menuicon

MATH1013 Quizzes

Quiz 9: Separable differential equations
Question 1 Questions
Find the solution to the differential equation
dy dx = 2y.
a)
y = x2 + C
 b)
y = e1 2x
c)
y = Ae2x
 d)
y = 1 2Aex

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
dy dx = 2y, dy y =2dx, ln|y| = 2x + C, |y| = e2x+C = Be2xwhereB = eC, y = Ae2x.
Choice (d) is incorrect
Find the solution to the differential equation
dP dt = 3PwhereP = 4whent = 0.
a)
P = Ae3t
 b)
P = 3Aet
c)
P = 12Aet
 d)
P = 4e3t

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
dP dt = 3P, dP P =3dt, ln|P| = 3t + C, |P| = e3t+C = Ae3whereA = eC, therefore, P = Be3t. P = Be0 = B = 4. At t=0 P = 4e3t. Hence
Find the solution to the differential equation
dy dx = xy.
a)
y = 1 2x2 + C
 b)
y = Be1 2x2
c)
y = e1 2x2
 d)
y = 1 2Bex2

Choice (a) is incorrect
Choice (b) is correct!
dy dx = xy, dy y =xdx, ln|y| = 1 2x2 + C, |y| = e1 2 x+C = Ae1 2 xwhereA = eC, y = Be1 2 x.
Choice (c) is incorrect
Choice (d) is incorrect
Find the solution to the differential equation
dP dt = 4P + 5.
a)
P = Ae4t + 5t
 b)
P = Ae4t 1.25
c)
P = 5e4t + 5t
 d)
P = Ae4t + 1.25t

Choice (a) is incorrect
Choice (b) is correct!
dP dt = 4P + 5 = 4(P + 1.25), dP (P + 1.25) =4dt, ln|P + 1.25| = 4t + C, |P + 1.25| = e4t+C = Be4twhereB = eC, P + 1.25 = Ae4t, P = Ae4t 1.25.
Choice (c) is incorrect
Choice (d) is incorrect
Find the solution to the differential equation
dP dt = 2P 2PtwhereP = 5whent = 0.
a)
P = 5e22t
 b)
P = 5e2tt2
c)
P = 5e22t2
 d)
P = P2 P2t

Choice (a) is incorrect
Choice (b) is correct!
dP dt = P(2 2t), dP P =(2 2t)dt, ln|P| = 2t t2 + C, |P| = e2tt2+C = Ae2tt2 whereA = eC, P = Be2tt2 . P = Be0 = B = 5. At t=0 P = 5e2tt2 . Hence
Choice (c) is incorrect
Choice (d) is incorrect
Find the solution to the differential equation
dN dt N k = 0wherekisaconstant.
a)
N = Be t k where B is a constant.
 b)
N = Bekt where B is a constant.
c)
N = Be t k where B is a constant.
 d)
N = Bekt where B is a constant.

Choice (a) is correct!
dN dt = N k , dN N =1 k, ln|N| = t k + C, |N| = Aet k whereA = eC, N = Bet k whereBisaconstant.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the family of curves which satisfy the differential equation dy dx = x y.
a)
x2 + y2 = r2
 b)
y = x2 + C
c)
y = 1 2x2 + C
 d)
This differential equation cannot be solved using the techniques learned in this course.

Choice (a) is correct!
dy dx = x y, ydy = xdx, 1 2y2 = 1 2x2 + C, 1 2x2 + 1 2y2 = CnotingthatC > 0. therefore, x2 + y2 = 2C = r2
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the family of curves which satisfy
dy dx = 3x2(y 1).
a)
y = Bex3 1 where B is a constant.
 b)
y2 y x3 C = 0 where C is a constant.
c)
y2 y + x3 C = 0 where C is a constant.
 d)
y = Bex3 + 1 where B is a constant.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
dy dx = 3x2(y 1), dy y 1 =3x2dx, ln|y 1| = x3 + C, |y 1| = Aex3 whereA = eC, y = Bex3 + 1whereBisaconstant.
Using Newton’s law of cooling, it is found that the differential equation  describing the heating of a potato in an oven is
dT dt = k(T 200).
What is the temperature inside the oven ?
a)
200koC
 b)
200oC
c)
21.1oC
 d)
The temperature cannot be determined from this equation.

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Use Newton’s law of cooling to determine the temperature with respect to time of a potato cooked in an oven at 200oC, where the temperature of the potato is 20oC shortly before being placed in the oven.
a)
T = 20ekt 200
 b)
T = 180ekt 200
c)
T = 21 ekt
 d)
200 180ekt

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
dT dt = k(T 200), dT T 200 =kdt, ln|T 200| = kt + C, |T 200| = AektwhereA = eC, T = Bekt + 200whereBisaconstant. At t = 0, T = 20, therefore 20 = B + 200 and B = 180, hence T = 200 180ekt.