Multiple Choice
Question 1
Which of the following is a separable differential equation?
A
dy/dx = x y
B
dy/dx = x + y
C
dy/dx = sin(x + y)
D
dy/dx = x/(x + y)
Solution: dy/dx = xy separates as dy/y = x dx. The others cannot be written as a product of a function of x times a function of y.
Question 2
A solution of a differential equation that contains an arbitrary constant is called:
A
the general solution
B
a particular solution
C
an equilibrium solution
D
a singular solution
Solution: The general solution includes an arbitrary constant and represents the whole family of solution curves.
Question 3
The order of a differential equation is:
A
the highest derivative that appears in it
B
the number of terms
C
the power of y
D
the value of the constant
Solution: The order is determined by the highest-order derivative present (e.g., dy/dx is first order; d^2y/dx^2 is second order).
Question 4
Solve dy/dx = 6x^2 y.
A
y = C e^(2x^3)
B
y = C e^(6x^3)
C
y = C e^(2x^2)
D
y = 2x^3 + C
Solution: Separate: dy/y = 6x^2 dx. Integrate: ln|y| = 2x^3 + C1, so y = C e^(2x^3).
Question 5
Solve the initial-value problem dy/dx = x/y with y(1) = 1. (Take the branch through (1,1).)
A
y = x
B
y = x^2
C
y = sqrt(x)
D
y = 1/x
Solution: Separating gives y^2 = x^2 + C. Applying y(1) = 1: 1 = 1 + C, so C = 0 and y^2 = x^2. Through (1,1) the solution branch is y = x. (Adapted from the Vidaara NCERT/JEE bank.)
Question 6
Solve dy/dx = 2xy with y(0) = 3.
A
y = 3 e^(x^2)
B
y = 3 e^(2x)
C
y = e^(x^2) + 3
D
y = 3x^2
Solution: Separate: dy/y = 2x dx, so ln|y| = x^2 + C1 and y = C e^(x^2). Using y(0) = 3 gives C = 3, so y = 3 e^(x^2).
Question 7
A particular solution of a differential equation is one that:
A
satisfies a given initial condition (a specific value of C)
B
contains an arbitrary constant
C
is always y = 0
D
has no derivative
Solution: The general solution contains an arbitrary constant; applying an initial condition determines that constant, yielding a particular solution.
Question 8
A substance decays with a half-life of 10 years. The decay constant k in y = y0 e^(-kt) is:
A
(ln 2)/10
B
10 ln 2
C
ln 2
D
0.5
Solution: Half-life: (1/2)y0 = y0 e^(-k*10), so e^(-10k) = 1/2, giving 10k = ln 2 and k = (ln 2)/10. (Adapted from the Vidaara NCERT/JEE bank.)
Question 9
A bacteria culture doubles every 3 hours and follows exponential growth. The growth constant k is:
A
(ln 2)/3
B
3 ln 2
C
ln 2
D
2/3
Solution: Doubling time: 2 y0 = y0 e^(3k), so e^(3k) = 2 and k = (ln 2)/3.
Question 10
If a quantity grows exponentially and triples in 5 years, its value after 10 years (starting from A0) is:
A
9 A0
B
6 A0
C
3 A0
D
A0 e^5
Solution: Tripling every 5 years: after 10 years (two tripling periods) the amount is 3 * 3 = 9 times the initial value, 9 A0.
Question 11
In the logistic model dP/dt = kP(1 - P/M), the constant M represents:
A
the carrying capacity
B
the initial population
C
the growth rate
D
the doubling time
Solution: M is the carrying capacity, the limiting population that the model approaches as t increases.
Question 12
In a logistic model, as t approaches infinity, the population P approaches:
A
the carrying capacity M
B
0
C
infinity
D
M/2
Solution: Logistic solutions level off at the carrying capacity, so P approaches M as t approaches infinity.
Question 13
The velocity of an object under gravity (ignoring air resistance) satisfies dv/dt = -g. If v(0) = v0, then v(t) equals:
A
v0 - g t
B
v0 + g t
C
-g t^2
D
v0 e^(-gt)
Solution: Antidifferentiating dv/dt = -g gives v = -g t + C. Using v(0) = v0 gives C = v0, so v(t) = v0 - g t.
Question 14
Is y = Cx^2 a solution of the differential equation x(dy/dx) = 2y?
A
Yes, for every constant C
B
No
C
Only for C = 1
D
Only at x = 0
Solution: y = Cx^2 gives dy/dx = 2Cx, so x(dy/dx) = 2Cx^2 = 2y. The equation holds for all C, so this is the general solution.
Question 15
Solve dy/dx = 3x^2 with y(1) = 5.
A
y = x^3 + 4
B
y = x^3 + 5
C
y = x^3
D
y = 3x^3 + 2
Solution: Antidifferentiate: y = x^3 + C. Using y(1) = 5: 1 + C = 5, so C = 4 and y = x^3 + 4.
Question 16
A colony of bacteria numbers 1000 initially and grows exponentially, reaching 4000 in 2 hours. Its size after 4 hours is:
A
16000
B
8000
C
12000
D
6000
Solution: It quadruples every 2 hours (1000 -> 4000). After another 2 hours (4 total), it quadruples again: 4000 -> 16000.
Question 17
A logistic population P(t) with carrying capacity 1000 currently has P = 400 and is increasing. As time goes on, P will:
A
keep increasing and level off near 1000
B
decrease to 0
C
increase without bound
D
oscillate
Solution: Below carrying capacity, a logistic population increases and asymptotically approaches the carrying capacity of 1000.
Question 18
Solve dy/dx = y^2 (for y not 0).
A
y = -1/(x + C)
B
y = e^x + C
C
y = x^2 + C
D
y = 1/(x + C)
Solution: Separate: dy/y^2 = dx. Integrate: -1/y = x + C1, so y = -1/(x + C).
Question 19
Water leaks from a tank so that the volume satisfies dV/dt = -0.2V. This describes:
A
exponential decay of the volume
B
linear decrease
C
exponential growth
D
constant volume
Solution: dV/dt = -0.2V is the exponential-decay model; the solution is V = V0 e^(-0.2t).
Question 20
For dy/dx = x + y, what is the slope of the solution curve at the point (1, 2)?
A
3
B
2
C
1
D
-1
Solution: The slope is dy/dx evaluated at the point: 1 + 2 = 3.