MATH201 Lecture 25: 25. Solving Simple Heat Equations.pdf

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turquoiseminnow67

18 Jun 2015

School

University of Alberta

Department

Mathematics

Course

MATH201

Professor

van

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Document Summary

First write down formulas for general solutions and at the same time identify di erent ranges of . 0 < x < , y(0) = 0, y ( ) + y( ) = 0. y = c1 e(2+ 4 . Substituting into the second equation and assuming c1(cid:2) 0, we. (cid:20) e2 4 contradiction means c1(cid:2) 0 does not hold, that is c1 = 0. For this problem we can calculate have to rst calculate y =(cid:0)2 + 4 (cid:0)3 + 4 . =(cid:0)3 4 h(cid:0)3 + 4 (cid:1) e 4 (cid:0)3 + 4 (cid:1) e 4 . In general, we know that (cid:1) e 4 . I c2 = 0. a c1 + b c2 = 0; c c1 + d c2 = 0 (cid:1) e 4 (4) (5) (6) (7) (8) ) = 0. no < 4 that is an eigenvalue. 1 has only zero solution if and only if det(cid:18) a b det.

Related Questions

here is the problem it is already solved and this is the yahooanswer solution and my teachers solution as well. All i need is aGraph and for someone to explain how my teachergot 0,0 and 0,-2 as intresection points.

teachers solution:find all 4 points of intersections: (0,0),(0,-2), (-4sqrt(17sqrt(2)-24), 10sqrt(2)-14),(-4sqrt(17sqrt(2)-24), 10sqrt(2)-14).

yahoo answers:sin(theta) = r^2/4 ==> sin(theta)>= 0 . Hence 0 =< theta =< pi

r = 1 - sin(theta) ; therefore 0 =< r =< 1, because now weknow 0 =< sin(theta) =< 1

Therefore y = r sin(theta) >= 0 so the intersection points are in theupper half plane.

Also, sin(theta) = r^2/4 =< 1/4 < 1/sqrt(2) = sin pi/4 =sin(3pi/4),

so 0 =< theta < pi/4 or 3pi/4 < theta =< pi.

---
At an intersection point, sin(theta) = r^2/4 = 1 - r

r^2 + 4r - 4 = 0

Roots: r = -- 2 +/- 2 sqrt(2); but r >= 0, so r = 2 sqrt(2) - 2= 2(sqrt(2) - 1)

sin(theta) = 1 - r = 3 - 2 sqrt(2)

There is just one value of y, namely

y = r sin(theta) = 2 (sqrt(2) - 1)(3 - 2 sqrt(2) ) = 2 (5 sqrt(2) -7) = 10 sqrt(2) - 14

( or y = 10 (sqrt(2) - 1.4 ) = 0.142135...

However, there are two intersection points, with x^2 = r^2 -y^2

x = +/- sqrt( r^2 - y^2 )

x^2 = r^2 - y^2 = 4 (sin theta) -- y^2 = 4 (3 - 2 sqrt(2)) -- [2(5sqrt(2) - 7)]^2

x^2 = 12 - 8 sqrt(2) -4 [ 99 - 70 sqrt(2)] = 272 sqrt(2) - 384 (=17 x 2^4 sqrt(2) - 3 x 2^7)

x = +/- 4 sqrt(17 sqrt(2) - 24) [ = +/- 0.8161427...]

Q29. Solve the system of equations using Cramer's Rule if it is applicable.

a. x = 9, y = 2, z = 5; (9, 2, 5)
b. x = 8, y = 4, z = 5; (8, 4, 5)
c. x = 4, y = 5, z = 4; (4, 5, 4)
d. x = 8, y = -4, z = -5; (8, -4, -5)

Q30. Solve the system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent.

a. x = 2, y = 3, z = -7; (2, 3, -7)
b. x = 1, y = 3, z = -4; (1, 3, -4)
c. x = 5, y = 13/2, z = -3; (5, 13/2, -3)
d. x = 1, y = -3, z = -16; (1, -3, -16)

Q33. Write the partial fraction decomposition of the rational expression (4x³ + 4x²)/(x² + 5)².
a. (4x - 4)/(x² + 5) + (-20x + 20)/(x² + 5)²
b. (4x + 4)/(x² + 5) + (-20x - 20)/(x² + 5)²
c. (4x + 4)/(x² + 5) + (20x - 20)/(x² + 5)²
d. (4x + 4)/(x² + 5) + (20x + 20)/(x² + 5)²

Q34. Solve the linear programming problem. Minimize z = 11x + 6y + 7 subject to: x = 0, y = 0, x + y = 1.
a. minimum: 7
b. minimum: 18
c. minimum: 24
d. minimum: 13

Q35. Solve the system of equations by elimination.

a. x = 10, y = 0; (10, 0)
b. x = 10, y = 10; (10, 10)
c. x = 0, y = 10; (0, 10)
d. x = 0, y = 0; (0, 0)

Q36. Perform the row operations on the given augmented matrix.

a.
b.
c.
d.

Q37. Solve the system of equations using Cramer's Rule if it is applicable.

a. x = -4, y = 3; (-4, 3)
b. x = -3, y = -4; (-3, -4)
c. x = 4, y = 3; (4, 3)
d. x = 3, y = 4; (3, 4)

Q38. Write the partial fraction decomposition of the rational expression x/(x² - 9x + 20).
a. -4/(x - 4) + 5/(x - 5)
b. -5/(x - 4) + 4/(x - 5)
c. 4/(x - 4) + -5/(x - 5)
d. -4/(x - 4) + -5/(x - 5)

Q39. Solve the system of equations.

a. x = -4, y = 4, z = -3; (-4, 4, -3)
b. x = -3, y = 4, z = -4; (-3, 4, -4)
c. x = -4, y = -3, z = 4; (-4, -3, 4)
d. inconsistent

Q40. Find the value of the determinant.

a. 224
b. 24
c. -72
d. 72