MB4018 Differential equations

Transcription

1 MB4018 Differential equations Part II Prof. Natalia Kopteva Spring 2015 MB4018 (Spring 2015) Differential equations Part II 0 / 69

2 Section 1 Second-Order Linear ODEs (involve second-order derivatives) A second-order ODE is called linear if it can be written as y + p(x) y + q(x) y = R(x) ( ) for some functions p(x) and q(x), called the coefficients. and some function R(x) called the right-hand side. Otherwise it is called nonlinear. E.g., y + x y = cos x is linear; while y y + x y + y 3 = 0 is nonlinear In the linear case ( ): if R(x) = 0, it is called homogeneous; if R(x) 0, it is called nonhomogeneous. Section 1 Second-Order Linear ODEs 1 / 69

3 A general solution of ( ) is a formula that describes all solutions of ( ) as particular cases. Example: for the ODE 2x 2 y x y 2 y = 4, the general solution is y = C 1 x 2 + C 2 1 x 2 (where C 1, C 2 are arbitrary constants). NOTE: A general solution of a linear second-order ODE must involve two arbitrary constants A particular solution is found by obtaining values for the 2 arbitrary constants from 2 initial or boundary conditions, e.g., y(x 0 ) = A, y (x 0 ) = B Initial conditions y(x 0 ) = A, y(x 1 ) = B Boundary conditions Section 1 Second-Order Linear ODEs 2 / 69

4 1.1 Homogeneous ODEs y + p(x) y + q(x) y = 0 ( ) (it s ( ) with R(x) = 0) SUPERPOSITION PRINCIPLE: If y 1 (x) and y 2 (x) are any 2 solutions of ( ), then A y 1 (x) + B y 2 (x) is also a solution of ( ) for any real A and B. Proof: y 1 + p(x) y 1 + q(x) y 1 = 0 y 2 + p(x) y 2 + q(x) y 2 = 0 (A y 1 + B y 2 ) + p(x) (A y 1 + B y 2 ) + q(x) (A y 1 + B y 2 ) = Section 1 Second-Order Linear ODEs 3 / 69

5 A general solution of a homogeneous linear second-order ODE always has the form: y = C 1 y 1 (x) + C 2 y 2 (x) where C 1, C 2 are arbitrary constants, y 1 (x), y 2 (x) are two different particular solutions of the ODE such that y 1(x) y 2 (x) constant. Such particular solutions are called linearly independent. CONCLUSION: It suffices to find 2 linearly independent particular solutions to construct a general solution for equation ( ). Example: for the ODE 2x 2 y x y 2 y = 0, and y 1 (x) = x 2, y 2 (x) = 1 x are two particular solutions, so a general solution is y = C 1 x 2 + C 2 1 x, where C 1, C 2 are arbitrary constants Section 1 Second-Order Linear ODEs 4 / 69

6 Problems for Consider the initial-value problem x 2 y 7xy + 15y = 0, y(1) = 1, y (1) = 0. 1 Check that the functions x 5 and x 3 are particular solutions of the above ODE. 2 Check that x 5 and x 3 are linearly independent. 3 Hence construct a general solution of the above ODE. 4 Solve the above initial-value problem, i.e. find a unique solution of the ODE that satisfies the 2 initial conditions. 2 Consider the boundary-value problem x 2 y + 3xy 3y = 0, y(1) = 1, y(2) = Check that the functions x and x 3 are particular solutions of the above ODE. 2 Check that x and x 3 are linearly independent. 3 Hence construct a general solution of the above ODE. 4 Solve the above boundary-value problem, i.e. find a unique solution of the ODE that satisfies the 2 boundary conditions. Section 1 Second-Order Linear ODEs 5 / 69

7 1.2 Homogeneous ODEs with Constant Coefficients If the coefficients p(x) and q(x) in ( ) are constant, this ODE is called a linear ODE with constant coefficients. Consider the homogeneous and nonhomogeneous cases separately Consider a homogeneous linear second-order ODE with constant coefficients: y + b y + c y = 0 To find a solution, we make a conjecture that it has the form y = e r x ( ) with some (unknown at this stage) constant r. To find r: substitute our guess in ( ): y = e rx y = r e r x y = r 2 e r x so substitution in ( ) yields: r 2 e r x + b r e r x + c e r x = 0 r 2 + b r + c = 0 ( ) this quadratic equation is called auxiliary (characteristic) for ODE ( ). Its roots are r = b ± b 2 4c 2. Section 1 Second-Order Linear ODEs 6 / 69

8 Case 1: let b 2 4c > 0. Then ( ) has 2 distinct real roots r 1 and r 2. So ODE ( ) has 2 different particular solutions y 1 = e r1x and y 2 = e r2x. So general solution is y = C 1 e r1x + C 2 e r 2x Section 1 Second-Order Linear ODEs 7 / 69

9 Case 2: let b 2 4c = 0. Then ( ) has only one root r 1 = b 2 so ODE ( ) has a particular solution y 1 = e r 1x. But to get a general solution, we need another particular solution y 1! First, rewrite ( ) as y + b y + b2 4 y = 0. Make another solution guess y 2 = x e r1x. To check, whether y 2 is indeed a solution, substitute it into our ODE: y 2 = (1 + x r 1) e r 1x y 2 = (2 r 1 + x r1 2) er 1x Hence y 2 + b y 2 + b2 4 y 2 = [ (2 r 1 + x r1 2) + b (1 + x r 1) + b2 4 }{{ x] } er1x = 0 =0 as r 1 = b 2 (Ex.) Hence y 2 (x) is indeed another particular solution of ( ) so the general solution y = C 1 y 1 (x) + C 2 y 2 (x) becomes y = (C 1 + C 2 x) e r1x. Section 1 Second-Order Linear ODEs 8 / 69

10 Case 3: let b 2 4c < 0 similar to Case 1: ( ) has 2 distinct roots, but they are complex: r 1 = α + i β and r 2 = α i β. As in Case 1, the ODE ( ) has a general solution y = C 1 e r1x + C 2 e r2x = C 1 e (α+i β) x + C 2 e (α i β) x where C 1, C 2 are arbitrary complex constants. To restrict this formula to real functions and constants, recall that e (α+i β) x = e α x e i β x = e α x( cos(β x) + i sin(β x) ) e (α i β) x = e α x e i β x = e α x( cos(β x) i sin(β x) ) These imply: y(x) = e α x( (C 1 + C 2 ) cos(β x) + i (C }{{} 1 C 2 ) sin(β x) ) }{{} C 1 C 2 Finally, a general solution of ( ) is written as y = e α x( C 1 cos(β x) + C 2 sin(β x)) (where C 1, C 2 are arbitrary real constants). Section 1 Second-Order Linear ODEs 9 / 69

11 For a homogeneous linear second-order ODE with constant coefficients: y + b y + c y = 0 ( ) Summary: Roots of ( ) General Solution of ( ) (1) b 2 4c > 0 2 real roots: r 1, r 2 y = C 1 e r 1x + C 2 e r 2x (2) b 2 4c = 0 1 real root r 1 = b 2 y = (C 1 + C 2 x) e r 1x (3) b 2 4c < 0 2 complex roots: r 1,2 = α ± i β y = e α x( C 1 cos(β x) + C 2 sin(β x) ) where r 2 + b r + c = 0 ( ) Section 1 Second-Order Linear ODEs 10 / 69

12 Examples: 1 y + y 2 y = 0. S: r 2 + r 2 = 0 r = 1± ( 2) 2 r 1 = 2, r 2 = 1 y 1 = e 2x, y 2 = e x y = C 1 e 2x + C 2 e x 2 16 y 8 y + y = 0. S: 16 r 2 8 r + 1 = 0 r = 8± = 1 4 single real root y = (C 1 + C 2 x) e x/4 3 y + 4 y + 13 y = 0 S: r r + 13 = 0 r = 4± = 2 ± 9 r = 2 ± 3 i: r 1 = i, r 2 = 2 3 i (i.e. α = 2, β = 3) y = e 2 x( C 1 cos(3 x) + C 2 sin(3 x) ) Section 1 Second-Order Linear ODEs 11 / 69

13 Problems for Find the general solutions of the following differential equations: 1 y y = 0 2 y 2y 3y = 0 3 y + 4y = 0 4 4y + 4y + y = 0 2 Find the solution of the following initial-value problems 1 y + 4y = 0, y(0) = 0, y (0) = 1; 2 y + 4y + 5y = 0, y(0) = 1, y (0) = 0; 3 y 6y + 13y = 0, y(π/2) = 0, y (π/2) = 2; 4 y + y = 0, y(π/3) = 2, y (π/3) = 4. Section 1 Second-Order Linear ODEs 12 / 69

14 1.3 Physical Example: Damped Mass-Spring System A mass m is attached to a spring with spring constant k, and to a dashpod with damping coefficient c. Its displacement at time t (w.r.t. the rest point) is x(t). RECALL: By Newton s Second Law: mass acceleration = force m a = } k {{ x } } {{ c v } damping force is proportional to v Here a = d 2 x dt 2 Hook s force is proportional to x So x(t) satisfies the ODE: is the acceleration, v = dx dt is the velocity. m d 2 x dt 2 + c dx dt + k x = 0 (with positive constants m, c, k > 0). Section 1 Second-Order Linear ODEs 13 / 69

15 (a) c = 0, i.e. no damping: called simple harmonic motion, m r r + k = 0 has complex roots r 1,2 = ±i k x(t) = C 1 cos( m t) + C k 2 sin( m t) k m (b) c > 0, c 2 < 4mk, i.e. Damped Oscillator: m r 2 + c r + k = 0 ( ) has roots r 1,2 = c± c 2 4mk 2m i.e. r 1,2 = c 2m ± i β with β = 4mk c 2 2m x(t) = e 2m c t ( C 1 cos(β t) + C 2 sin(β t) ) Section 1 Second-Order Linear ODEs 14 / 69

16 (c) c 2 = 4mk, i.e. Critically Damped case: ( ) has a single root: r 1 = c 2m x(t) = (C 1 + C 2 t) e 2m c t no oscillations (d) c 2 > 4mk, i.e. Overdamped case: ( ) has 2 real roots r 1 < r 2 < 0 x(t) = C 1 e r1t + C 2 e r 2t decay, no oscillations Section 1 Second-Order Linear ODEs 15 / 69

17 Problems for The motion of a certain spring-mass system is governed by the differential equation x x + x = 0 where x = x(t) is measured in feet and t in seconds. If x(0) = 2 and x (0) = 0, determine the position of mass at any time The motion of a certain spring-mass system is governed by the differential equation x + c x + x = 0 where x = x(t) is measured in feet and t in seconds, and c is the damping coefficient. How large should one choose the damping coefficient c so that there are no oscillations? Section 1 Second-Order Linear ODEs 16 / 69

18 1.4 Nonhomogeneous Linear ODEs Consider a nonhomogeneous linear second-order ODE y + p(x) y + q(x) y = R(x) ( ) The corresponding homogeneous equation: y h + p y h + q y h = 0 ( ) If y p (x) is any particular solution of ( ), while y h (x) is any solution of ( ), then y p (x) + y h (x) is also a solution of ( ). y y p + p y p + q y p = R(x) Proof: h + p y h + q y h = 0 (y p + y h ) + p (y p + y h ) + q (y p + y h ) = R(x) Section 1 Second-Order Linear ODEs 17 / 69

19 Hence, general solution of nonhomogeneous equation ( ) = particular general solution solution of ( ) + of homogeneous equation ( ) }{{} this must involve 2 arbitrary constants To find a general solution of nonhomogeneous equation ( ): Step 1: Find a general solution of the corresponding homogeneous equation ( ) (e.g., as in 1.2). Step 2: Add a particular solution of ( ). To find this, there are various methods; two of them are Method of Undetermined Coefficients (see 1.5) Variation of Parameters (see 1.6) Section 1 Second-Order Linear ODEs 18 / 69

20 1.5 Method of Undetermined Coefficients Method of Undetermined Coefficients applies to the constant-coefficient y + b y + c y = R(x) ( ) R(x) Try y p (x) = α α x + β α x 2 + β x + γ A A x + B A x 2 + B x + C polynomial of degree n another poly of the same degree n α e kx A e kx (α + β x) e kx (A x + B) e kx α cos(kx) + β sin(kx) A cos(kx) + B sin(kx) (α x + β) cos(kx) + (γ x + δ) sin(kx)(a x + B) cos(kx) + (C x + D) sin(kx) }{{} Make this guess, substitute in ( ), then choose A, B, C... in y p (x) so that y p + b y p + c y p = R(x)... Section 1 Second-Order Linear ODEs 19 / 69

21 Examples: 1 y + y = x 2 Step 1: y h + y h = 0 homogeneous r = 0 r = ±i y h = C 1 cos x + C 2 sin x Step 2: We guess that y p = A x 2 + B x + C (see the Table) y p = 2A x + B y p = 2A Substitute in the ODE: }{{} 2A + (A x 2 + B x + C) = x }{{} 2 =y p =y p A x 2 + B x + (2A + C) = 1 x 2 A = 1, B = 0, (2A + C) = 0 C = 2A = 2 y p = 1 x x = x 2 2 Finally, a general solution: y = (x 2 2) + C 1 cos x + C 2 sin x Section 1 Second-Order Linear ODEs 20 / 69

22 2 y y 2y = 10 cos t subject to y(0) = 0 and y (0) = 2. Plan: (i) find a general solution of the ODE; (ii) use the initial conditions. (i) Step 1: y h y h 2y h = 0 homogeneous r 2 r 2 = 0 r 1 = 2, r 2 = 1 y h = C 1 e 2t + C 2 e t Step 2: Our guess is y p = A cos t + B sin t y p = A sin t + B cos t y p = A cos t B sin t So y p y p 2y p = cos t ( A B 2 A ) + sin t ( B + A 2 B ) = cos t ( 3 A B ) + sin t ( A 3 B ) = 10 cos t } 3 A B = 10 A 3 B = 0 B = 1, A = 3 y p = 3 cos t sin t General solution: y = ( 3 cos t sin t) + C 1 e 2t + C 2 e t Section 1 Second-Order Linear ODEs 21 / 69

23 Example 2 (continued): (ii) Use initial conditions y(0) = 0: 0 = ( 3 cos }{{ 0} sin }{{} 0) + C C 2 1 C 1 + C 2 = 3 =1 =0 y (0) = 2: Note that y = (3 sin t cos t) + 2 C 1 e 2t C 2 e t 2 = (3 sin }{{} 0 cos }{{ 0} ) + 2 C 1 1 C C 1 C 2 = 3 } C =0 =1 1 + C 2 = 3 2 C 1 C 2 = 3 C 1 = 2, C 2 = 1 y = ( 3 cos t sin t) + 2 e 2t + e t Section 1 Second-Order Linear ODEs 22 / 69

24 Remark 1 Remark 1: Consider y + b y + c y = R 1 (x) + R 2 (x) ( ) If y 1 + b y 1 + c y 1 = R 1 (x) and y 2 + b y 2 + c y 2 = R 2 (x) then y 1 + y 2 is a particular solution of ( ). Section 1 Second-Order Linear ODEs 23 / 69

25 Further Examples: 3 y + 4y = sin t + 2 e t. Step 1: y h + 4y h = 0 homogeneous r = 0, r = ±2 i y h = C 1 cos(2t) + C 2 sin(2t) Step 2: R(x) Try y p (x) = Our Table sin t A cos t + B sin t e t C e t Remark 1 sin t + 2 e t A cos t + B sin t + C e t So y p = A cos t + B sin t + C e t y p = A cos t B sin t+c e t y p +4y p = ( A cos t B sin t +C e t) +4 ( A cos t +B sin t +C e t) 3 A = 0 = 3 A cos t + 3 B sin t + 5 C e t = sin t + 2 e t 3 B = 1 A = 0, B = 1 3, C = 2 5 y p = 1 3 sin t e t 5 C = 2 General solution: y = 1 3 sin t e t + C 1 cos(2t) + C 2 sin(2t) Section 1 Second-Order Linear ODEs 24 / 69

26 (Very Important) Remark 2 If the guess from the Table happens to be a particular solution of the homogeneous equation ( ), then use y p = x (guess from Table ). If the new guess also happens to be a particular solution of the homogeneous equation ( ), then use y p = x 2 (guess from Table ) Section 1 Second-Order Linear ODEs 25 / 69

27 Further Examples: y 2 y 3 y = e x Step 1: y h = C 1 e 3x + C 2 e x (check!) Step 2: The Table suggests: y p = A e x, but it is a particular case of y h : it s clear from the y h formula (set C 1 = 0 and C 2 = A); alternatively, one can see this directly: (A e x ) 2 (A e x ) 3 (A e x ) = 0 e x. Clearly, y p = A e x doesn t work! By Remark 2, try y p = x A e x : y p = A (1 x) e x, y p = A (x 2) e x y p 2 y p 3 y p = A ( (x 2) 2 (1 x) 3 x ) e x = A ( 4) e x = e x 4A = 1 A = 1 4 y p = 1 4 x e x y = 1 4 x e x + C 1 e 3x + C 2 e x Section 1 Second-Order Linear ODEs 26 / 69

28 5 y + 4y + 4y = e 2t Step 1: r 2 + 4r + 4 = 0 r = 2 single root y h = (C 1 + C 2 t) e 2t Step 2: The Table suggests y p = A e 2t But (A e 2t ) + 4(A e 2t ) + 4(A e 2t ) = 0, e 2t (can be checked directly; of from the y h formula...) By Remark 2, try y p = t A e 2t again won t work as By Remark 2, now try (t A e 2t ) + 4(t A e 2t ) + 4(t A e 2t ) = 0, e 2t (can be checked directly; of from the y h formula...) y p = t 2 A e 2t y p = (2t 2t 2 ) A e 2t, y p = (2 8t + 4t 2 ) A e 2t y p + 4y p + 4y p = ( (2 8t + 4t 2 ) + 4 (2t 2t 2 ) + 4 t 2) A e 2t = 2 A e 2t = e 2t A = 1 2 y p = 1 2 t2 e 2t y = 1 2 t2 e 2t + (C 1 + C 2 t) e 2t Section 1 Second-Order Linear ODEs 27 / 69

29 6 y + 4y = 8 cos(2t) Step 1: y h = C 1 cos(2t) + C 2 sin(2t) (see Example 3). Step 2: The Table suggests y p = A cos(2t) + B sin(2t) But will NOT work as it s a particular case of the y h formula... By Remark 2, try y p = t (A cos(2t) + B sin(2t) ) y p + 4y p = 4 B cos(2t) + 4 ( A) sin(2t) = 8 cos(2t) B = 2, A = 0 so y p = 2t sin(2t) Finally, we get the Answer: y = 2t sin(2t) + C 1 cos(2t) + C 2 sin(2t) NOTE: this phenomenon is referred to as resonance (see the prime text...) Section 1 Second-Order Linear ODEs 28 / 69

30 Problems for Find general solutions of the following differential equations: 1 y 2y 3y = 2x 2 y 2y 3y = 11 + cos x y y = 5 sin(3x) 4 y y = e x 5 y y = e x 6 y y = 3 7 y y = 7x 8 y y = x e 2x y 2y + 3y = x 3 10 y 4y = e 2x 11 y 3y = 2e 2x sin x Section 1 Second-Order Linear ODEs 29 / 69

31 2 Find unique solutions of the following initial-value problems 1 y + y 2y = e x + e 2x, y(0) = 0, y (0) = 4/3 2 y + 4y = 3x cos x, y(0) = 1, y(π/4) = 0 3 y 4y + 4y = e 2x, y(0) = y (0) = 0 3 Find unique solutions of the following initial-value problems 1 4y 4y + y = 5 2 e x/2, y(0) = 1, y (0) = 3 2 y + 3y = 10 sin x 6, y(0) = 3, y (0) = 0 3 y 2y 3y = 8e 3x, y(0) = 4, y (0) = 2 4 y 2y + 5y = 25 x 2, y(0) = 1.6, y (0) = 4. Section 1 Second-Order Linear ODEs 30 / 69

32 1.6 Variation of Parameters Applies to y + p(x) y + q(x) y = R(x) ( ) NOTE: the coefficients are NOT necessarily constant! Step 1: Find a general solution of the corresponding homogeneous equation y h + p y h + q y h = 0 ( ) (e.g., as in 1.2): So we get y = C 1 y 1 (x) + C 2 y 2 (x) ( ) Step 2: Replace the constants C 1 and C 2 in (***) by the unknown functions u(x) and v(x): y = u(x) y 1 (x) + v(x) y 2 (x) ( ) Step 3: Next, find u(x) and v(x) such that ( ) gives a particular solution of ( ) (more detail on next page) Step 4: Finally, add ( ) to ( ), to get a general solution: y = u(x) y 1 (x) + v(x) y 2 (x) + C 1 y 1 (x) + C 2 y 2 (x) Section 1 Second-Order Linear ODEs 31 / 69

33 Step 3 more DETAIL: substitute ( ) in our ODE ( ) y = u(x) y 1 (x) + v(x) y 2 (x) ( ) (i) Evaluate y (x): y = [u y 1 + v y 2 ] + [u y 1 + v y 2 ] always ASSUME u y 1 + v y 2 = 0 This greatly simplifies y : y = u y 1 + v y 2 (ii) Evaluate y (x): y = [u y 1 + v y 2 ] + [u y 1 + v y 2 ] (iii) Substitute y (x), y (x) and y(x) in in our ODE ( ); combining this with y 1 + py 1 + qy 1 = 0 and y 2 + py 2 + qy 2 = 0 yields u y 1 + v y 2 = R(x) CONCLUSION: Step 3 reduces to solving the system of 2 equations for u and v : u y 1 + v y 2 = 0 ; u y 1 + v y 2 = R(x) Section 1 Second-Order Linear ODEs 32 / 69

34 CONCLUSION (from previous page): Step 3 reduces to solving the system of 2 equations for u and v : u y 1 + v y 2 = 0 ; u y 1 + v y 2 = R(x) This matrix form of this system is [ ] [ ] [ ] y1 y 2 u 0 y 1 y 2 v = R(x) and the solution is u (x) = y 2 R(x) W (x), v (x) = y 1 R(x) W (x), where W (x) = y 1 y 2 y 2 y 1 It remains to apply integration: y2 R(x) u(x) = W (x), v(x) = y1 R(x) W (x), where W (x) = y 1 y 2 y 2 y End of Step Section 1 Second-Order Linear ODEs 33 / 69

35 EXAMPLES: Using variation of parameters, find the general solutions of the following ODEs. 1 y 5y + 6y = 2e x. Answer: y(x) = e x + C 1 e 2x + C 2 e 3x. NOTE: Variation of parameters is not the best method for this example. But this method works for any R(x). 2 y + 2y + y = e x ln x. [ x 2 3x 2 ] Answer: y(x) = ln x C 1 + C 2 x e x. Section 1 Second-Order Linear ODEs 34 / 69

36 Problems for 1.6 Find general solutions of the following differential equations using variation of parameters : 1 y 5y + 6y = 2e x Answer: y = e x + C 1 e 2x + C 2 e 3x 2 y y = sin 2 x 3 y 4y + 4y = e2x 1 + x 4 y + 4y + 4y = x 2 e 2x ; Answer: y = e 2x ln x + C 1 e 2x + C 2 x e 2x 5 y 2y + y = ex 1 + x 2 ; Answer: y = 1 2 ex ln(1 + t 2 ) + x e x tan 1 x + C 1 e x + C 2 x e x Section 1 Second-Order Linear ODEs 35 / 69

37 1.7 Euler Equations Euler Equations are equations of the type x 2 y + b x y + c y = R(x) ( ) where x > 0, and b and c are constants. NOTE: This is a linear second-order ODE. The coefficients are NOT constant, but 1.1, 1.4 and 1.6 apply A simple TRICK reduces this ODE to the constant-coefficient case: Now, a calculation using d dx = dt d dx = 1 x d dt and then (CHECK!!!) d 2 y dt 2 Let t = ln x x = e t dx d dt and dt + (b 1) dy dt + c y = R(e t ) dx = d dx ln x = 1 x yields Section 1 Second-Order Linear ODEs 36 / 69

38 NOTE: depending on R(x), it may be convenient to apply this trick to the original nonhomogeneous equation or to its homogeneous version only EXAMPLES: 1 Find general solutions for the following homogeneous Euler equations: 1 x 2 y 7xy + 15y = 0 Answer: y(x) = C 1 x 5 + C 2 x 3 (see Problems for 1.1). 2 x 2 y + xy + y = 0 Answer: y(x) = C 1 cos(ln x) + C 2 sin(ln x). Section 1 Second-Order Linear ODEs 37 / 69

39 NOTE: depending on R(x), it may be convenient to apply the substitution t = ln x to the original nonhomogeneous equation (as we shall demonstrate in Examples ) or to its homogeneous version only (as we shall show in Example 2.3; then variation of parameters may be used...) EXAMPLES: 2 Find general solutions for the following nonhomogeneous Euler equations: 1 x 2 y 7xy + 15y = x 2 Answer: y(x) = 1 3 x 2 + C 1 x 5 + C 2 x 3. 2 x 2 y 2y = 3x 2 1 Answer: y(x) = x 2 ln x + C 1 x 2 + C 2 x x 2 y 2xy + 2y = 4x 2 Answer: y(x) = 4x 2 ln x + C 1 x + C 2 x 2. Solution: Here we shall use the substitution t = ln x find a solution of the corresponding homogeneous equation. Then we shall use variation of parameters to find a particular solution of the nonhomogeneous equation. (Hint: don t forget to rewrite as y +... for variation of parameters). Section 1 Second-Order Linear ODEs 38 / 69

40 Problems for Find general solutions for the following homogeneous Euler equations: 1 x 2 y + 4xy + 2y = 0 Answer: y(x) = C 1 x 1 + C 2 x 2. 2 x 2 y 4xy 6y = 0 Answer: y(x) = C 1 x 6 + C 2 x 1. 3 x 2 y + 3xy y = 0 Answer: y(x) = C 1 x 1 cos( 1 2 ln x) + C 2x 1 sin( 1 2 ln x). 4 x 2 y + 5xy + 4y = 0 Answer: y(x) = C 1 x 2 + C 2 x 2 ln x. 5 x 2 y xy + 5y = 0 Answer: y(x) = C 1 x cos(2 ln x) + C 2 x sin(2 ln x). 6 x 2 y + 7xy + 10y = 0 Answer: y(x) = C 1 x 3 cos(ln x) + C 2 x 3 sin(ln x). Section 1 Second-Order Linear ODEs 39 / 69

41 2 Find general solutions for the following nonhomogeneous Euler equations: 1 x 2 y + 4xy + 2y = x 3 5 Answer: y(x) = 1 20 x C 1x 1 + C 2 x 2. 2 x 2 y 4xy 6y = 3 sin(2 ln x) Answer: y(x) = cos(2 ln x) 20 sin(2 ln x) + C 1x 6 + C 2 x 1. 3 x 2 y 3xy + 4 y = x 2 ln x Answer: y(x) = 1 6 x 2 (ln x) 3 + C 1 x 2 + C 2 x 2 ln x. 3 Find general solutions for the following nonhomogeneous Euler equations using variation of parameters. 1 x 2 y 7xy + 15y = x 2 Answer: y(x) = 1 3 x 2 + C 1 x 5 + C 2 x 3. 2 x 2 y + 4xy + 2y = x 3 5 Answer: y(x) = 1 20 x C 1x 1 + C 2 x 2. Section 1 Second-Order Linear ODEs 40 / 69

42 Section 3 Numerical Solution of First-Order ODEs Consider the Initial-Value Problem y = f (x, y), y(x 0 ) = y 0 ( ) this can be interpreted as that at each x, the slope of the curve y(x) is f (x, y) If we can NOT solve ( ) explicitly, we can find approximate values y n y(x n ) at the points x n = x 0 + n h: Section 3 Numerical Solution of First-Order ODEs 41 / 69

43 3.1 Euler Method Integrate y = f (x, y) over the interval [x n, x n+1 ]: xn+1 x n y dx }{{} = x n+1 x n =y(x) x n+1 =y(x n+1 ) y(x n ) xn f (x, y) dx So y(x n+1 ) y(x n ) = x n+1 x n f (x, y(x)) dx h f (x n, y(x n )) here we used the Rectangular Rule of Numerical Integration Now, we replace the exact values y(x n ) by approximate values y n, and also replace by =, and so get the definition of a numerical method: y(x n+1 ) y n+1 = y n + h f (x n, y n ) ( ) called the Euler Method. NOTE: one can rewrite ( ) as y(x 0 ) = y 0 (use Initial Condition) y(x 0 + h) y 1 = y 0 + h f (x 0, y 0 ) y(x 0 + 2h) y 2 = y 1 + h f (x 1, y 1 ) y(x 0 + 3h) y 3 = y 2 + h f (x 2, y 2 ) i.e. the computation goes as y 0 y 1 y 2 y 3 Section 3 Numerical Solution of First-Order ODEs 42 / 69

44 INTERPRETATION: note that y n+1 y n x n+1 x n = y n+1 y n h = f (x n, y n ), }{{} by the Euler method ( ) i.e. the slope of the computed solution on each (x n, x n+1 ) is f (x n, y n ): Section 3 Numerical Solution of First-Order ODEs 43 / 69

45 Example: consider the IVP: y = x + y subject to y(0) = 0. Exercise: show that the exact solution is y(x) = e x x 1. Euler Method: Choose h = 0.2 so x n = x 0 + h n = 0.2 n ; y(0) = y 0 = 0 Using f (x, y) = x + y one gets y(x n+1 ) = y(0.2[n + 1]) y n+1 = y n + h f (x n, y n ) = y n (x n + y n ) n x n = 0.2n computed y n exact y(x n ) error y(x n ) y n Error y(x n ) y n : at each step, the Euler method picks up an error of order h 2, but the error accumulate from step to step, so at x = x n, the error is of order n h 2 = (nh) h = x n h h. }{{} length of the interval Section 3 Numerical Solution of First-Order ODEs 44 / 69

46 Problems for Use the Euler method with the following step sizes h to approximate y(0.4) given that y = cos y and y(0) = 0: 1 h = 0.2 Answer: y 2 = h = 0.1 Answer: y 4 = h = 0.05 Answer: y 8 = Section 3 Numerical Solution of First-Order ODEs 45 / 69

47 3.2 The improved Euler Method (Predictor-Corrector) y(x n+1 ) y(x n ) = x n+1 x n f (x, y(x)) dx 1 2 }{{} h [ f (x n, y(x n )) + f (x n+1, y(x n+1 )) ] Recall from 23.1 here we use the Trapezoidal Rule of Numerical Integration. If we replace the exact values y(x n ) by approximate values y n, and also replace by =, then we get a numerical method: y n+1 = y n h [ f (x n, y n ) + f (x n+1, y n+1 ) ]. This method is expected to be more accurate that the Euler method. However, we have the unknown value y n+1 on the right-hand side!! To get an easier-to-implement method, replace this y n+1 by the Euler approximation from 23.1 (that we now denote yn+1 ), so we arrive at Improved Euler Method y(x 0 ) = y 0 ; y n+1 = y n + h f (x n, y n ) = predictor stage; y(x n+1 ) y n+1 = y n h [ f (x n, y n ) + f (x n+1, y n+1 )] = corrector stage Section 3 Numerical Solution of First-Order ODEs 46 / 69

48 Example: consider the IVP: y = x + y subject to y(0) = 0. Improved Euler Method: Choose h = 0.2 so x n = x 0 + h n = 0.2 n. y(0) = y 0 = 0 Note that f (x, y) = x + y so yn+1 = y n + h f (x n, y n ) = y n (x n + y n ) Next we get: y(x n+1 ) y n+1 = y n h [ f (x n, y n ) + f (x n+1, y n+1 )] = y n [ (x n + y n ) + (x n+1 + y n+1 )] n x n predictor yn corrected y n exact y(x n ) error y(x n ) y n Section 3 Numerical Solution of First-Order ODEs 47 / 69

49 Error y(x n ) y n : at each step, the Improved Euler method picks up an error of order h 3, but the error accumulate from step to step, so at x = x n, the error is of order n h 3 = (nh) h 2 = x n h 2 h 2. }{{} length of the interval E.g.: at x n = 1 the error is of order x n h 2 = 1 h 2 = h 2 (x n = 1 is a representative choice as it s not too big, not too small). So the Improved Euler Method is a second-order method (while the Euler method of 23.1 is a first-order method) In real-world applications, one uses more accurate methods, e.g., Fourth-Order Runge-Kutta methods (see the prime text for further details). Its error at x = x n is of order n h 5 = (nh) h 4 = x n h 4 ; }{{} length of the interval in particular, at x n = 1, the error is of order h Section 3 Numerical Solution of First-Order ODEs 48 / 69

50 Another Example: consider the IVP: y = x y subject to y(0) = 1 on [0, 1] with the step size h = 0.2 using the Euler and the Improved Euler methods. Exercise: show that the exact solution is y(x) = x e x. Solution: We have f (x, y) = x y, x 0 = 0, and y(0) = y 0 = 1 also h = 0.2 so x n = x 0 + h n = 0.2 n for n = 0,, 5. (a) Euler: y(x n+1 ) y n+1 = y n + h f (x n, y n ), So y n+1 = y n (x n y n ) subject to y 0 = 1. (b) Improved Euler: y n+1 = y n + h f (x n, y n ) = y n (x n y n ) (as above). Next we get: y(x n+1 ) y n+1 = y n h [ f (x n, y n ) + f (x n+1, y n+1 )] = y n [ ( x n }{{} 0.2 n y n ) + ( x }{{} n+1 yn+1 )] 0.2 (n+1) Section 3 Numerical Solution of First-Order ODEs 49 / 69

51 Numerical Results: Euler Improved Euler n x n y n error y(x n ) y n y n error y(x n ) y n } {{ }} {{} the error is over 10% the error is less than 1%! Section 3 Numerical Solution of First-Order ODEs 50 / 69

52 Problems for Use the improved Euler method with the following step sizes h to approximate y(0.4) given that y = cos y and y(0) = 0: 1 h = 0.2 Answer: y 2 = h = 0.1 Answer: y 4 = h = 0.05 Answer: y 8 = Write down the iterative scheme of the Improved Euler method applied to the initial value problem y = x + y 2, y(0) = 2 with step size h = 0.2. Evaluate the approximations of y(0.2) and y(0.4) obtained using this scheme. Answer: y(0.2) and y(0.4) Write down the iterative scheme of the Improved Euler method applied to the initial value problem y = x 3 + y, y(0) = 3 with step size h = 0.1. Evaluate the approximations of y(0.1) and y(0.2) obtained using this scheme. Answer: y(0.2) Section 3 Numerical Solution of First-Order ODEs 51 / 69

53 4 Write down the iterative scheme of the Improved Euler method applied to the initial value problem y = ln(2y x), y(0) = 2 with step size h = 0.2. Evaluate the approximations of y(0.2) and y(0.4) obtained using this scheme. Answer: y(0.4) Write down the iterative scheme of the Improved Euler method applied to the initial value problem y = e x y 2, y(0) = 1 with step size h = 0.1. Evaluate the approximations of y(0.1), y(0.2) and y(0.3) obtained using this scheme. Answer: y(0.3) Write down the iterative scheme of the Improved Euler method applied to the initial value problem y = e xy, y(0) = 0 with step size h = 0.2. Evaluate the approximations of y(0.2), y(0.4) and y(0.6) obtained using this scheme. Answer: y(0.6) Section 3 Numerical Solution of First-Order ODEs 52 / 69

54 7 Write down the iterative scheme of the Improved Euler method applied to the initial value problem y = x y 2, y(0) = 2 with step size h = 0.1. Evaluate the approximations of y(0.1), y(0.2) and y(0.3) obtained using this scheme. Answer: y(0.3) Write down the iterative scheme of the Improved Euler method applied to the initial value problem y = cos(x + y), y(0) = 0 with step size h = 0.1. Evaluate the approximations of y(0.1) and y(0.2) obtained using this scheme. Answer: y(0.2) Section 3 Numerical Solution of First-Order ODEs 53 / 69

55 Section 2 Boundary Value Problems 2.1 Examples Consider a second-order linear ODE y + p(x) y + q(x) y = R(x) ( ) RECALL: A particular solution is found by obtaining values for the 2 arbitrary constants in its general solution from 2 initial or boundary conditions, e.g., y(x 0 ) = A, y (x 0 ) = B Initial conditions y(x 0 ) = A, y(x 1 ) = B Boundary conditions NOTE: An Initial-Value Problem (i.e. an ODE subject to initial conditions) typically has a unique solution. For Boundary-Value Problems (i.e. and ODE subject to boundary conditions) the situation is much more complex. We shall illustrate this considering a few examples. Section 2 Boundary Value Problems 54 / 69

56 Boundary Conditions Terminology Typically there are 2 boundary conditions for a second-order ODE: at some x 0 and x 1 > x0. There are 3 types of boundary conditions: y(x k ) = A y (x k ) = A αy (x k ) + βy(x k ) = A called a Dirichlet boundary condition called a Neumann boundary condition called a mixed boundary condition Section 2 Boundary Value Problems 55 / 69

57 Example 1: [Homogeneous ODE] y + y = 0 (it general solution is y(x) = C 1 cos x + C 2 sin x) subject to the following boundary conditions 1 y(0) = y(1) = 0 Answer: UNIQUE solution y(x) = 0 2 y(0) = y(π) = 0 Answer: y(x) = C sin x ( solutions) 3 y(0) = 2, y(π) = 0 Answer: NO solutions 4 y (0) = y( π 2 ) = 0 Answer: y(x) = C cos x ( solutions) 5 y (0) = y( π 2 + ε) = 0 for ε 0, 0 Answer: UNIQUE solution y(x) = 0 Section 2 Boundary Value Problems 56 / 69

58 Example 1 : [NON-Homogeneous ODE] y + y = 3 (it general solution is y(x) = 3 + C 1 cos x + C 2 sin x) subject to the following boundary conditions 1 y(0) = y(1) = 0 Answer: UNIQUE solution y(x) = 3 cos x 3 cos 1 sin 1 sin x 2 y(0) = y(π) = 0 Answer: NO solutions 3 y(0) = 2, y(π) = 0 Answer: NO solutions 4 y (0) = y( π 2 ) = 0 Answer: NO solutions 5 y (0) = y( π 2 + ε) = 0 for ε 0, 0 Answer: UNIQUE solution... Section 2 Boundary Value Problems 57 / 69

59 Example 2: [Another Homogeneous ODE] y 2y 3y = 0 (it general solution is y(x) = C 1 e x + C 2 e 3x ) subject to the following boundary conditions 1 y(0) = 5, y(+ ) = 0 Answer: UNIQUE solution y(x) = 5e x 2 y (0) = 3, y(+ ) = 0 Answer: UNIQUE solution y(x) = 3e x 3 y (0) 2y(0) = 7, y (+ ) = 0 Answer: UNIQUE solution y(x) = 7 3 e x Section 2 Boundary Value Problems 58 / 69

60 Example 3: [similar-looking Homogeneous ODE] y + 4y + 5y = 0 (it general solution is y(x) = C 1 e 2x cos x + C 2 e 2x sin x) subject to the following boundary conditions 1 y(0) = 5, y(+ ) = 0 Answer: ( solutions) y(x) = 5e 2x cos x + Ce 2x sin x 2 y (0) = 0, y(+ ) = 0 Answer: ( solutions) y(x) = Ce 2x [cos x + 2 sin x] 3 y (0) 2y(0) = 7, y (+ ) = 0 Answer: ( solutions) y(x) = Ce 2x cos x + (7 3C)e 2x sin x Section 2 Boundary Value Problems 59 / 69

61 Problems for Solve all above examples in detail. 2 (I) Find general solutions of the following differential equations. (II)Describe all solutions of each ODE subject to the given boundary conditions. How many solutions does this boundary value problem have? 1 y + 5y + 6y = 0 subject to y (0) = 0, y (+ ) = 0 2 y + 5y + 6y = 3e 2x subject to y (0) = 0, y(+ ) = 0 3 y + 4y = 3e 2x subject to y (0) = 0, y(+ ) = y + y = 0 subject to y (0) = y (π) = 0 5 y + 4y = 0 subject to y (0) = y (π) = 0 6 y + 5y = 0 subject to y (0) = y (π) = 0 7 y + 9y = 0 subject to y (0) = y (π) = 0 Section 2 Boundary Value Problems 60 / 69

62 Answers Problems for (I) Find general solutions of the following differential equations. (II)Describe all solutions of each ODE subject to the given boundary conditions. How many solutions does this boundary value problem have? 1 y + 5y + 6y = 0, y (0) = y (+ ) = 0 Answer: y(x) = C [ e 2x 2 3 e 3x] 2 y + 5y + 6y = 3e 2x, y (0) = y(+ ) = 0 Answer: y(x) = 3x e 2x e 3x + C [ e 2x 2 3 e 3x] 3 y + 4y = 3e 2x, y (0) = y(+ ) = 0 Answer: NO solutions y + y = 0, y (0) = y (π) = 0 Answer: y(x) = C 1 cos x 5 y + 4y = 0, y (0) = y (π) = 0 Answer: y(x) = C 1 cos(2x) 6 y + 5y = 0, y (0) = y (π) = 0 Answer: y(x) = 0 7 y + 9y = 0, y (0) = y (π) = 0 Answer: y(x) = C 1 cos(3x) Section 2 Boundary Value Problems 61 / 69

63 2.2 BVPs with λ In this section, we shall consider Boundary Value Problems that involve a parameter λ. TASK: solve the given BVP (= ODE + 2 boundary conditions) for all real values of λ NOTE: such problems are called Eigenproblems; they are very important in the theory of Partial Differential Equations. For each real λ there is at least one solution y(x) = 0, called the trivial solution. For certain values of λ, this solution is NOT unique and there are other, nontrivial solutions. Such values of λ are called Eigenvalues, while the corresponding nontrivial solutions are called Eigenfunctions Section 2 Boundary Value Problems 62 / 69

64 EXAMPLE: (I) For each value of λ, describe all solutions of the following boundary value problem: y + λy = 0 subject to y(0) = y(π) = 0 (II) Specify the values of λ for which nontrivial solutions occur. HINT: one typically considers 3 cases λ = 0, λ < 0 and λ > 0... ANSWER: (I) For λ = n 2, where n = 1, 2, 3,, the general solution is y(x) = C sin(nx); For all other values of λ, there is a unique solution y(x) = 0. (II) Nontrivial solutions occur for λ = n 2, where n = 1, 2, 3,. Section 2 Boundary Value Problems 63 / 69

65 Problems for 2.2 TASK: (I) For each value of λ, describe all solutions of each of the following boundary value problems. (II) Specify the values of λ for which nontrivial solutions occur. 1 y + λy = 0 subject to y(0) = y (1) = 0 2 y + λy = 0 subject to y (0) = y (1) = 0 3 y + λy = 0 subject to y (0) = y(π) = 0 4 y + λy = 0 subject to y(0) = y(1) = y + λy = 0 subject to y( π) = y(π) and y ( π) = y (π) 6 y + λy = 0 subject to y (0) = 0 and y (1) + λ y(1) = 0 Section 2 Boundary Value Problems 64 / 69

66 1 y + λy = 0 subject to y(0) = y (1) = 0 ANSWER: (I) For λ = ( π 2 + πn)2, where n = 0, 1, 2, 3,, the general solution is y(x) = C sin([ π 2 + πn]x); For all other values of λ, there is a unique solution y(x) = 0. (II) Nontrivial solutions occur for λ = ( π 2 + πn)2, where n = 0, 1, 2, 3,. 2 y + λy = 0 subject to y (0) = y (1) = 0 ANSWER: (I) For λ = 0 the general solution is y(x) = C; for λ = (πn) 2, where n = 1, 2, 3,, the general solution is y(x) = C cos(πnx); For all other values of λ, there is a unique solution y(x) = 0. (II) Nontrivial solutions occur for λ = 0 and λ = (πn) 2, where n = 1, 2, 3,. Section 2 Boundary Value Problems 65 / 69

67 3 y + λy = 0 subject to y (0) = y(π) = 0 ANSWER: (I) For λ = ( n)2, where n = 0, 1, 2, 3,, the general solution is y(x) = C cos([ n]x); For all other values of λ, there is a unique solution y(x) = 0. (II) Nontrivial solutions occur for λ = ( n)2, where n = 0, 1, 2, 3,. 4 y + λy = 0 subject to y(0) = y(1) = 0 ANSWER: (I) For λ = (πn) 2, where n = 1, 2, 3,, the general solution is y(x) = C sin(πnx); For all other values of λ, there is a unique solution y(x) = 0. (II) Nontrivial solutions occur for λ = (πn) 2, where n = 1, 2, 3,. Section 2 Boundary Value Problems 66 / 69

68 5 y + λy = 0 subject to y( π) = y(π) and y ( π) = y (π) ANSWER: (I) For λ = 0 the general solution is y(x) = C; for λ = n 2, where n = 1, 2, 3,, the general solution is y(x) = C 1 cos(nx) + C 2 sin(nx); For all other values of λ, there is a unique solution y(x) = 0. (II) Nontrivial solutions occur for λ = 0 and λ = n 2, where n = 1, 2, 3,. 6 y + λy = 0 subject to y (0) = 0 and y (1) + λ y(1) = 0 ANSWER: (I) For λ = 0 the general solution is y(x) = C; for λ = [ π 4 + πn]2, where n = 1, 2, 3,, the general solution is y(x) = C cos([ π 4 + πn]x); For all other values of λ, there is a unique solution y(x) = 0. (II) Nontrivial solutions occur for λ = 0 and λ = [ π 4 + πn]2, where n = 1, 2, 3,. Section 2 Boundary Value Problems 67 / 69

69 FINAL EXAM Q 3+4 PROBLEM TYPES that may be given: Euler Equations ( 1.7 combined with ) Variation of Parameters ( 1.6 combined with ) Boundary-Value Problem / Initial-Value Problem (All of the above combined with 2.1 / given initial conditions.) For each value of λ, describe all solutions of a given boundary value problem. Specify the values of λ for which nontrivial solutions occur. ( 2.2 combined with ) Euler method / Improved Euler method ( 3.1 / 3.2) FINAL EXAM Q / 69

70 FINAL EXAM Q 3+4 What formulas will be given USEFUL FORMULAE: The method of Variation of Parameters for the equation y + p(x)y + q(x)y = R(x) yields a particular solution in the form u y 1 + v y 2, where y 1 and y 2 are linearly independent solutions of the corresponding homogeneous equation, while u and v satisfy u y 1 + v y 2 = 0, u y 1 + v y 2 = R(x). Method of Undetermined Coefficients Right-Hand Side Possible Particular Solution α A α x + β A x + B α x 2 + β x + γ A x 2 + B x + C α e kx A e kx (α + β x) e kx (A x + B) e kx α cos(kx) + β sin(kx) A cos(kx) + B sin(kx) (α x + β) cos(kx) + (γ x + δ) sin(kx) (A x + B) cos(kx) + (C x + D) sin(kx) FINAL EXAM Q / 69