aw ) ( W ) εσu + du d at the limit ields a governing equation for temperature distribution: d u bu d 4 b εσ a (,L) subject to boundar conditions: ( ) u u du d L The equation is a non-linear nd order ODE. This is an autonomous equation which can be reduced to the st order equation b the change of variable u v u vv Then the equation becomes 4 dv vv bu where v du Separate variables 4 vdv bu du and integrate to get a general solution v b 5 u + c 5 du d Appl the second boundar condition v[ u( L) ] ( L) L and notation u L u for the fin s midpoint temperature to determine the constant of integration b 5 u L + c 5

43 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, Then v b 5 ( u u ) 5 L Because for the interval (,L) temperature of the fin is decreasing and v is in a direction of the temperature gradient, then the previous equation ields b v ( u u ) 5 L 5 which is followed with the bac-substitution to du b ( u u ) 5 L d 5 and after separation of variables du d b 5 ( u u L ) 5 Definite integration of this equation for the change of temperature from ( ) u when the space variable changes from to, ields u to u( ) du ( ) u b 5 ( u u L ) 5 This is an implicit equation for the value of the temperature at. The value of the midpoint temperature u L can be determined from the solution of the equation ul du L u b 5 ( u u L ) 5 which can be solved numericall. Then for fied values of the coordinate temperature values ( ) from the numerical solution of equation ( ). u can be found Consider the particular case with the following values of parameters: W a.m, ε. 8,, σ 5.67 m K 8 W m K 4, L.5m, u 33K Then from equation ( ), the following temperature distribution follows with the midpoint temperature u L 59.9K ( Maple file: fin3.mws) u u( ) u L L

44 368 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 3. Reduction of the order of a linear equation if one solution is nown a) If an non-trivial solution ( ) differential equation is nown a ( ) ( n ) + a ( ) + a ( ) + n n of a linear n th order homogeneous then the order of the homogeneous equation can be reduced b one order b the change of dependent variable with v followed b the change of variable v u These two substitutions can be combined in one change of variable b ud which preserves linearit and homogeneit of the equation. The order of the non-homogeneous equation ( ) ( ) ( ) ( ) ( n) a + + a + a f n n can be reduced b one order b the change of dependent variable with the same substitution v, but the resulting equation will be non-homogeneous. This method was used b Euler for solution of linear ODE s b sstematic reduction of order. b) Reduction formula for a nd order linear homogeneous ODE: ( ) ( ) ( ) a + a + a Let ( ) be a non-trivial solution, then it satisfies ( ) ( ) ( ) a + a + a n Let u (variation of parameter) then u + u u + u + u Substitute into the equation and collect terms in the following wa ( ) ( ) ( ) ( ) ( ) a u + a + a u + a a an + + u The last term is equal to zero because ( ) equation ( ) ( ) a u + a + a u is a solution of the homogeneous Now this equation does not include the unnown function u eplicitl, therefore, b substitution u v it can be reduced to a st order equation u v u v ( ) ( ) a v + a + a v

45 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, ( ) a ( ) a + a v + v The integrating factor for this equation is µ ( ) + d d + d ( ) a e e e e a a a a d a a a a a d d d ln a ln a a e e e e e then the general solution for v is v c e a d a Then the formal solution for the function u is a d a e u vd + c c d + c then the second solution can be written as a d a e u c d + c Choose arbitrar constants as c,c then reduction formula a d a e d (3) which is called the reduction formula. Chec if the solutions, are linearl independent: ( ) W, u u u + u a d a e > Therefore, the solutions are linearl independent and constitute the fundamental set for a nd order linear ODE.

46 37 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 V..9. ORTHOGONAL TRAJECTORIES. Orthogonal trajectories There are man mathematical models of engineering processes where families of orthogonal curves appear. The most tpical are: isotherms (curves of constant temperature) and adiabats (heat flow curves) in planar heat transfer sstems; streamlines (lines tangent to the velocit vector) and potential lines of the incompressible flow of irrotational fluid; magnetic field ; level curves and lines of steepest descent, etc. Famil of trajectories A one-parameter famil of planar curves is defined, in general, b the implicit equation F (,,c), c () For each value of the parameter c, there corresponds one particular curve (a trajector). For eample, equation c describes the famil of ellipses shown in the figure > g:{seq(*^+*+3*^i,i-..)}: > implicitplot(g,-3..3,-3..3,numpoints, scalingconstrained,view[-3..3,-3..3]); F (,,c) Slope of tangent line At each point of the curve, we can define a slope or tangent line to the curve b differentiation of equation () with respect to and solving it for its derivative F(,,c) f (,, c) () Orthogonal lines Let two lines L and L be defined b equations L : m + b m L : m + b m Then line L is orthogonal to line L if and onl if m (3) m Orthogonal trajectories Two curves are orthogonal at the point of intersection if the tangent lines to the curves at this point are orthogonal F F Two families of curves are called orthogonal families, if the curves from the different families are orthogonal at an point of their intersection

47 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 37 Algorithm: The following algorithm can be applied for finding the famil of curves F orthogonal to the given famil of curves F (shown with an eample): Let F : ce, c. Find the orthogonal famil F ) Find the slope of the tangent lines to curves from F : Differentiate F : c e and solve it for (if c appears in the equation, replace it b the solution of equation (,,c) c e + ce F for c, ( ) ) e 4+ + e ( ) + 4 ) Determine the equation for the orthogonal slope as the negative reciprocal to the previous equation: ) Solve the differential equation (the general solution will define an orthogonal famil): Rewrite the equation in the standard form of a linear equation 4 + Find the integrating factor 4 d 4ln e e + 4 µ Then the general solution is: d ) Answer: F : ce c F : > F:{seq(4*+^++(i/)*ep(*),i-8..8)}: > p:implicitplot(f,-3..3,-.., colorblue,scalingconstrained,numpoints): > F:{seq((j/)/^4-^/6+/4,j-8..8)}: > p:implicitplot(f,-3..3,-.., colorred,scalingconstrained,numpoints): > displa({p,p}); F F

48 37 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7 V... EXERCISES ) Solve the following differential equations and setch the solution curves: a) ( ) ( ) + + d + + d 3 b) ( + ) d + ( + ) d c) e + d + ln d d) ( + + ) d + ( + + ) d e) ( 3 ) + f) e sin d ( e cos ) d g) + + ln ) Solve the differential equation and setch the solution curves: + d + a) ( ) ( ) d b) + c) + 3) Solve the differential equation and setch the solution curves: a) ( ) + b) ( ) ( ) d + d c) ( ) d + ( ) d d) e) sec + subject to ( ) π 4) Solve the differential equation b conversion to polar coordinates and setch the solution curves: a) + 5) Solve the differential equation and setch the solution curves: + + e a) ( ) subject to ( ) b) + 4e sin c)

49 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, ) Show that if u ( ) is a particular solution of the Riccati equation ( ) + Q( ) R( ) P + Then the substitution z+ u reduces the Riccati equation to a Bernulli equation z ( Q + Pu) z + Pz 7) Solve the differential equation and setch the solution curves: a) (particular solution u ) b) ( ) c) e 8) Find a famil of orthogonal curves defined b the equation: + 3 c and setch the graph of curves ) Solve b reduction of order + (hint: is a solution) ) Solve the BVP and setch the solution curves: + (,π ) a) [ ] + π [ ] π b) 6 + (,) 4 [ ] [ ] c) e + (,) [ ] [ ] 5

50 374 Chapter V ODE V. ODE Basics V. First Order Ordinar Differential Equations September 4, 7