MATH 211; DIFFERENTIAL EQUATIONS
LQ-3 REVIEWER
Test 1: FAMILIARIZATION OF SPECIAL
DIFFERENTIAL FORMULAS x3 dx + (ey dx + xey dy)
+ 2(x e 2y dx + x2e2y dy) + e 4y dy = 0
Identify the symbol, letter or number that makes
the equation/formula false and replace the x3 dx + d(xey) + d(x2 e 2y) + e 4y dy = 0 Answer
same that will make the equation/formula true.
Write your answers in your Answer Sheet.
x4 e4 y c
2x 2 2y + xey + x2e 2y + =
1. 2xe (dx – x dy) = d (x e ) x/y 4 4 4
2. 3x3 du = d(x4) 3/4 x4 + 4xey + 4x2e 2y + e 4y = c Answer
x
3. x 2 ln ydx dy = d (x3 ln y) 3/2
y when x = 1, y = 0.
2 ye 2 x ( ydx xdy) e2x c = x4 + 4xey + 4x2e 2y + e 4y
4. = d x/1
y4 y2 c = 1 + 4(1)(1) + 4(1)(1) + (1) c = 10
Answer.
2 xydx x 2 dy y
5. = d arctan 2/4 TEST 2 ANSWERS:
x y 2 2
x2
1. M(x, y) = x3 + x e 2y + ey
TEST 1 ANSWERS:
2. N(x, y) = ey(e 3y + x2ey + x)
1. x/y 2. ¾ 3. 3/2 4. x/1 5. 2/4
3. x3 dx + d(xey) + d(x2 e 2y) + e 4y dy = 0
4. x4 + 4xey + 4x2e 2y + e 4y = c.
Test 2: INTEGRATING FACTOR
5. c = 10
FOUND BY INSPECTION
ODE: (A3 + AB2 + B) dA + (B3 + A2B + A) dB = 0
Test 3: SUBSTITUTION SUGGESTED
REQUIRED: BY THE EQUATION
Formulate the equation into
M(x,y) dy + N(x,y) dy = 0 and find: ODE: (rA + sB) dA + tA dB = 0
1. M(x,y)
2. N(x,y) REQUIRED:
3. Formulate the special differentials. Formulate the equation into
4. Find the GS using the constant c/4. M(x,y) dy + N(x,y) dy = 0 and find:
5. Find the value of c when x = x1 and y = y1 1. M(x,y) and N(x,y)
2. P(u) and Q(u)
DATA: A = x B = ey 3. Formulate the integrating factor, t
4. Find the GS using the constant c/5.
SOLUTION: 5. Find the value of c when x = x1 and y = y1
(A3 + AB2 + B) dA + (B3 + A2B + A) dB = 0
DATA: A = x B = ey r = 1 s = –2 t=3
(x3 + x e 2y + ey) dx + (e 3y + x2ey + x) eydy = 0
SOLUTION:
(x3 + x e 2y + ey) dx + ey(e 3y + x2ey + x) dy = 0
(rA + sB) dA + tA dB = 0
M(x, y) = x3 + x e 2y + ey Answer
(x – 2ey) dx + 3x d(ey) = 0
N(x, y) = ey(e 3y + x2ey + x) Answer
(x – 2ey) dx + 3x eydy = 0
3 2y y
x dx + x e dx + e dx
M(x, y) = x – 2ey Answer
4y 2 2y y
+e dy + x e dy + xe dy = 0
LQ-3 REVIEWER
Test 1: FAMILIARIZATION OF SPECIAL
DIFFERENTIAL FORMULAS x3 dx + (ey dx + xey dy)
+ 2(x e 2y dx + x2e2y dy) + e 4y dy = 0
Identify the symbol, letter or number that makes
the equation/formula false and replace the x3 dx + d(xey) + d(x2 e 2y) + e 4y dy = 0 Answer
same that will make the equation/formula true.
Write your answers in your Answer Sheet.
x4 e4 y c
2x 2 2y + xey + x2e 2y + =
1. 2xe (dx – x dy) = d (x e ) x/y 4 4 4
2. 3x3 du = d(x4) 3/4 x4 + 4xey + 4x2e 2y + e 4y = c Answer
x
3. x 2 ln ydx dy = d (x3 ln y) 3/2
y when x = 1, y = 0.
2 ye 2 x ( ydx xdy) e2x c = x4 + 4xey + 4x2e 2y + e 4y
4. = d x/1
y4 y2 c = 1 + 4(1)(1) + 4(1)(1) + (1) c = 10
Answer.
2 xydx x 2 dy y
5. = d arctan 2/4 TEST 2 ANSWERS:
x y 2 2
x2
1. M(x, y) = x3 + x e 2y + ey
TEST 1 ANSWERS:
2. N(x, y) = ey(e 3y + x2ey + x)
1. x/y 2. ¾ 3. 3/2 4. x/1 5. 2/4
3. x3 dx + d(xey) + d(x2 e 2y) + e 4y dy = 0
4. x4 + 4xey + 4x2e 2y + e 4y = c.
Test 2: INTEGRATING FACTOR
5. c = 10
FOUND BY INSPECTION
ODE: (A3 + AB2 + B) dA + (B3 + A2B + A) dB = 0
Test 3: SUBSTITUTION SUGGESTED
REQUIRED: BY THE EQUATION
Formulate the equation into
M(x,y) dy + N(x,y) dy = 0 and find: ODE: (rA + sB) dA + tA dB = 0
1. M(x,y)
2. N(x,y) REQUIRED:
3. Formulate the special differentials. Formulate the equation into
4. Find the GS using the constant c/4. M(x,y) dy + N(x,y) dy = 0 and find:
5. Find the value of c when x = x1 and y = y1 1. M(x,y) and N(x,y)
2. P(u) and Q(u)
DATA: A = x B = ey 3. Formulate the integrating factor, t
4. Find the GS using the constant c/5.
SOLUTION: 5. Find the value of c when x = x1 and y = y1
(A3 + AB2 + B) dA + (B3 + A2B + A) dB = 0
DATA: A = x B = ey r = 1 s = –2 t=3
(x3 + x e 2y + ey) dx + (e 3y + x2ey + x) eydy = 0
SOLUTION:
(x3 + x e 2y + ey) dx + ey(e 3y + x2ey + x) dy = 0
(rA + sB) dA + tA dB = 0
M(x, y) = x3 + x e 2y + ey Answer
(x – 2ey) dx + 3x d(ey) = 0
N(x, y) = ey(e 3y + x2ey + x) Answer
(x – 2ey) dx + 3x eydy = 0
3 2y y
x dx + x e dx + e dx
M(x, y) = x – 2ey Answer
4y 2 2y y
+e dy + x e dy + xe dy = 0
