Find the Fourier series of the odd-periodic extension of the function f (x) = 1 for x ( 1, 0). Solution: The Fourier series is.

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1 Review for Final Exam. Monday /09, :45-:45pm in CC-403. Exam is cumulative, -4 problems. 5 grading attempts per problem. Problems similar to homeworks. Integration and LT tables provided. No notes, no books, no calculators. Heat Eq. and Fourier Series (Chptr.6. Eigenvalue-Eigenfunction BVP (Chptr. 6. Systems of linear Equations (Chptr. 5. Laplace transforms (Chptr. 4. Second order linear equations (Chptr.. First order differential equations (Chptr.. Fourier Series Find the Fourier series of the odd-periodic extension of the function f (x = for x (, 0. Solution: The Fourier series is f (x = a 0 + [ a n cos L n= ] + b n sin. L Since f is odd and periodic, then the Fourier Series is a Sine Series, that is, a n = 0. b n = L f (x sin dx = L f (x sin dx. L L L 0 L b n = L 0 ( sin(nπx dx = ( ( nπ cos(nπx 0, b n = [ ] cos(nπ nπ b n = nπ [ ( n ].

2 Fourier Series Find the Fourier series of the odd-periodic extension of the function f (x = for x (, 0. Solution: Recall: b n = nπ If n = k, then b k = kπ [ ( n ]. [ ( k ] = 0. If n = k, [ b (k = ( k ] 4 = (k π (k π. We conclude: f (x = 4 π k= (k sin[(k πx]. Fourier Series Find the Fourier series of the odd-periodic extension of the function f (x = x for x (0,. Solution: The Fourier series is f (x = a 0 + [ a n cos L n= ] + b n sin. L Since f is odd and periodic, then the Fourier Series is a Sine Series, that is, a n = 0. b n = L L L f (x sin dx = L L L 0 f (x sin dx, L =, L b n = 0 ( x sin dx.a

3 Fourier Series Find the Fourier series of the odd-periodic extension of the function f (x = x for x (0,. Solution: b n = I = 0 sin dx sin The other integral is done by parts, x sin dx, I = x nπ cos 0 x sin dx. dx = nπ cos, u = x, v = sin u =, v = nπ cos ( cos dx. nπ Fourier Series Find the Fourier series of the odd-periodic extension of the function f (x = x for x (0,. Solution: I = x nπ I = x nπ cos ( cos nπ ( + sin nπ cos dx.. So, we get [ ] b n = nπ cos [ x ] + 0 nπ cos ( sin 0 nπ 0 b n = 4 [ ] [ 4 ] cos(nπ + nπ nπ cos(nπ 0 We conclude: f (x = 4 π n= b n = 4 nπ. n sin.

4 Fourier Series Find the Fourier series of the even-periodic extension of the function f (x = x for x (0,. Solution: The Fourier series is f (x = a 0 + [ a n cos L n= ] + b n sin. L Since f is even and periodic, then the Fourier Series is a Cosine Series, that is, b n = 0. a 0 = a n = L L L f (x dx = 0 f (x cos L a n = 0 ( x dx = base x height L dx = L f (x cos L 0 ( x cos dx. a 0 =. dx, L =, Fourier Series Find the Fourier series of the even-periodic extension of the function f (x = x for x (0,. Solution: a n = I = 0 cos dx cos dx = nπ sin The other integral is done by parts, x cos dx, 0 x cos dx., u = x, v = cos u =, v = nπ sin I = x nπ sin nπ sin dx.

5 Fourier Series Find the Fourier series of the even-periodic extension of the function f (x = x for x (0,. Solution: Recall: I = x nπ sin nπ sin dx. I = x ( nπ sin + cos. So, we get nπ [ ] a n = nπ sin [ x ] 0 nπ sin ( cos 0 nπ 0 a n = n π [ cos(nπ ] a n = 4 n π [ ( n ]. Fourier Series Find the Fourier series of the even-periodic extension of the function f (x = x for x (0,. Solution: Recall: b n = 0, a 0 =, a n = 4 n π [ ( n ]. If n = k, then a k = If n = k, then we obtain a (k = 4 (k π [ ( k ] = 0. 4 (k π [ ( k ] = We conclude: f (x = + 8 π k= 8 (k π. ( (k πx (k cos.

6 Review for Final Exam. Heat Eq. and Fourier Series (Chptr.6. Eigenvalue-Eigenfunction BVP (Chptr. 6. Systems of linear Equations (Chptr. 5. Laplace transforms (Chptr. 4. Second order linear equations (Chptr.. First order differential equations (Chptr.. Eigenvalue-Eigenfunction BVP. Find the positive eigenvalues and their eigenfunctions of y + λ y = 0, y(0 = 0, y(8 = 0. Solution: Since λ > 0, introduce λ = µ, with µ > 0. y(x = e rx implies that r is solution of p(r = r + µ = 0 r ± = ±µi. The general solution is y(x = c cos(µx + c sin(µx. The boundary conditions imply: 0 = y(0 = c y(x = c sin(µx. 0 = y(8 = c sin(µ8, c 0 sin(µ8 = 0. µ = nπ ( nπ, 8, λ = yn (x = sin, n =,, 8 8

7 Eigenvalue-Eigenfunction BVP. Find the positive eigenvalues and their eigenfunctions of y + λ y = 0, y(0 = 0, y (8 = 0. Solution: The general solution is y(x = c cos(µx + c sin(µx. The boundary conditions imply: 0 = y(0 = c y(x = c sin(µx. 0 = y (8 = c µ cos(µ8, c 0 cos(µ8 = 0. 8µ = (n + π (n + π, µ =. 6 Then, for n =,, holds [ (n + π ], ( (n + πx λ = yn (x = sin 6 6. Eigenvalue-Eigenfunction BVP. Find the non-negative eigenvalues and their eigenfunctions of y + λ y = 0, y (0 = 0, y (8 = 0. Solution: Case λ > 0. Then, y(x = c cos(µx + c sin(µx. Then, y (x = c µ sin(µx + c µ cos(µx. The B.C. imply: 0 = y (0 = c y(x = c cos(µx, y (x = c µ sin(µx. 0 = y (8 = c µ sin(µ8, c 0 sin(µ8 = 0. 8µ = nπ, µ = nπ 8. Then, choosing c =, for n =,, holds ( nπ, λ = yn (x = cos 8 8.

8 Eigenvalue-Eigenfunction BVP. Find the non-negative eigenvalues and their eigenfunctions of y + λ y = 0, y (0 = 0, y (8 = 0. Solution: The case λ = 0. The general solution is The B.C. imply: y(x = c + c x. 0 = y (0 = c y(x = c, y (x = 0. Then, choosing c =, holds, 0 = y (8 = 0. λ = 0, y 0 (x =. A Boundary Value Problem. Find the solution of the BVP y + y = 0, y (0 =, y(π/3 = 0. Solution: y(x = e rx implies that r is solution of p(r = r + µ = 0 r ± = ±i. The general solution is y(x = c cos(x + c sin(x. Then, y (x = c sin(x + c cos(x. The B.C. imply: = y (0 = c y(x = c cos(x + sin(x. 0 = y(π/3 = c cos(π/3 + sin(π/3 c = sin(π/3 cos(π/3. 3/ c = / = 3 y(x = 3 cos(x + sin(x.

9 Review for Final Exam. Heat Eq. and Fourier Series (Chptr.6. Eigenvalue-Eigenfunction BVP (Chptr. 6. Systems of linear Equations (Chptr. 5. Laplace transforms (Chptr. 4. Second order linear equations (Chptr.. First order differential equations (Chptr.. Systems of linear Equations. Summary: Find solutions of x = A x, with A a matrix. First find the eigenvalues λ i and the eigenvectors v (i of A. (a If λ λ, real, then {v (, v ( } are linearly independent, and the general solution is x(x = c v ( e λ t + c v ( e λ t. (b If λ λ, complex, then denoting λ ± = α ± βi and v (± = a ± bi, the complex-valued fundamental solutions x (± = (a ± bi e (α±βit x (± = e αt (a ± bi [ cos(βt + i sin(βt ]. x (± = e αt [ a cos(βt b sin(βt ] ±ie αt [ a sin(βt+b cos(βt ]. Real-valued fundamental solutions are x ( = e αt [ a cos(βt b sin(βt ], x ( = e αt [ a sin(βt + b cos(βt ].

10 Systems of linear Equations. Summary: Find solutions of x = A x, with A a matrix. First find the eigenvalues λ i and the eigenvectors v (i of A. (c If λ = λ = λ, real, and their eigenvectors {v (, v ( } are linearly independent, then the general solution is x(x = c v ( e λt + c v ( e λt. (d If λ = λ = λ, real, and there is only one eigendirection v, then find w solution of (A λi w = v. Then fundamental solutions to the differential equation are given by x ( = v e λt, x ( = (v t + w e λt. Then, the general solution is x = c v e λt + c (v t + w e λt. Systems of linear Equations. [ ] [ ] Find the solution to: x 3 4 = A x, x(0 =, A =. Solution: p(λ = ( λ 4 ( λ = (λ (λ + 8 = λ 8, p(λ = λ 9 = 0 λ ± = ±3. Case λ + = 3, [ ] 4 A 3I = 4 [ ] 0 0 v = v v (+ = [ ] Case λ = 3, [ ] 4 4 A + 3I = [ ] 0 0 v = v v ( = [ ]

11 Systems of linear Equations. [ ] [ ] Find the solution to: x 3 4 = A x, x(0 =, A =. [ ] [ ] Solution: Recall: λ ± = ±3, v (+ =, v ( =. [ ] [ ] The general solution is x(t = c e 3t + c e 3t. The initial condition implies, [ ] [ ] [ ] [ ] [ ] 3 c = x(0 = c + c = c [ ] [ ] [ ] [ ] c 3 c = = [ ] 5. c ( + c 3 We conclude: x(t = 5 [ ] e 3t + [ ] e 3t. 3 3 [ ] 3. Review for Final Exam. Heat Eq. and Fourier Series (Chptr.6. Eigenvalue-Eigenfunction BVP (Chptr. 6. Systems of linear Equations (Chptr. 5. Laplace transforms (Chptr. 4. Second order linear equations (Chptr.. First order differential equations (Chptr..

12 Laplace transforms. Summary: Main Properties: L [ f (n (t ] = s n L[f (t] s (n f (0 f (n (0; (8 e cs L[f (t] = L[u c (t f (t c]; (3 L[f (t] = L[e ct f (t]. (4 (s c Convolutions: L[(f g(t] = L[f (t] L[g(t]. Partial fraction decompositions, completing the squares. Laplace transforms. Use L.T. to find the solution to the IVP y + 9y = u 5 (t, y(0 = 3, y (0 =. Solution: Compute L[y ] + 9 L[y] = L[u 5 (t] = e 5s s, and recall, L[y ] = s L[y] s y(0 y (0 L[y ] = s L[y] 3s. L[y] = 3 (s + 9 L[y] 3s = e 5s s L[y] = (3s + (s e 5s s(s + 9. s (s (s e 5s s(s + 9.

13 Laplace transforms. Use L.T. to find the solution to the IVP y + 9y = u 5 (t, y(0 = 3, y (0 =. Solution: Recall L[y] = 3 s (s (s e 5s s(s + 9. L[y] = 3 L[cos(3t] + 3 L[sin(3t] + e 5s s(s + 9. Partial fractions on H(s = s(s + 9 = a s (bs + c + (s + 9 = a(s (bs + cs s(s, + 9 = as + 9a + bs + cs = (a + b s + cs + 9a a = 9, c = 0, b = a b = 9. Laplace transforms. Use L.T. to find the solution to the IVP y + 9y = u 5 (t, y(0 = 3, y (0 =. Solution: So, L[y] = 3 L[cos(3t] + 3 L[sin(3t] + e 5s H(s, and H(s = s(s + 9 = [ 9 s s ] s = ( L[u(t] L[cos(3t] e 5s H(s = ( e 5s L[u(t] e 5s L[cos(3t] 9 e 5s H(s = ( L[u 5 (t] L [ u 5 (t cos(3(t 5 ]. 9 L[y] = 3 L[cos(3t]+ 3 L[sin(3t]+ 9 ( L[u 5 (t] L [ u 5 (t cos(3(t 5 ].

14 Laplace transforms. Use L.T. to find the solution to the IVP Solution: y + 9y = u 5 (t, y(0 = 3, y (0 =. L[y] = 3 L[cos(3t]+ 3 L[sin(3t]+ 9 ( L[u 5 (t] L [ u 5 (t cos(3(t 5 ]. Therefore, we conclude that, y(t = 3 cos(3t + 3 sin(3t + u [ ] 5(t cos(3(t 5. 9 Review for Final Exam. Heat Eq. and Fourier Series (Chptr.6. Eigenvalue-Eigenfunction BVP (Chptr. 6. Systems of linear Equations (Chptr. 5. Laplace transforms (Chptr. 4. Second order linear equations (Chptr.. First order differential equations (Chptr..

15 Second order linear equations. Summary: Solve y + a y + a 0 y = g(t. First find fundamental solutions y(t = e rt to the case g = 0, where r is a root of p(r = r + a r + a 0. (a If r r, real, then the general solution is y(t = c e r t + c e r t. (b If r r, complex, then denoting r ± = α ± βi, complex-valued fundamental solutions are y ± (t = e (α±βit y ± (t = e αt [ cos(βt ± i sin(βt ], and real-valued fundamental solutions are y (t = e αt cos(βt, y (t = e αt sin(βt. If r = r = r, real, then the general solution is y(t = (c + c t e rt. Second order linear equations. Remark: Case (c is solved using the reduction of order method. See page 70 in the textbook. Do the extra homework problems Sect. 3.4: 3, 5, 7. Summary: Non-homogeneous equations: g 0. (i Undetermined coefficients: Guess the particular solution y p using the guessing table, g y p. (ii Variation of parameters: If y and y are fundamental solutions to the homogeneous equation, and W is their Wronskian, then y p = u y + u y, where u = y g W, u = y g W.

16 Second order linear equations. Knowing that y (x = x solves x y 4x y + 6y = 0, with x > 0, find a second solution y not proportional to y. Solution: Use the reduction of order method. We verify that y = x solves the equation, x ( 4x (x + 6x = 0. Look for a solution y (x = v(x y (x, and find an equation for v. y = x v, y = x v + xv, y = x v + 4xv + v. x (x v + 4xv + v 4x (x v + xv + 6 (x v = 0. x 4 v + (4x 3 4x 3 v + (x 8x + 6x v = 0. v = 0 v = c + c x y = c y + c x y. Choose c = 0, c =. Hence y (x = x 3, and y (x = x. Second order linear equations. Find the solution y to the initial value problem y y 3y = 3 e t, y(0 =, y (0 = 4. Solution: ( Solve the homogeneous equation. r ± = y(t = e rt, p(r = r r 3 = 0. { [ ] [ ] r+ = 3, ± 4 + = ± 6 = ± r =. Fundamental solutions: y (t = e 3t and y (t = e t. ( Guess y p. Since g(t = 3 e t y p (t = k e t. But this y p = k e t is solution of the homogeneous equation. Then propose y p (t = kt e t.

17 Second order linear equations. Find the solution y to the initial value problem y y 3y = 3 e t, y(0 =, y (0 = 4. Solution: Recall: y p (t = kt e t. This is correct, since te t is not solution of the homogeneous equation. (3 Find the undetermined coefficient k. y p = k e t kt e t, y p = k e t + kt e t. ( k e t + kt e t (k e t kt e t 3(kt e t = 3 e t ( + t + t 3t k e t = 3 e t 4k = 3 k = 3 4. We obtain: y p (t = 3 4 t e t. Second order linear equations. Find the solution y to the initial value problem y y 3y = 3 e t, y(0 =, y (0 = 4. Solution: Recall: y p (t = 3 4 t e t. (4 Find the general solution: y(t = c e 3t + c e t 3 4 t e t. (5 Impose the initial conditions. The derivative function is y (t = 3c e 3t c e t 3 4 (e t t e t. = y(0 = c + c, } c + c =, 3 c = 4 = y (0 = 3c c 3 4. [ ] [ ] [ ] c =. 3 c

18 Second order linear equations. Find the solution y to the initial value problem y y 3y = 3 e t, y(0 =, y (0 = 4. Solution: Recall: [ ] [ ] c = 3 c y(t = c e 3t + c e t 3 4 t e t, and [ ] [ c c ] = 4 Since c = and c =, we obtain, [ ] [ ] 3 = 4 [ ]. y(t = ( e 3t + e t 3 4 t e t. Review for Final Exam. Heat Eq. and Fourier Series (Chptr.6. Eigenvalue-Eigenfunction BVP (Chptr. 6. Systems of linear Equations (Chptr. 5. Laplace transforms (Chptr. 4. Second order linear equations (Chptr.. First order differential equations (Chptr..

19 First order differential equations. Summary: Linear, first order equations: y + p(t y = q(t. Use the integrating factor method: µ(t = e R p(t dt. Separable, non-linear equations: Integrate with the substitution: that is, h(u du = h(y y = g(t. u = y(t, du = y (t dt, g(t dt + c. The solution can be found in implicit of explicit form. Homogeneous equations can be converted into separable equations. Read page 49 in the textbook. No modeling problems from Sect..3. First order differential equations. Summary: Bernoulli equations: y + p(t y = q(t y n, with n R. Read page 77 in the textbook, page in the Lecture Notes. A Bernoulli equation for y can be converted into a linear equation for v = y n. Exact equations and integrating factors. N(x, y y + M(x, y = 0. The equation is exact iff x N = y M. If the equation is exact, then there is a potential function ψ, such that N = y ψ and M = x ψ. The solution of the differential equation is ψ ( x, y(x = c.

20 First order differential equations. Advice: In order to find out what type of equation is the one you have to solve, check from simple types to the more difficult types:. Linear equations. (Just by looking at it: y + a(t y = b(t.. Bernoulli equations. (Just by looking at it: y + a(t y = b(t y n. 3. Separable equations. (Few manipulations: h(y y = g(t. 4. Homogeneous equations. (Several manipulations: y = F (y/t. 5. Exact equations. (Check one equation: N y + M = 0, and t N = y M. 6. Exact equation with integrating factor. (Very complicated to check. First order differential equations. Find all solutions of y = x + xy + y. xy Solution: The sum of the powers in x and y on every term is the same number, two in this example. The equation is homogeneous. y = x + xy + y xy v(x = y x (/x (/x y = + ( y x + ( y x ( y x. y = + v + v. v y = x v, y = x v + v x v + v = + v + v. v x v = + v + v v v = + v + v v v x v = + v v.

21 First order differential equations. Find all solutions of y = x + xy + y. xy Solution: Recall: v = + v. This is a separable equation. v v(x + v(x v (x = v(x dx x + v(x v (x dx = x + c. Use the substitution u = + v, hence du = v (x dx. (u dx ( du = u x + c du = u dx x + c u ln u = ln x + c + v ln + v = ln x + c. v = y + y(x ln + y(x = ln x + c. x x x First order differential equations. Find the solution y to the initial value problem y + y + e x y 3 = 0, y(0 = 3. Solution: This is a Bernoulli equation, y + y = e x y n, n = 3. Divide by y 3. That is, y y 3 + y = ex. Let v = y. Since v = y y 3, we obtain v + v = e x. We obtain the linear equation v v = e x. Use the integrating factor method. µ(x = e x. e x v e x v = ( e x v =.

22 First order differential equations. Find the solution y to the initial value problem y + y + e x y 3 = 0, y(0 = 3. Solution: Recall: v = y and ( e x v =. e x v = x + c v(x = (x + c e x y = (x + c ex. y = e x (x + c y ± (x = ± e x x + c. The initial condition y(0 = /3 > 0 implies: Choose y +. 3 = y +(0 = c = 9 y(x = e x. c x + 9 First order differential equations. Find all solutions of xy + y + x y y + x y = 0. Solution: Re-write the equation is a more organized way, [x y + x] y + [xy + y] = 0. } N = [x y + x] x N = 4xy +. M = [xy x N = y M. + y] y M = 4xy +. The equation is exact. There exists a potential function ψ with y ψ = N, x ψ = M. y ψ = x y + x ψ(x, y = x y + xy + g(x. xy + y + g (x = x ψ = M = xy + y g (x = 0. ψ(x, y = x y + xy + c, x y (x + x y(x + c = 0.

Boundary Value Problems (Sect. 10.1). Two-point Boundary Value Problem.

Boundary Value Problems (Sect. 10.1). Two-point BVP. from physics. Comparison: IVP vs BVP. Existence, uniqueness of solutions to BVP. Particular case of BVP: Eigenvalue-eigenfunction problem. Two-point