MATH 211: DIFFERENTIAL EQUATIONS
LONG QUIZ NO. 6A – PART 1; REVIEWER
Preliminaries: Your LQ-6A consists of 4 rounds:
Round 1: Determination of roots of Auxiliary Equation and formulation of Yc.
Round 2: R(x) Algebraic Function
Round 3: R(x) Exponential Function
Round 4: R(x) Combination of Rounds 2 and 3.
Conditions:
1. Answers will be submitted via text.
2. You will be given an Answers Sheet for each round.
3. Verification of grades will be done in each round.
4. You cannot proceed to rounds 2, 3 and 4 unless you will be cleared of round 1.
5. If you got a perfect score in rounds 1, 2 and 3 on 1st submission,you may not solve
anymore round 4 and your Rating = 1.0
6. If you got a passing scores in all rounds, your grade will be the highest of the 4 rounds.
7. You will be given 2 chances to correct your answers but with point deductions.
m = 1, –1, –2 Answer.
PROBLEM: Given the ODE
which translates into a Complementary Solution of
y’’’ + 2y” – y’ – 2y = R(x) (1) –x –2x
CS: Yc = c1ex + c2e + c3e (3)
REQUIRED:
A. ROUND 1. If R(x) = 0 B. ROUND 2:
Find:
1. The roots of the Auxiliary Equation. 1. Solving for the coefficients of and formulation of
2. The General Solution Ys.
3
R(x) = –4x3 + 20x – 5 (4)
B. ROUND 2. If R(x) = –4x + 20x – 5
Using (1) with y Ys and (4) gives
Find:
1. The coefficients of Ys. Ys’’’ + 2Y”s – Y’s – 2Ys = –4x3 + 20x – 5 (5)
2. The value of arbitrary constants in the GS of In the case of 3rd degree algebraic function
when x = 0, y = 1, y’ = 0 and y” = 1.
Ys = Ax3 + Bx2 + Cx + D (6)
C. ROUND 3. If R(x) = –6e2x and by successive differentiation, we have
Find: Y’s = 3Ax2 + 2Bx + C (7)
1. The coefficients of Ys.
2. The value of arbitrary constants in the GS of Y”s = 6Ax + 2B (8)
Item A when x = 0, y = 3, y’ = 2 and y” = –3. Y’’’s = 6A (9)
D. ROUND 4. If R(x) = 12e2x –4x2 – 4x + 2 Substituting (6), (7), (8) and (9) into (5) gives:
Find: 6A + 2(6Ax + 2B) – (3Ax2 + 2Bx + C)
1. The coefficients of Ys. – 2(Ax3 + Bx2 + Cx + D) (10)
2. The value of arbitrary constants in the GS of
Item A when x = 0, y = 10, y’ = 4 and y” = 17. = – 4x3 + 0 x2 + 20 x – 5
Note that the missing term in R(x) is affixed with a
SOLUTION:
zero coefficient to complete the descending
exponents of “x”.
A. ROUND 1:
We will now derive the values of the coefficients A,
R(x) = 0 B, C and D in (10). The rule is
y’’’ + 2y” – y’ – 2y = 0 (2) “Equate coefficients of similar exponent of “x”.”
Converting (1) into Auxiliary Equation and getting its so, we have
roots, we have
3 2
Coeff. of x3: –2A = –4 A=2
LONG QUIZ NO. 6A – PART 1; REVIEWER
Preliminaries: Your LQ-6A consists of 4 rounds:
Round 1: Determination of roots of Auxiliary Equation and formulation of Yc.
Round 2: R(x) Algebraic Function
Round 3: R(x) Exponential Function
Round 4: R(x) Combination of Rounds 2 and 3.
Conditions:
1. Answers will be submitted via text.
2. You will be given an Answers Sheet for each round.
3. Verification of grades will be done in each round.
4. You cannot proceed to rounds 2, 3 and 4 unless you will be cleared of round 1.
5. If you got a perfect score in rounds 1, 2 and 3 on 1st submission,you may not solve
anymore round 4 and your Rating = 1.0
6. If you got a passing scores in all rounds, your grade will be the highest of the 4 rounds.
7. You will be given 2 chances to correct your answers but with point deductions.
m = 1, –1, –2 Answer.
PROBLEM: Given the ODE
which translates into a Complementary Solution of
y’’’ + 2y” – y’ – 2y = R(x) (1) –x –2x
CS: Yc = c1ex + c2e + c3e (3)
REQUIRED:
A. ROUND 1. If R(x) = 0 B. ROUND 2:
Find:
1. The roots of the Auxiliary Equation. 1. Solving for the coefficients of and formulation of
2. The General Solution Ys.
3
R(x) = –4x3 + 20x – 5 (4)
B. ROUND 2. If R(x) = –4x + 20x – 5
Using (1) with y Ys and (4) gives
Find:
1. The coefficients of Ys. Ys’’’ + 2Y”s – Y’s – 2Ys = –4x3 + 20x – 5 (5)
2. The value of arbitrary constants in the GS of In the case of 3rd degree algebraic function
when x = 0, y = 1, y’ = 0 and y” = 1.
Ys = Ax3 + Bx2 + Cx + D (6)
C. ROUND 3. If R(x) = –6e2x and by successive differentiation, we have
Find: Y’s = 3Ax2 + 2Bx + C (7)
1. The coefficients of Ys.
2. The value of arbitrary constants in the GS of Y”s = 6Ax + 2B (8)
Item A when x = 0, y = 3, y’ = 2 and y” = –3. Y’’’s = 6A (9)
D. ROUND 4. If R(x) = 12e2x –4x2 – 4x + 2 Substituting (6), (7), (8) and (9) into (5) gives:
Find: 6A + 2(6Ax + 2B) – (3Ax2 + 2Bx + C)
1. The coefficients of Ys. – 2(Ax3 + Bx2 + Cx + D) (10)
2. The value of arbitrary constants in the GS of
Item A when x = 0, y = 10, y’ = 4 and y” = 17. = – 4x3 + 0 x2 + 20 x – 5
Note that the missing term in R(x) is affixed with a
SOLUTION:
zero coefficient to complete the descending
exponents of “x”.
A. ROUND 1:
We will now derive the values of the coefficients A,
R(x) = 0 B, C and D in (10). The rule is
y’’’ + 2y” – y’ – 2y = 0 (2) “Equate coefficients of similar exponent of “x”.”
Converting (1) into Auxiliary Equation and getting its so, we have
roots, we have
3 2
Coeff. of x3: –2A = –4 A=2
