الجامعة اإلسالمية كلية العلوم غزة قسم الرياضيات المعادالت التفاضلية العادية Elementary differential equations and boundary value problems المحاضرون أ.د. رائد صالحة د. فاتن أبو شوقة 1
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بسم هللا الرحمن الرحيم The Islamic University of Gaza Department of Mathematics Ordinary Differential Equations nd Semester 017-018 Instructors: Prof. Raid Salha, Dr. Faten Abu-Shoga Course Description: First order differential equations, Second order linear equations, Higher order linear equations, Series solutions of second order linear equations, The Laplace transform. Course Objectives: At the end of this course, the student should be able 1- To understand the definition of differential equations and how to classify them. - To solve several types of first order differential equations. 3- To solve linear differential equations with constant coefficients. 4- To use series to solve second order linear equations with nonconstant coefficients. 5- To use Laplace transformation in solving initial value problems. Text Book: Elementary Differential Equations and Boundary Value Problems Eight Edition By W. E. Boyce and R. C. Diprima 7
Methods of Teaching: Lectures, discussions, solving selected problems. Evaluation and Grading: Midterm Exam 40% Quizzes 10% Final Exam 50% Total 100% Office Hours: Saturday, Monday, Wednesday: 11-1 توزيع المادة الدراسية على أسابيع الفصل الدراسي Week Sections to be covered Week Sections to be covered 1st week Chapter 1 9 th week 4.1, 4. nd week.1,. 10 th week 4.3, 4.4 3 rd week.4,.6 11 th week 5.1, 5. 4 th week.8 1 th week 5., 5.3 5 th week 3.1, 3. 13 th week 5.4, 5.5 6 th week 3., 3.4 14 th week 6.1, 6. 7 th week 3.5, 3.6 15 th week General Review 8 th week 3.7 16 th week Final Exam Midterm Exam 8
Definition: The differential equation (DE) is an equation contains one or more derivatives of unknown function. Examples: The following are DE 1. y y 1. dp r p k dt 3. xy (sin x ) y 3y cos x 4. F ma (Newton s Law) Example: Consider the DE (1) 9
() To determine the constant c, we need a condition, as for example Subsitute (3) into (), we get (3) Then c= -50. Definition : Condition (3) is called an initial condition and the DE (1) with the initial condition (3) is called an initial value problem (IVP). Example : Consider the IVP, 10
is the general solution. Usingthe initial condition, Then Is the solution of the IVP. 1. If the unknown function depends on a single independent variable, then only ordinary derivatives appear in the differential equation and it is called an ordinary differential equation.. If the unknown function depends on several independent variables, then only partial derivatives appear in the differential equation and it is called a partial differential equation. 11
Examples: Classify the following differential equation Order of the DE. The order of a DE is the order of the highest derivative that appears in the DE. Example: Find the order of the following differential equations 1. y y 1. (4) 1
. The general linear ordinary DE of order n is (5) An equation that is not of the form (5) is nonlinear equation. For example the DE is non linear because of the term. 13
Chapter First Order Differential Equations.1 Linear Equations; Method of Integrating Factors We will usually the general first order linear equations in the form y p( t ) y g ( t ) (1) where pt () and gt () are given functions of the independent variable t. How we can solve the DE (1)? To solve the DE (1),we multiply it by a certain function () t to get ( t ) y ( t ) p( t ) y ( t ) g ( t ) () such that ( ( t ) y ) ( t ) g ( t ) (3) From equation () and (3), we get the following This implies that ( t ) y ( t ) y ( t ) y p( t ) ( t ) y ( t ) p( t ) ( t ) 14
() t () t pt () Now, If () t is positive for all t, then ln ( t ) p( t ) dt k By choosing the arbitrary constant k to be zero, we get that ( t ) exp p( t ) dt Note: () t is positive for all t as we assumed and it is called an integrating factor. Now using equation (3), we get that ( t ) y ( t ) g ( t ) dt c This implies that the general solution of the DE (1) is given by 1 y ( t ) g ( t ) dt c (4) () t Example1: Solve the initial value problem (IVP) ty y t y 4, (1). 15
Note: If the integral ( t ) g ( t ) dt cannot be evaluated in terms of the usual elementary functional, so we leave the integral unevaluated and the general solution of the DE (1) is given by Where t o 1 t y ( s ) g ( s ) ds c (5) () t t o is some convenient lower limit of integration. Example: Solve the initial value problem (IVP) 1 t y y e, y (0) 1. 16
H.W. Problems 1-0 Pages 39-40. Example3(Q.8, page 39). Solve the DE (1 t ) y 4 ty (1 t ). Example4 (Q.0,page 40). Solve the (IVP) ty ( t 1) y t, y (ln ) 1, t 0. 17
. Separable Equations Consider the first Order DE y f ( x, y ) (1) If equation (1) is nonlinear, we can write it in the form 18
dy M ( x, y ) N ( x, y ) 0 () dx This form in equation () is not unique. If M is a function of x only and N is a function of y only then equation () becomes dy M ( x ) N ( y ) 0 (3) dx This equation is called separable because it is written in the form M ( x ) dx N ( y ) dy 0 (4) A separable equation in (4) can be solved by integrating the functions M and N. Example1: Solve the Differential equation x y 1 y. 19
Example: Solve the initial value problem (IVP) y 3x 4x, y (0) 1. ( y 1) H.W. Problems 1-0 Pages 47-48. 0
Example3 (Q.11, page 48). Solve the IVP x xdx ye dy 0, y (0) 1 Example4 (Q.0,page 48). Solve the (IVP) 1 1 y (1 x ) dy sin x dx 0, y (0) 1 1
Homogeneous Equations dy If the right side of the DE f ( x, y ) dx can be expressed as a function of the ratio y x only, then the equation is said to be homogeneous. Such DE can always be transformed into separable DE by a change of the dependent variable. Example5 (Q.30, page 49). Solve the DE dy y 4 x. dx x y
Example6 (Q.33, page 50). Solve the DE dy 4y 3 x. dx x y 3
Example7 (Q.37, page 50). Solve the DE dy x 3y dx xy..4 Differences Between Linear and Nonlinear Equations Theorem.4.1 If the functions p and g are continuous on an open interval I : containing the point 0 t t t, then there exist a unique function y () t satisfies the differential equation 4
y p( t ) y g ( t ) (1) for each t I, and that also satisfies the initial condition y ( t ) y () 0 0 Where y 0 is an arbitrary prescribed initial value. Note: 1. (Existence and Uniqueness) Theorem.4.1 states that the given (IVP) has a solution and it is unique.. It also states that the solution exists through any interval I containing the initial point t 0 in which p and g are continuous. Example 1: Use Theorem.4.1 to find an interval in which the (IVP) ty y t y 4, (1) has a unique solution. Now, we generalize Theorem.4.1 to the case of nonlinear DE. Theorem.4. Let the functions f and f y be continuous in some rectangle t, y, containing the point ( t0, y 0), Then in some interval t 0 h t t 0 h contained in t, there is a unique solution y () t of the IVP 5
y f ( t, y ), y ( t ) y (3) 0 0 Example : Use Theorem.4. to show the (IVP) y 3x 4x, y (0) 1 ( y 1) has a unique solution. Example3: Find an interval in which the solution of the (IVP) exists (ln t ) y y cot t, y () 3. 6
Example 4: State where in the ty plane the hypothesis of Theorem.4. are satisfied. dy 1t dt 3y y Bernoulli Equations: The DE of the form n y p( t ) y g ( t ) y, n 0,1 (4) is called a Bernoulli equation which is nonlinear. Note: 7
n then DE (4) has the form y p( t ) y g ( t ), 1. If 0, which is a linear DE. n then DE (4) has the form y p( t ) y g ( t ) y,. If 1, y p( t ) g ( t ) y 0, which can be rewritten in the form which is a linear DE. 3. If n 1or n 0, then the DE has the form n 1n y y p( t ) y g ( t ), (5) Let 1n dz n dy n dy 1 dz z y (1 n) y y dt dt dt 1 n dt Substitute in DE (5) becomes 1 dz p( t ) z g ( t ), (6) 1 n dt Note that DE (6) is a linear DE, where the dependent variable is z. We solve it for Z, then we replace z by y. Example 5. Solve the DE 1 n y, then we solve it for x y ( x 1) y 6y 3 8
Example 6. Solve the DE 0, 0. 3 t y ty y t 9
Discontinuous coefficients Example 7. Solve the DE y y g ( t ), y (0) 0, Where 1, 0 t 1, gt () 0, t 1. H.W. Problems 1-1, 4, 5, 8, 33. 30
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.6 Exact Equations and Integrating Factors Consider the DE M ( x, y ) N ( x, y ) y 0. (1) Suppose that we identify a function ( x, y) such that ( x, y ) ( x, y ) M ( x, y ), N ( x, y ) x y () and such that ( x, y ) c. Defines y ( x) implicitly as a differentiable function of x. Then ( x, y ) ( x, y ) dy M ( x, y ) N ( x, y ) y x y dx d x, ( x). dx So the DE in (1) becomes d x, ( x) 0 dx (3) In this case DE(1) is said to be exact DE. The solution of DE(1)or the equivalent DE(3) are given implicitly by ( x, y ) c (4) 3
Example 1. The following DE is an exact DE 0 x y xy y Theorem.6.1 Let the functions M, N, M y and N x be continuous in rectangular region R : x, y. Then the DE is an exact DE in R if and only if M ( x, y ) N ( x, y ) y M ( x, y ) N ( x, y ) y 0 x (5) At each point of R. 33
Example : Solve the DE y x xe x x e y y y cos (sin 1) 0 Example 3 Solve the DE ( ) ( ) 0. xy y x y x y 34
Integrating Factors Example 4 Determine whether the DE is exact or not (3 ) ( ) 0. xy y x xy y Note: It is sometimes possible to convert a DE that is not exact into an exact equation by multiplying the equation by suitable integrating factor. Consider the DE M ( x, y ) dx N ( x, y ) dy 0. (6) Multiply DE(6) by a function ( x, y) ( x, y ) M ( x, y ) dx ( x, y ) N ( x, y ) dy 0. (7) Assume DE (7) is an exact, then ( M ) ( N ) This implies that y y x x y (8) M M N N (9) x 35
Note: The most important situation in which simple integrating factors can be found when is a function of one of the variables x and y. Case 1: If is a function of x, Then Eq(8) becomes d M N N dx y x (9) This implies that d M dx y N N x (10) If M y N N x is a function of x only, then there is an integrating factor that depends on x only and it can be found by solving DE (10) which is both linear and separable and it's solution is given by M y N x ( x ) exp dx. (11) N Example 5 Find an integrating factor for the DE (3 ) ( ) 0. xy y x xy y then solve it. 36
Case : If is a function of y only. How we can find an integrating factor?. See problem 3 page 100. Example 6 Find an integrating factor for the DE x x e dx ( e cot y y csc y ) dy 0. then solve it. 37
Note: If the DE is not exact, then we have the following. 1.There is no integrating factor for the DE.. There is one integrating factor for the DE. 3. There is more than one integrating factor for the DE. Example 7 Use the integrating factor ( x, y ) xy ( x y ) 1 To solve the DE (3 ) ( ) 0. xy y x xy y 38
H.W. Problems 1-3 except (17, 4, 31) page 99-101. 39
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Example 8: Solve the IVP (9x y 1) dx (4 y x ) dy 0, y (1) 0. Example 9 : Find the value of b for which the given DE is exact then solve the DE ( xy xy y e x ) dx bx e dy 0. 41
Example 10 Solve the IVP 3 x 3 x dx 4 ( ) 3 4y dy 0. y y y Miscellaneous Problems (Page 131) H.W. Problems 1-3 Page 131. 4
Some special second order equations 1- Equations with the dependent variable missing Consider the second order DE y f ( t, y ) (1) Let v y this implies v y and this transform DE(1) to the first order DE v f ( t, v ) () We solve DE(1) for v then we find y by integrating v. Example1: Solve the DE 1 0, 0. t y ty t 43
- Equations with the independent variable missing Consider the second order DE y f ( y, y ) (3) Let v dv y this implies v f ( y, v ) (4) dt If we think of y as an independent variable, then by the chain rule, we have dv dv dy dv v dt dy dt dy dv This implies that DE (4) can be written as v f ( y, v ) dy (5) dy We solve DE (5) for v as a function of y then we replace v by y dt which gives us a separable DE with dependent variable y and independent variable t. We solve this separable DE to find y. Example: Solve the DE yy y t ( ) 0, 0. 44
Example3: Solve the DE ( ), 0. t y y t 45
Example 4:(Q 47 Page 133) Solve the DE y y e t y ( ), 0. H.W. Problems 36-51 Page 133. 46
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Chapter 3 Second Order Differential Equations 3.1 Homogeneous Equations with constant coefficients A second order linear DE is written in the form P( t ) y Q( t ) y R ( t ) y G ( t ) (1) y p( t ) y q( t ) y g ( t ) () where Q ( t ) R ( t ) p( t ), q( t ) P( t ) P( t ) and Gt () gt (). Pt () Note: If the DE can't be written in the form (1) and () then it is a nonlinear DE. An initial value problem (IVP) consists of a DE such as in (1) and () with a pair of initial conditions y ( t ) y y ( t ) y (3) 0 0, 0 0 Where y 0 and y 0 are given numbers. Definition: A second order linear equation is said to be homogeneous if the term Gt () in DE (1) or the term gt () in DE () is zero for all t. Otherwise the DE is called a nonhomogeneous. 48
Note: In this chapter, we will concentrate our attention on DEs in which the functions PQand, R in DE (1) are constants, that is in the form Where ab, and c are constants. ay by cy 0 (4) To solve DE (4), suppose y rt e is a solution of the DE (4). Now substitute in DE (4), we get r t r t y re, y r e r t r t r t ar e bre ce e ar br c rt ( ) 0 0 rt Since e 0 then ar br c 0 (5) Equation (5) is called the Characteristic equation for the DE (4). Assume the two roots of equation (5) are two different real roots, r 1 and r, r 1 r. Then y 1 e rt 1 and DE (4). The general solution of the DE (4) is given by y rt e are two solutions of the y c y c y c e c e r1t rt (6) 1 1 1 49
Example 1: Solve the DE y y 0. Example : Find the general solution of the DE y 5y 6y 0. 50
Example 3: Find the solution of the IVP y 5y 6y 0, y (0), y (0) 3. Example 4: Find the solution of the IVP 4y 8y 3y 0, y (0), y (0) 3. 51
Example 5: Solve the IVP y 5y 3y 0, y (0) 1, y (0) 0. Example 6. Find a DE whose general solution is t 1 y c e c e t 5
3. Fundamental Solutions of Linear Homogeneous Equations: Theorem 3..1: Consider the initial value problem y p( t ) y q( t ) y g ( t ), y ( t 0) y 0, y ( t 0) y 0 53
where p( t ), q( t ) and gt () are continuous on an interval I that contains the point t 0. Then there is exactly one solution of this IVP, and the solution exists throughout the interval I. Note: 1. (Existence and Uniqueness) Theorem 3..1 states that the given (IVP) has a solution and it is unique. It also states that the solution exists through any interval I containing the t initial point 0 in which pq, and g are continuous. Example 1: Find the longest interval in which the (IVP) ( t 3 t ) y ty ( t 3) y 0, y (1), y (1) has a unique solution. 54
Theorem 3..: If y 1and y are two solutions of the DE y p( t ) y q( t ) y 0 Then the linear combination c1y 1 cy is also a solution for any values of the constants c1 and c on an interval I that contains the point t 0. Definition: Let y 1 and y be two solutions of the y p( t ) y q( t ) y 0. The wronskian determinate (or simply the Wronskian) of the solutions y1 y y 1 and y is given by W ( y 1, y ) y 1y y y 1. y y 1 Theorem 3..3: If y 1 and y are two solutions of the DE y p( t ) y q( t ) y 0 and the Wronskian W ( y 1, y ) y 1y y y 1 is not zero at the point t 0 where the initial conditions y ( t ) y y ( t ) y. 0 0, 0 0 Then there are constants c1 and c for which y c1y 1 cy and the initial conditions. Theorem 3..4: If y 1and y are two solutions of the DE the DE y p( t ) y q( t ) y 0 and if there is a point 0 Wronskian of y 1 and y is nonzero, then the family of solutions t where the 55
y c1y 1 cy c With arbitrary coefficients 1 and c includes every solution of the DE. Definition: The solution y c1y 1 cy is called the general solution of the DE y p( t ) y q( t ) y 0. The solutions y 1 and y with a nonzero Wronskian are said to form a fundamental set of solutions of the DE. Example : Suppose y 1 e rt 1 and y e rt are two solutions of y p( t ) y q( t ) y 0. Show that they form a fundamental set of solutions if r 1 r. 56
Example 3: Show that solutions of y1 t 1 and t y 3ty y 0, t 0. y 1 t form a fundamental set of Example 4(Q.1, page 151). Find the longest interval in which the (IVP) ( x ) y y ( x )(tan x ) y 0, y (3) 1, y (3) has a unique solution. 57
Example 5: If the Wronskian of f and g is gt (). t te and if f () t t find Example 6: Is the functions y1 set of solutions of the DE x and y x xe form a fundamental ( ) ( ) 0, 0. x y x x y x y x 58
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3.3 Linear independence and Wronskian Definition: Two functions f () t and gt () are said to be linearly dependent on an interval I if there exist two constants k 1and k no both zero, such that 1 For all t k f ( t ) k g ( t ) 0 (1) I. The functions f () t and gt () are said to be linearly independent on an interval I if they are not linearly dependent. Theorem 3.3.1 If f () t and gt () are differentiable functions on an open interval I and if W ( f, g )( t 0) 0 for some point t 0 I then f () t and gt () are linearly independent on an interval I. Moreover, if f () t and gt () are linearly dependent on an interval I then W ( f, g )( t ) 0 for every t I. Theorem 3.3. (Abel's Theorem) If y 1 and y are two solutions of the DE y p( t ) y q( t ) y 0 () Where pt () and qt () are continuous on an open interval I, then the Wronskian W ( y 1, y 1)( t ) is given by W ( y 1, y 1)( t ) c exp p( t ) dt (3) Where c is a constant that depends on y 1 and y but not on t. 0, c 0, Note: W ( y 1, y 1)( t ) 0, c 0. 60
Example 1: Assume that y1 t 1 and y 1 t are solution of the DE t y 3ty y 0, t 0. Use Abel's Theorem to find the constant c in (3). Theorem 3.3.3: Let y 1 and y be solutions of the DE y p( t ) y q( t ) y 0 where pt () and qt () are continuous on an open interval I, y 1 and zero for all t then y are linearly dependent on I if and only if W ( y 1, y 1)( t ) is I. Alternatively, y 1and y are linearly independent on I if and only if W ( y 1, y 1)( t ) is never zero for all t I. 61
Summary: We can summarize the facts about fundamental set of solutions, Wronskian and linear independence in the following Let y 1and y be solutions of the DE y p( t ) y q( t ) y 0 where pt () and qt () are continuous on an open interval I. Then the following statements are equivalent. 1- The functions y 1and y are a fundamental set of solutions on I. - The functions y 1 and y are lineary independent on I W ( y 3-1, y 1)( t 0) 0 for some t 0 I. 4- W ( y 1, y 1)( t ) 0 for all t I. Example : Determine whether the following functions are linearly independent or linearly dependent f ( x ) e 3x and g ( x ) e 3( x 1) 6
Example 3: Find the Wronskian of two solutions of the given DE Example 4(Q.1, page 158) 63
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3.4 Complex roots of the Characteristic Equations (1) Consider the DE ay by cy 0 where ab, and c are constants. The Characteristic equation of the DE(1) ar br c 0 () Suppose b 4ac 0. Then the roots of Eq.() are conjugate complex numbers, we denote them by 1 Then the two solutions of DE (1), are ( i ) t ( i ) t 1 r i, r i (3) y e, y e (4) Euler,s formula ti e cost i sint (5) ti e cost i sint Now, using Euler,s formula, we have y e e e e t i t (6) ( i ) t t ti t 1 (cos sin ) y e e e e t i t (7) ( i ) t t ti t (cos sin ) Note, the two solutions y 1 and y are complex-valued functions. We want a real valued solutions. To get real solutions from y 1, and y, we use their sum and difference 65
t t t y y e (cos t i sin t ) e (cos t i sin t ) e cos t 1 t t t y y e (cos t i sin t ) e (cos t i sin t ) ie sin t 1 t t Let u( t ) e cos t, v ( t ) e sin t (8) e cos t e sin t W u v t e e cos t e sin t e sin t e cos t t t t (, )( ) 0, t t t t If 0. This implies that ut () and vt () form a fundamental set of solutions of DE(1). From the above, we conclude that, the general solution of DE(1) is t y c e cos t c e sin t (9) t 1 where c1 and c are arbitrary constants. Example 1: Find the general solution of y y y 0. 66
Example : Find the general solution of y 9y 0. Example 3: Find the solution of the IVP 16y 8y 145y 0, y (0), y (0) 1. 67
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3.5 Repeated roots; Reduction of order (1) Consider thede ay by cy 0 where aband, c are constants. The Characteristic equation of the DE(1) ar br c 0 () Suppose b 4ac 0. Then Eq.() has two equal roots r b r (3) a 1 Then we have one solution of DE (1), y 1 e b t a To find a second solution y that is linearly independent with the first solution y 1, we assume the general solution of DE(1) y v ( t ) y v ( t ) e 1 b t a b b t t a b a y v e v e a b b b t t t a b a b a y v e v e v e a 4a Substitute in DE(1), we get 69
b b b b b t t t t t a b a b a a b a a v e v e v e b v e v e a 4a a c v t e b t a ( ) 0. b b av ( b b ) v ( c) v 0 4a a v 0 v c t c 1 b b b b t t t t a a a a 1 1 1 y v ( t ) y v ( t ) e ( c t c ) e c te c e y 1 b t a e and y te b t a b b t t a a e te b t a W ( y 1, y ) b b b e 0. b t t t a a bt a e e e a a This implies that y 1 and y form a fundamental set of solutions of DE(1). Example 1: Find the general solution of the DE y 4y 4y 0. 70
Example : Find the solution of the IVP 1 y y 0.5y 0, y (0), y (0). 3 Summary Now, we can summarize the results that we have obtain for the second order linear homogeneous DEs with constant coefficients as follows: Consider the DE ay by cy 0 Let r 1 and r be two roots of the corresponding characteristic equation ar br c 0 71
1- If r 1 and r are different real roots r 1 r solution is given by y c e c e r t 1 r t 1,then the general - If r 1 and r are complex conjugate roots r i, r i,then the general solution is 1 t t y c e cos t c e sin t. given by 1 3- If r r1 r, then the general solution is given by Reduction of order y c e c te rt rt 1. Suppose we know one solution y () t 1, not every where zero of the DE y p( t ) y q( t ) y 0 (4) To find a second solution y v ( t ) y 1( t ) (5) y v y vy, y v y v y vy Then 1 1 1 1 1 Substitute y, y, y in DE(4) and colleting, we have y v ( y py ) v ( y py qy ) v 0 1 1 1 1 1 y v ( y py ) v 0 (6) 1 1 1 7
Equation (6) is a first order equation for the function v and can be solved as a linear DE or a separable DE. After finding v, then v can be obtained by an integration. Note: This method is called the method of reduction of order because the main step is to solve DE (6) which is a first order DE for v rather than solving DE (4) which is second order DE for y. Example 3: Given that y1 1 t is a solution of t y 3ty y 0, t 0, Find a second linearly independent solution. 73
H.W. 1-14, 3-30 Page 17-174. 74
3.6 Nonhomogeneous Equations, Method of undetermined coefficients Consider the nonhomogeneous DE y p( t ) y q( t ) y g ( t ) (1) where p, q and g are given continuous functions on an open interval I. The DE y p( t ) y q( t ) y 0 () in which gt ( ) 0 and p, q as in DE(1) is called the corresponding homogeneous DE to DE(1). Theorem 3.6.1 If Y 1 and Y are two solutions of the nonhomogeneous DE(1), then their difference Y1 Y is a solution of the corresponding homogeneous DE(). In addition if y 1 and y are a fundamental set of solutions of DE(), then Y ( t ) Y ( t ) c y ( t ) c y ( t ) 1 1 1 Where c 1 and c are constants. Theorem 3.6. The general solution of the nonhomogeneous DE(1) can be written in the form y ( t ) c y ( t ) c y ( t ) Y ( t ) (3) 1 1 75
Where y 1 and y are a fundamental set of the corresponding homogeneous DE(), c 1 and c are arbitrary constants and Y is some specific solution of the nonhomogeneous DE(1). Note: The solution of the nonhomogeneous DE(1) can be obtained using the following three steps. y c y c y 1. Find the complementary solution c 1 1 which is the general solution of corresponding homogeneous DE().. Find a particular solution y p of the nonhomogeneous DE(1). 3. Find the general solution of the nonhomogeneous DE(1) by adding y c and y. From the first two steps c p p The method of undetermined coefficients y y y This method usually is used only for problems in which the homogeneous DE has constant coefficients and the nonhomogeneous terms that consist of polynomials, exponential functions, sines and cosines. Example 1: Find a particular solution of the DE y y y e t 3 4 3. 76
Example : Find a particular solution of the DE y 3y 4y sin t. Example 3: Find a particular solution of the DE y y y t 3 4 4 1. 77
Note: g () t y e t Ae sin t or cos t A sin t B cos t n n 1 n n 1 0 1 n 0 1 p t a t a t a A t A t A n Remark 1: When gt () is a product of any two or all three of these types of functions, we use the product the same functions. Example 4: Find a particular solution of the DE t y 3y 4y 8e cos t. 78
Remark : g ( t ) g ( t ) g ( t ) and suppose that Y () t and Y () t 1 Suppose that 1 are particular solutions of the DEs ay by cy g ( t ). ay by cy g ( t ). respectively, then Y () p t + Y () 1 p t is a particular solution of the DE ay by cy g ( t ). 1 p p Example 5: Find a particular solution of the DE y y y e t e t t t 3 4 3 sin 8 cos. 79
Example 6: Find a particular solution of the DE y 3y 4y e t. Remark 3: If the form of the particular solution that we choose is a solution of the corresponding homogeneous DE, then we multiply it s by t, s 1,. Example 7(Q. 14 page 184): Solve the following DE y y t e y y t 4 3, (0) 0, (0) 1. 80
Example 8(Q. 6 page 184): Determine a sutible form of the particular solution of the following DE y y y te t te t t t 5 3 cos cos. 81
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3.7 Variation of parameters This method is more general the method of undetermined coefficients, since it can enable us to get a particular solution of an arbitrary second order linear nonhomogeneous DE. y p( t ) y q( t ) y g ( t ) (1) where p, q and g are given continuous functions on an open interval I. If gt () is not one of the functions as in Section 3.6 ( exponential, polynomial, sines, cosines, their product or sums) then we can't use the method of undetermined coefficients, therefore we will use the method of variation of parameters. Suppose y1, y are two linearly independent solutions of the corresponding homogeneous DE(). y p( t ) y q( t ) y 0 () y c y ( t ) c y ( t ), Then c 1 1 where c 1 and c are constants is the solution of DE(). In the method of Variation to find a particular solution of DE(1), we replace the constants c 1 and c by two functions u () t 1 and u () t to get y u( t ) y ( t ) u ( t ) y ( t ) particular solution p 1 1 83
where, y ( t ) g ( t ) y ( t ) g ( t ) u ( t ) dt, u ( t ) dt. W ( y, y )( t ) W ( y, y )( t ) 1 1 1 1 The general solution of the nonhomogeneous DE(1) can be written in the y ( t ) c y ( t ) c y ( t ) y ( t ) (3) form 1 1 where y 1 and y are a fundamental set of the corresponding homogeneous DE(), c 1 and c are arbitrary constants and y () t is some specific solution of the nonhomogeneous DE(1). Example 1: Find a particular solution of the DE y 4y 3csc t. p p 84
Example : Solve the following DE y y tan t, 0 t Example 3: Verify that y1 x and y x x ln are solutions of the corresponding homogeneous DE of the following nonhomogeneous DE 3 4 ln, 0. x y xy y x x x 85
Example 4(Q. 9 page 19): Use the method of reduction of order to solve the DE 4, 0,. t y ty y t t y 1 t 86
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Chapter 4 Higher Order Linear Equations 4.1 General Theory of the nth Order Linear Equations Definition 1: An nth order linear DE is an equation written in the form P ( t ) y P ( t ) y P ( t ) y G ( t ) (1) ( n) ( n1) 0 1 P ( t ), P ( t ),, P ( t ) and () Where 0 1 n functions on some interval I : n Gt are continuous real-valued t and P () t 0 is nowhere zero in I. Then by dividing DE(1) by P () t 0 ( n) ( n1) y p t y pn t y g t ( ) ( ) ( ) 1 () Definition : An initial value problem (IVP) consists of DE() with n initial conditions where y ( t ) y, y ( t ) y, y ( t ) y,, y ( t ) y ( n 1) ( n1 0 0 0 0 0 0 0 0 (3) where t 0 is a point in I. Theorem 4.1.1: If the functions 1 p ( t ), p ( t ),, p ( t ) and () n gt are continuous realvalued functions on an open interval I, then there exists exactly one 88
solution y () t of the DE() that also satisfies the initial conditions (3). This solution exists through the interval I. Definition 3: A homogeneous equation is an equation as in () where the term gt ( ) 0 i.e. ( n) ( n1) y p t y pn t y Otherwise the DE is called a nonhomogeneous. Theorem 4.1.: ( ) ( ) 0 1 (4) If the functions p1, p,, pn are continuous on an open interval I, if the functions y 1, y,, y n are solutions of DE(4) and if W ( y, y,, y ) 0. For at least one point in I then every solution of 1 n DE can be expressed as a linear combination of the solutions y 1, y,, y n. Definition 4: A set of solutions y 1, y,, y n of DE (4) where the wronskian is nonzero is called a fundamental set of solutions. The solution y c1y 1 cy cny n is called the general solution of the DE(4). f Definition 5: The functions 1, f,, f n on I k if there exist a set of constants 1, k,, k n that k1f 1 k f k f 0. n n are said to be linearly dependent not all zeros, such 89
f The functions 1, f,, f n are said to be linearly independent if they are not linearly dependent. Remark: y 1, y,, y n is a fundamental set of solutions of DE(4) if and only if they are linearly independent Example 1(Q5, Page ): Determine intervals where the solutions of the DEs exist. (4) ( x 1) y ( x 1) y (tan x ) y 0. 90
Example (Q15, Page ):Verify that the give functions are solutions of the DE and determine their wronskian 3 xy y 0, 1, x, x. 91
4. Homogeneous Equations with Constant Coefficients Consider the nth order linear homogeneous DE Where 0 1 a, a,, n a y a y a y (1) ( n) ( n1) 0 1 n 0 a are constants. y e rt be a solution of DE(1) then n n1 a0r a1r a n 0 () is called the characteristic equation of the DE(1). Eq() has n roots r1, r,, r n. 1- Real and unequal roots. The general solution 1 y c e c e c e (3) r t r t rnt 1 n Example 1: Find the general solution of the DE 7 6 0, (0) 1, (4) y y y y y y y (0) 0, y (0), y (0) 1 9
.Complex Roots: If the characteristic equation has complex roots, they must occur in conjugate pairs i Example : Find the general solution of the DE y (4) y 0. Repeated Roots: If the characteristic equation has s repeated roots, e, te, t e,, t e rt rt rt s 1 rt If the complex root i. i is repeated s times. is repeated s times, then its conjugate 93
Example 3: Find the general solution of the DE 0. (4) y y y Example 4: Find the general solution of the DE y (4) y 0. 94
Example 5 (Q.7, Page 30): Find the roots of 1 3 1 Example 6: Find the general solution of the DE y 5y 3y y 0. 95
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4.3 The Method of Undetermined Coefficients This method as we have discussed in Section 3.6 and the main difference that we use it for higher order DE. Example 1: Find the general solution of the DE t y 3y 3y y 4 e. Example : Find a particular solution of the DE 3sin 5cos. (4) y y y t t 97
Example 3: Find a particular solution of the DE y y t t e t 4 3cos. 98
Example 4: Without finding the constants, find a formula for a particular solution of the nonhomogeneous DE if y c c e c te c sin t c cost (a) t t c 1 3 4 5 (i) g ( t ) 3e t t cost (ii) t g ( t ) 3e t cost 99
(b) y c c t c t ( c c t )sin3 t ( c c t )cos3t c 1 3 4 5 6 7 g ( t ) ( t 1) t sin3t Example 5 (Q.5, Page 35): Determine the general solution of the DE 4. (4) t y y t e 100
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4.4 The Method of Variation of Parameters This method is a generalization of the method for second order DE from Section 3.7. Consider the DE ( n) ( n1) y p t y pn t y g t ( ) ( ) ( ) 1 (1) To find a particular solution, at first we solve the corresponding homogeneous DE ( n) ( n1) y p t y pn t y Let c 1 1 n n ( ) ( ) 0 1 () y c y c y c y (3) Be the general solution of the DE() y u y u y u y (4) Then p 1 1 n n Where 1 u, u,, u n are functions of t and W m ( t ) g ( t ) u m ( t ) dt, m 1,,, n W () t Where W ( t ) W ( y 1, y,, y n )( t ) W () m t is the determinate obtained from W () t by replacing the mth column by the column (0,0,,1). 10
Example 1(Q1, Page 40): Find the general solution of the DE y y tan t, 0 t. Example : Solve the DE y y t. 103
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Chapter 5 Series solutions of Second Order Linear Equations 5.1 Review of Power Series Definition: A power series a ( 0) n n x x is said to converge at a point m n 0 n x 0 if lim an ( x x 0) exists for that x 0. m n 0 Definition: The series n 0 n 0 a ( x x ) n 105 is said to converge absolutely at a n n a ( x x ) a x x converges. point x 0 if the series n 0 n 0 n0 n0 Remark: If the series converges absolutely, then the series also converges, but the converse is not necessarily true. Definition: A nonnegative number is called the radius of convergence of a series a ( 0) n n x x if it converges absolutely for 0 n 0 diverges for x x 0. Remark: 1- If the series converges only at x 0 then 0. - If the series converges for all x then. x x and
3- If 0 then the interval x x 0 is called the interval of convergence. Definition: If a function f is continuous and has derivatives of all order x x. Then the power series generated by f about x x 0 for 0 given by f ( x ) f x x x ( n ) 0 n ( ) ( 0) n 0 n! f ( x )! 0 f ( x 0) f ( x 0)( x x 0) ( x x 0) is called a Taylor series expansion of f about x x 0. If x 0 0 then it is called a Maclaurin series. Definition: A function f that has Taylor series expansion about x x 0 with radius of convergence 0 Examples: The Maclaurin series of is said to be analytic at x x 0. 1. e x n x. n 0 n! sin x n 0 n ( 1) x (n 1)! n1. cos x n n ( 1) x 4. ( n)! n 0 a ax 1 x n 0 n 106
Shift of Index of Summation n n n0 n a x a x n n n1 nan x ( n 1) an 1 x n1 n0 n n ( 1) n ( )( 1) n n0 n n a x n n a x n n n n n an x x 0 n n an x x 0 n0 n ( 4)( 3) ( ) ( )( 1) ( ) n 5. Series Solutions Near an Ordinary Point Part I P( x ) y Q( x ) y R ( x ) y 0 (1) Suppose PQ, and R are polynomials and that they have no common factors. Suppose that we want to solve DE(1) in the neighborhood of a point x 0. Definition: A point x 0 such that 0 Px ( ) 0 is called an ordinary point. 107
Since P is continuous, it follows that there is an interval about x 0 in which Px ( ) 0. In that interval, we can divide DE(1) by Px ( ) to get DE() y p( x ) y q( x ) y 0 () Definition: A point x 0 such that Px ( 0) 0 is called a singular point. Example 1: Find a series solution of the DE y y 0 Example : Find a series solution of the DE y xy 0. x. 108
Example 3: Find a series solution in powers of ( x 1) of the DE y xy 0. x. Example 4 (Q 9 page 59): Find a series solution of the DE (1 x ) y 4xy 6y 0. 109
Example 5 (Q 15 page 59): Find a series solution of the IVP y xy y 0, y (0), y (0) 1. 110
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5.3 Series Solutions Near an Ordinary Point Part II Consider the DE P( x ) y Q( x ) y R ( x ) y 0 (1) Suppose PQ, and R are polynomials. In the neighborhood of a point x 0, Suppose that n 0 () n 0 y ( x ) a ( x x ) n is a solution of the DE(1). Now, by differentiating Equation () m times and setting 0, it follows Since y ( x) m a x m ( m )! m ( 0), 0,1,, is a solution of DE (1), we get that P( x ) ( x ) Q( x ) ( x ) R( x ) ( x ) 0 From which we get that x x Q ( x ) R ( x ) ( x ) ( x ) ( x ) P ( x ) P ( x ) p( x ) ( x ) q( x ) ( x ) (3) ( x ) p( x ) ( x ) q( x ) ( x ) 0 0 0 0 0! a p( x ) ( x ) q( x ) ( x ) 0 0 0 0 11
To find a 3 we differentiate Equation (3) and set x x 0. Remark: To determine the derivatives of the functions px ( ) and qx ( ). Example 1 (Q page 65): a 's we need to compute infinitely many n If y ( x) is a solution of the DE Find ( x ), ( x ), ( x ). (4) 0 0 0 y (sin x ) y (cos x ) y 0, y (0) 0, y (0) 1, Example : What is the radius of convergence of the Taylor series for (1 ) x? 1 x about 0 0 113
Example 3: What is the radius of convergence of the Taylor series for 1 ( x x ) about x 0 0? about x 0 1? Example 4 (Page 64): Determine a lower bound for the radius of convergence of series solution of the DE 1 about x 0 0? about x 0? (1 x ) y xy 4x y 0 114
Example 5 (Page 64): Can we determine a series solution about x 0 0 for the DE y x y x y (sin ) (1 ) 0and if so, what is the radius of convergence? Example 6 (Q7. Page 65): Determine a lower bound for the radius of convergence of series solution of the DE about x 0 0? about x 0? 3 (1 x ) y 4xy y 0 115
5.4 Regular Singular Points Consider the DE P( x ) y Q( x ) y R ( x ) y 0 (1) Definition: The singular point x 0 of the DE(1) is called a regular singular point if both the functions Qx ( ) ( x ) Px ( ) x 0 and ( x ) R( x) Px ( ) x 0 () Are analytic functions, (i. e. the functions in () have a convergent Taylor series expansion about x x 0 ). Remark 1: If P( x ), Q( x ) and R( x ) are polynomials, then a singular point x 0 is regular singular point if the limits Qx ( ) lim ( x x 0), x0 Px ( ) x and lim ( x ) xx 0 R( x) Px ( ) x 0 are both finite. 116
Remark : If x 0 is not a regular singular point of DE(1), the it is called an irregular singular point. Example 1: Find and classify the singular points of the DE. (1 x ) y xy ( 1) y 0 Example : Find and classify the singular points of the DE. x ( x ) y 3 xy ( x ) y 0 117
Example 3: Find and classify the singular points of the DE. ( x ) y (cos x ) y (sin x ) y 0 118
Example 4: Find and classify the singular points of the DE. ( x x ) y ( x 1) y y 0 5.5 Euler equations Definition: The DE of the form 0 (1) x y xy y Where and are real constants is called an Euler equation. Note: The point x 0 0 is a regular singular point of the DE(1). In any interval not including the origin, Eq(1) has a general solution of the form y c1y 1 cy () where y 1 and y a re linearly independent There are two cases when we solve DE(1). Case 1, For x 0. 119
Assume y x r is a solution of DE(1). This implies that y rx y r r x r1 r, ( 1), Then by substituting in DE(1) we get r r 1 r x ( r( r 1) x ) x ( rx ) x 0. r x r( r 1) r 0. This implies that r ( 1) r 0. (3) Eq(3) has two roots and they are given by ( 1) ( 1) 4 r1, r. Real distinct roots: If Eq.(3) has two distinct real roots r 1, r, then y x y x r1 r, are two solutions of DE(1) and the general solution of 1 DE(1) is given by y c x c x r 1 r 1 Example 1: Solve the DE. x y 3xy y 0, x 0 10
Equal Real roots: r r If Eq.(3) has two equal real roots r 1 r r, then y 1 x, y x ln x are two solutions of DE(1) and the general solution of DE(1) is given by y c x c x x r r 1 ln Example : Solve the DE 5 4 0, 0. x y xy y x 11
Complex roots: If Eq.(3) has two complex roots r 1 i, r i, then y x cos ln x, y x sin ln x are two solutions of DE(1) and the 1 general solution of DE(1) is given by y c1x cos ln x cx sin ln x Example 3: Solve the DE 0, 0. x y xy y x Case, For x 0. This case as the first case but we replace x by x. Remark: The solution of the Euler equation of the form ( x x ) y ( x x ) y y 0 (4) 0 0 ( x x ) is the same as for DE(1) but we replace x by 0 transform DE(4) to the form of DE(1) by putting t x x 0. Also, we can 1
Example 4(Q.10 Page 78): Solve the DE. ( x ) y 5( x ) y 8y 0 Transform the Euler equation to an equation with constant coefficients Let x e z, z ln x, x 0. dy dy dx dy z dy e dz dx dz dx dx dy 1 dy dx x dz x d y 1 1 1 1 1 d y ( ) ( ) dy d y dy dx x dz x x dz x dz x dz Now by substituting in DE(1), we get 13
1 d y 1 dy 1 dy x x y 0 x dz x dz x dz d y dy dy dz dz dz y 0 d y dy ( 1) y 0 (4) dz dz Note that DE(4) has a constant coefficients. 14
Example 5(Q.5 Page 79): Transform the following DE into a DE with constant coefficients then solve it. x y 3xy 4y ln x Example 6(Q.8 Page 79): Transform the following DE into a DE with constant coefficients then solve it 4 sin(ln ).. x y xy y x 15
Chapter 6 The Laplace Transformation 6.1 Definition of The Laplace Transformation Definition: An improper integral over an unbounded interval is defined a limit of integrals over finite intervals, thus A f ( t ) dt lim f ( t ) dt a, (1) a A where A is a positive real number. If the integral from a to A exists for each A a and if limit as A exists then the integral is said to converge to that limiting value. Otherwise the integral is said to diverge. Example 1: 0 ct e dt 16
Example : 1 dt t Example 3: 0 p t dt, p 1 17
Definition: Let f () t be given for t 0 and suppose that f () t satisfies certain conditions. Then the Laplace transformation of f () t is denoted by L{ f ( t )} or Fs ( ) is defined by st L{ f ( t )} F ( s ) e f ( t ) dt () 0 whenever the improper integral converges. Example 4: Find the Laplace transformation of the function f ( t ) 1, t 0. Example 5: Find the Laplace transformation of the function f ( t ) e at, t 0. 18
Example 6: Find the Laplace transformation of the function f ( t ) sin at, t 0. Remark: The Laplace transformation is linear operator because if f 1 ( t ), f ( t ) are two functions whose Laplace transformation exist for s a1 and Respectively, then for s max{ a1, a} L{ c f ( t ) c f ( t )} c L{ f ( t )} c L{ f ( t )} 1 1 1 1 s a 19
Example 7: Find the Laplace transformation of the function f t e t t t ( ) 5 3sin 4, 0. Example 8(Q. 7, Page 313): Find the Laplace transformation of the function f ( t ) cosh bt. 130
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6. Solution of The Initial Value Problems In this section, we use the Laplace transformation in solving initial value problems. Theorem 6..1: Suppose that f is continuous and f is piecewise continuous on any interval 0 t A. Suppose further that there exist constants f ( t ) Ke at for t M. Then L{ f ( t )} sl{ f ( t )} f (0). Theorem 6..: Suppose that the functions K, a and M such that L{ f ( t )} exists for s a ( n 1) f are continuous and f, f,, piecewise continuous on any interval 0 t A there exist constants K, a and M such that at at ( n 1) at f ( t ) Ke, f ( t ) Ke,, f ( t ) Ke and moreover f ( n ). Suppose further that is for t M. Then ( n ) L{ f ( t )} exists for s a and moreover L f t s L f t s f s f sf f ( n ) n n 1 n ( n ) ( n 1) { ( )} { ( )} (0) (0) (0) (0). 13
Example 1: Use Laplace transformation to solve the IVP y y y 0, y (0) 1, y (0) 0 Example : Use Laplace transformation to solve the IVP y y sin t, y (0), y (0) 1 133
Example 3: Use Laplace transformation to solve the IVP 0, (0) 0, (0) 1, (0) 0, (0) 0 (4) y y y y y y Example 4(Q. 9, Page 3): 1 s Find the inverse Laplace transform of the function s 4s 5. 134
Example 5(Q. 16, Page 3): Use Laplace transformation to solve the IVP y y 5y 0, y (0), y (0) 1 Example 6(Q. 3, Page 3): Use Laplace transformation to solve the IVP t y y y 4 e, y (0), y (0) 1 135
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Islamic University of Gaza Department of Math. Second semester /4/014 Differential Equations Midterm exam Time: one hour Solve the following questions: Q1) (a) [6Marks] Solve the following initial value problem y cos x y 1 y, y(0) 1. (b) [8 Marks] Solve the following differential equation y (tan x ) y x sin x, 0 x. Q) (a) [6 Marks] Determine the interval in which the solution of the given initial value problem exists and unique cos x ( x ) y 3x y y, y ( 1) 1. x (b) [10Marks] Solve the differential equation ( x y 1) dx x ( x y ) dy 0, x 0. Q3)(a) [6 Marks] Find a differential equation whose general solution is (b) [10 Marks] If y1 y c e cos3t c e sin 3t. t t 1 x is one solution of the differential equation 0, 0, x y xy y x use the method of reduction of order to find a second linearly independent solution of the differential equation. Q4) [14 Marks] Find the general solution of the differential equation x e y 4y 4 y, x 0. x 138
Islamic University of Gaza Department of Math. Summer semester 1/8/009 Differential Equations Midterm exam Time: one hour Solve the following questions: Q1) Find the general solution of the differential equation t y y y e 3cost. Q) Solve the following differential equations: a) y y e y e ( y ). b) x dy ( x 3 xy y ) dx 0. Q3) a) Solve the initial value problem 3 (1 xy ) dy y dx 0, y ( ) 1. 4 b) If the wronskian of the functions f and g is 3e t and if f () t gt. t e, find () Q4) Use the method of reduction of the order to find a second solution of the differential equation 3 4 0, 0, if x y xy y x y ( x ) x 1. 139
Islamic University of Gaza Department of Math. Summer semester 4/8/010 Differential Equations Midterm exam Time: 75 min. Solve the following questions: Q1) (a) [5 Marks] Solve the following differential equation y y e t. (b)[7 Marks] Solve the following differential equation dy 4x 3y dx x y. Q) (a) [6 Marks] Solve the following differential equation (b) [6 Marks] Solve the initial value problem Q3) (a) [5 Marks] Suppose that. 3 x y y x y, x 0 3 y dx xy dy y ( 1) 0, (0) 1 y ( x ) x 1 3 is one solution of the differential equation 5 9 0, 0. x y x y y x Find a second linearly independent solution of the differential equation. (b) [3 Marks] Suppose that y 1 and y are two linearly independent solutions of the differential equation t and if W ( y, y )( t ) ct e. Find pt (). 1 y p( t ) y q( t ) y 0, Q4) [8 Marks] Use the method of variation of parameters to solve the differential equation y y y te t 5 6. 140
Islamic University of Gaza Department of Math. Second semester /5/014 Differential Equations Final exam Time: Two hours Solve the following questions: Q1 Q Q3 Q4 Q5 Q6 Q7 Total االسم 8 8 8 8 8 8 1 60 الرقم Q1) (a) [4 marks] Find the general solution of the differential equation yy ( y ) 0. (b) [4 marks] Find a function f ( x ) such that the differential equation y cos x (cos x ) f ( x ) y 0 is an exact and f (0) 0. Q) (a)[4 marks] Solve the initial value problems 3 0, ( ) 0. y x y xy y e (b) [4 marks] Solve the differential equation 3 (4 8 ) 0. y x y y Q3) Consider the differential equation 3 5 4 x x y y y y e xe cos x (a)[5 marks] Find the complementary solution, y c. (b) [3 marks] Find a formula for the particular solution, y p. (Don't evaluate the coefficients). 141
Q4) [8 marks] Find a series solution in powers of ( x ) for the following differential equation (Include at least three nonzero terms in each series) y xy y 0. Q5)(a)[4 marks] Solve the following initial value problem 6 0, (1) 0, (1). x y x y y y y (b)[4 marks] Classify the singular points of the following differential equation as regular or irregular. 4 3 ( x x ) y ( x 1) y y 0 Q6)[8 marks] Use the Laplace transformation to solve the initial value problem t y y y te, y (0) 1, y (0). Q7)(a)[ 4 marks] Is the following initial value problem has a unique solution. Explain your answer. y (b)[4 marks] Are the two functions of 3x, y (1) 1 y 1 y1 3 x, y. x 4 form a fundamental set solutions of the differential equation Explain your answer. 6 1 0, 0. x y xy y x (c)[4marks] Suppose f, g and h are differential functions such that f () 3 and (, )() 4 W g h. Find (, )() W f g f h. 14
Islamic University of Gaza Department of Math. Summer semester 7/8/013 Differential Equations Final exam Time: Two hours Q1 Q Q3 Q4 Q5 Q6 Total االسم 16 18 16 16 16 18 100 الرقم Solve the following questions: Q1) (a) [6 Marks] Determine the interval in which the solution of the given initial value problem exists y (tan t ) y sin t, y ( ) 0. b) [1 Marks] Suppose that equation y ( x ) x 1 is one solution of the differential x y y 0. Find a second linearly independent solution of the differential equation. Q) (a) [10 Marks] Find the general solution of the differential equation y y y t e t t 4 4, 0. (b) [6 Marks]Solve the following differential equation dy 3y x. dx xy Q3) a) [10 Marks] Consider the differential equation 3 4 t. (4) y y t e (i) [5 Marks] Find the solution of the corresponding homogeneous equation. (ii) [5 Marks] Use the method of undetermined coefficients to find a suitable form of the particular solution without evaluating the constants. 143
(b) [6 Marks] Solve the differential equation 4 4 0. x y xy y Q4) [16 Marks] Find two linearly independent power series solutions at x 0 0 to the differential equation (include at least three nonzero terms in each series) y xy y 0. Q5) (a) [6 Marks] Determine a lower bound for the radius of convergence of the series solution for the given differential equation about x 0 0. (1 x ) y 3xy y 0 (b) [10 Marks] Determine the singular points of the differential equation, x ( x ) y 3 xy ( x ) y 0 and classify them as regular or irregular. Q6) (a) [6 Marks] Use the definition to find the Laplace transformation of the function f () t t t e. (b) [1 Marks] Use the Laplace transform to solve the initial value problem y y y cos t, y (0) 1, y (0) 0 144
Islamic University of Gaza Department of Math. Summer semester 1/9/010 Differential Equations Final exam Time: Two hours Q1 Q Q3 Q4 Q5 Total االسم 15 1 11 10 1 60 الرقم Solve the following questions: Q1) (a) [6Marks] Solve the following differential equation y y e ( ) y. (b) [5 Marks] Solve the following initial value problem dy sin x x y, y () 1. dx x (c) [4 Marks] Determine the interval in which the solution of the given initial value problem exists x ( x 4) y 3x y 4y ln x, y (3) 1, y (3) 1. Q)Find the general solution of the following differential equations (a) [6 Marks] (b) [6 Marks]. (4) y y sin t y y y t e t t 4 4, 0. Q3)(a) [6 Marks] Solve the following differential equation. 3( x 1) y 6( x 1) y 6y 6ln( x 1) (b) [5Marks] Classify the singular points of the following differential equation as regular or irregular. ( x x 3) y xy 4y 0 145
Q4)[10 Marks] Find two linearly independent power series solutions at x 0 1 to the differential equation (include at least three nonzero terms in each series) y xy y 0. Q5) (a) [5 Marks] Use the definition to find of the Laplace transformation of the function f ( t ) cos t, t 0. (b) [7 Marks] Use the Laplace transform to solve the initial value problem y y y cos t, y (0) 1, y (0) 0. 146