Quiz 9: Separable differential equations

Last unanswered question  Question  Next unanswered question

Question 1

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

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Your answer is correct.
dy dx = 2y, dy y =2dx, ln|y| = 2x + C, |y| = e2x+C = Be2xwhereB = eC, y = Ae2x.
Not correct. Choice (d) is false.

Question 2

Find the solution to the differential equation
dP dt = 3PwhereP = 4whent = 0.
a)
P = Ae3t
  b)
P = 3Aet
c)
P = 12Aet
  d)
P = 4e3t

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer 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

Question 3

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

 

Not correct. Choice (a) is false.
Your answer 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.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

Question 4

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

 

Not correct. Choice (a) is false.
Your answer 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.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

Question 5

Find the solution to the differential equation
dP dt = 2P - 2PtwhereP = 5whent = 0.
a)
P = 5e2-2t
  b)
P = 5e2t-t2
c)
P = 5e2-2t2
  d)
P = P2 - P2t

 

Not correct. Choice (a) is false.
Your answer is correct.
dP dt = P(2 - 2t), dP P =(2 - 2t)dt, ln|P| = 2t - t2 + C, |P| = e2t-t2+C = Ae2t-t2 whereA = eC, P = Be2t-t2 . P = Be0 = B = 5. At t=0 P = 5e2t-t2 . Hence
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

Question 6

Find the solution to the differential equation
dN dt -N k = 0wherekisaconstant.
a)
N = Be t k where B is a constant.
  b)
N = Be-kt where B is a constant.
c)
N = Be- t k where B is a constant.
  d)
N = Bekt where B is a constant.

 

Your answer is correct.
dN dt = N k , dN N =1 k, ln|N| = t k + C, |N| = Aet k whereA = eC, N = Bet k whereBisaconstant.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

Question 7

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.

 

Your answer is correct.
dy dx = -x y, ydy = -xdx, 1 2y2 = -1 2x2 + C, 1 2x2 + 1 2y2 = CnotingthatC > 0. therefore, x2 + y2 = 2C = r2
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

Question 8

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.

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
dy dx = 3x2(y - 1), dy y - 1 =3x2dx, ln|y - 1| = x3 + C, |y - 1| = Aex3 whereA = eC, y = Bex3 + 1whereBisaconstant.

Question 9

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.

 

Not correct. Choice (a) is false.
Your answer is correct.
Not correct. Choice (c) is false.
Not correct. Choice (d) is false.

Question 10

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 = 20e-kt - 200
  b)
T = 180e-kt - 200
c)
T = 21 - e-kt
  d)
200 - 180e-kt

 

Not correct. Choice (a) is false.
Not correct. Choice (b) is false.
Not correct. Choice (c) is false.
Your answer is correct.
dT dt = -k(T - 200), dT T - 200 =-kdt, ln|T - 200| = -kt + C, |T - 200| = Ae-ktwhereA = eC, T = Be-kt + 200whereBisaconstant. At t = 0, T = 20, therefore 20 = B + 200 and B = -180, hence T = 200 - 180e-kt.
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