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Differential Equations: Class 12 Maths Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A wildlife researcher is tracking the growth of a deer population in a forest reserve. She observes that at any time t (in years), the rate of growth of the population P is proportional to the current population P. Initially (at t = 0), the population was 200 deer. After 2 years, the population grew to 300 deer.

A wildlife researcher is tracking the growth of a deer population in a forest reserve. She observes that at any time t (in years), the rate of growth of the population P is proportional to the current population P. Initially (at t = 0), the population was 200 deer. After 2 years, the population grew to 300 deer.

(i) Write the differential equation that models this situation. [1 mark]
(ii) Find the general solution of the differential equation. [1 mark]
(iii) Find the particular solution using the given initial conditions. Hence, find the population after 4 years. [2 marks]

Show answer
(i) Since the rate of growth of population P is proportional to P,

dP/dt = kP

where k is a positive constant of proportionality.

∴ The required differential equation is dP/dt = kP.

(ii) Separating variables,

dP/P = k dt

Integrating both sides,

∫ dP/P = ∫ k dt

⟹ log|P| = kt + C

⟹ P = e<super>kt + C</super> = Ae<super>kt</super>, where A = e<super>C</super> is an arbitrary constant.

∴ General solution is P = Ae<super>kt</super>.

(iii) Applying the initial condition: at t = 0, P = 200,

200 = Ae<super>0</super> = A

⟹ A = 200 ...(i)

Applying the second condition: at t = 2, P = 300,

300 = 200 e<super>2k</super>

⟹ e<super>2k</super> = 300/200 = 3/2

⟹ 2k = log(3/2)

⟹ k = (1/2) log(3/2) ...(ii)

∴ Particular solution is P = 200 e<super>(t/2) log(3/2)</super>.

This can be written as P = 200 · (3/2)<super>t/2</super>.

At t = 4:

P = 200 · (3/2)<super>4/2</super> = 200 · (3/2)<super>2</super>

⟹ P = 200 × 9/4 = 450

∴ The population after 4 years is 450 deer.
Q2Case-based4 marks

A scientist is studying the cooling of a cup of coffee. She models the temperature T (in °C) of the coffee at time t (minutes) using Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its temperature and the surrounding (room) temperature. The room temperature is 25°C.

A scientist is studying the cooling of a cup of coffee. She models the temperature T (in °C) of the coffee at time t (minutes) using Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its temperature and the surrounding (room) temperature. The room temperature is 25°C.

Based on this model, answer the following:

(i) Write the differential equation representing Newton's Law of Cooling. [1 mark]

(ii) What are the order and degree of this differential equation? [1 mark]

(iii) Find the general solution of the differential equation obtained in part (i). [2 marks]

Show answer
(i) According to Newton's Law of Cooling, the rate of change of temperature T is proportional to (T − 25), where 25°C is the room temperature.

∴ The differential equation is:

dT/dt = −k(T − 25)

where k > 0 is the constant of proportionality (the negative sign indicates cooling, i.e., T decreases as t increases).

(ii) The differential equation is dT/dt = −k(T − 25).

The highest order derivative present is dT/dt, which is of order 1.
The power (degree) of this highest order derivative is 1.

∴ Order = 1 and Degree = 1.

(iii) The differential equation is:

dT/dt = −k(T − 25)

Separating the variables:

dT/(T − 25) = −k dt

Integrating both sides,

∫ dT/(T − 25) = ∫ −k dt

⟹ log|T − 25| = −kt + C₁

⟹ |T − 25| = e<super>−kt + C₁</super>

⟹ T − 25 = A e<super>−kt</super> (where A = ±e<super>C₁</super> is an arbitrary constant)

∴ Required general solution is T = 25 + A e<super>−kt</super>
Q3Case-based4 marks

A city's urban planners are modelling the spread of a new public bicycle-sharing scheme. They observe that the rate at which new users (y) join the scheme at any time depends on both the current number of users and the total infrastructure capacity (x), and can be modelled by the differential equation:

(2xy) dy/dx = x² + y²

A city's urban planners are modelling the spread of a new public bicycle-sharing scheme. They observe that the rate at which new users (y) join the scheme at any time depends on both the current number of users and the total infrastructure capacity (x), and can be modelled by the differential equation:

(2xy) dy/dx = x² + y²

Based on the above information, answer the following questions:

(i) Show that the given differential equation is homogeneous. [1 mark]
(ii) Using the substitution y = vx, rewrite the differential equation as a separable equation in v and x. [1 mark]
(iii) Find the general solution of the differential equation. [2 marks]

OR

(iii) If initially (when x = 1) the number of users y = 1, find the particular solution. [2 marks]

Show answer
(i) The given differential equation can be written as:

dy/dx = (x² + y²) / (2xy)

Let F(x, y) = (x² + y²) / (2xy).

Replace x by λx and y by λy:

F(λx, λy) = ((λx)² + (λy)²) / (2·λx·λy) = (λ²x² + λ²y²) / (2λ²xy) = (x² + y²) / (2xy) = λ⁰ F(x, y)

∴ F(x, y) is a homogeneous function of degree zero.
∴ The given differential equation is homogeneous. [1]

(ii) Put y = vx ⟹ dy/dx = v + x·dv/dx.

Substituting into dy/dx = (x² + y²)/(2xy):

v + x·dv/dx = (x² + v²x²) / (2x·vx) = (1 + v²) / (2v)

x·dv/dx = (1 + v²)/(2v) − v = (1 + v² − 2v²)/(2v) = (1 − v²)/(2v)

∴ The separable equation is: 2v/(1 − v²) dv = dx/x [1]

(iii) Integrating both sides:

∫ 2v/(1 − v²) dv = ∫ dx/x

For the left side, put t = 1 − v² ⟹ dt = −2v dv, so 2v dv = −dt:

∫ (−dt)/t = −log|t| = −log|1 − v²|

∴ −log|1 − v²| = log|x| + log|C₁|

⟹ log|1 − v²| + log|x| = −log|C₁| = log|C| (absorbing sign into constant)

⟹ log|x(1 − v²)| = log|C|

⟹ x(1 − v²) = C

Back-substituting v = y/x:

x(1 − y²/x²) = C

⟹ x · (x² − y²)/x² = C

∴ Required general solution is x² − y² = Cx [2]

OR

(iii) From part (ii), integrating both sides:

∫ 2v/(1 − v²) dv = ∫ dx/x

⟹ −log|1 − v²| = log|x| + log|C|

⟹ x(1 − v²) = C

Back-substituting v = y/x:

x² − y² = Cx ...(i)

Applying the initial condition: when x = 1, y = 1:

(1)² − (1)² = C(1) ⟹ 0 = C ⟹ C = 0

Substituting C = 0 in (i):

x² − y² = 0

∴ Required particular solution is y² = x² (i.e., y = x, taking the positive branch since y, x > 0) [2]
Q4Case-based4 marks

A environmental scientist is modelling the spread of a pollutant in a river. The concentration y (in mg/L) of the pollutant at a distance x (in km) downstream satisfies the differential equation (x² + xy) dy = (x² + y²) dx, with y = 0 when x = 1.

A environmental scientist is modelling the spread of a pollutant in a river. The concentration y (in mg/L) of the pollutant at a distance x (in km) downstream satisfies the differential equation:

(x² + xy) dy = (x² + y²) dx

It is observed that when x = 1 km, the concentration y = 0 mg/L.

(i) Show that the differential equation is homogeneous. [1]
(ii) Using the substitution y = vx, reduce the equation to a separable form. [1]
(iii) Find the particular solution of the differential equation satisfying the given condition. [2]

Show answer
(i) Rewrite the equation as:

dy/dx = (x² + y²)/(x² + xy)

Let f(x, y) = (x² + y²)/(x² + xy).

Replace x by λx and y by λy:

f(λx, λy) = (λ²x² + λ²y²)/(λ²x² + λ²xy) = λ²(x² + y²)/λ²(x² + xy) = f(x, y)

∴ f(x, y) is a homogeneous function of degree 0.
∴ The given differential equation is homogeneous. [1]

(ii) Put y = vx ⟹ dy/dx = v + x·dv/dx.

Substituting into dy/dx = (x² + y²)/(x² + xy):

v + x·dv/dx = (x² + v²x²)/(x² + x·vx) = x²(1 + v²)/x²(1 + v) = (1 + v²)/(1 + v)

x·dv/dx = (1 + v²)/(1 + v) − v = (1 + v² − v − v²)/(1 + v) = (1 − v)/(1 + v)

∴ The separable form is:

(1 + v)/(1 − v) dv = dx/x [1]

(iii) Separating variables and integrating both sides:

∫(1 + v)/(1 − v) dv = ∫dx/x

Write (1 + v)/(1 − v) = −1 + 2/(1 − v):

∫[−1 + 2/(1 − v)] dv = ∫dx/x

−v − 2 log|1 − v| = log|x| + C ...(i)

Back-substituting v = y/x:

−y/x − 2 log|1 − y/x| = log|x| + C

−y/x − 2 log|(x − y)/x| = log|x| + C

−y/x − 2 [log|x − y| − log|x|] = log|x| + C

−y/x − 2 log|x − y| + 2 log|x| = log|x| + C

−y/x − 2 log|x − y| + log|x| = C ...(ii)

Applying the initial condition y = 0 when x = 1:

−0/1 − 2 log|1 − 0| + log|1| = C

0 − 2(0) + 0 = C ⟹ C = 0

Substituting C = 0 into (ii):

−y/x − 2 log|x − y| + log|x| = 0

∴ Required particular solution is: log|x| − 2 log|x − y| = y/x [2]
Q5Case-based4 marks

A water tank is being drained through a pipe. The rate of change of the volume V (in litres) of water in the tank with respect to time t (in minutes) satisfies the differential equation:

dV/dt + (2/t)·V = 6t, t > 0

Initially (at t = 1 minute), the tank contains 4 litres of water.

A water tank is being drained through a pipe. The rate of change of the volume V (in litres) of water in the tank with respect to time t (in minutes) satisfies the differential equation:

dV/dt + (2/t)·V = 6t, t > 0

Initially (at t = 1 minute), the tank contains 4 litres of water.

(i) Identify the integrating factor of the given differential equation. [1]
(ii) Find the general solution of the differential equation. [1]
(iii) Find the particular solution and hence determine the volume of water in the tank at t = 2 minutes. [2]

Show answer
(i) This is a linear DE of the form dV/dt + PV = Q, where P = 2/t and Q = 6t.

Integrating factor = e^(∫P dt) = e^(∫2/t dt) = e^(2 log t) = e^(log t²) = t².

∴ Integrating factor = t²

(ii) Multiplying both sides by the I.F. t²,

t² dV/dt + 2t·V = 6t³

⟹ d/dt (V·t²) = 6t³

Integrating both sides,

V·t² = ∫6t³ dt = (6t⁴)/4 + C = (3t⁴)/2 + C

∴ Required general solution is V·t² = (3/2)t⁴ + C

(iii) Applying the initial condition V = 4 when t = 1:

4·(1)² = (3/2)(1)⁴ + C

⟹ 4 = 3/2 + C

⟹ C = 4 − 3/2 = 5/2

∴ Required particular solution is V·t² = (3/2)t⁴ + 5/2

At t = 2:

V·(2)² = (3/2)(2)⁴ + 5/2

⟹ 4V = (3/2)(16) + 5/2

⟹ 4V = 24 + 5/2 = 48/2 + 5/2 = 53/2

⟹ V = 53/8

∴ Volume of water in the tank at t = 2 minutes is 53/8 litres.
Q6Case-based4 marks

A scientist is studying the spread of a rumour in a town. She models the situation using the differential equation

dy/dx = (y² + xy) / x²

where y represents the number of people (in hundreds) who have heard the rumour and x represents the time elapsed (in hours) since the rumour started.

A scientist is studying the spread of a rumour in a town. She models the situation using the differential equation

dy/dx = (y² + xy) / x²

where y represents the number of people (in hundreds) who have heard the rumour and x represents the time elapsed (in hours) since the rumour started.

(i) Show that the differential equation is homogeneous. [1 mark]
(ii) Using the substitution y = vx, reduce the equation to a separable form and hence find the general solution. [2 marks]
(iii) Given that at x = 1 hour, y = 1 hundred people, find the particular solution. Also find the number of people (in hundreds) who have heard the rumour after x = 2 hours. [1 mark]

Show answer
(i) Checking that the differential equation is homogeneous:

The given differential equation is

dy/dx = (y² + xy) / x²

Let F(x, y) = (y² + xy) / x².

Replace x by λx and y by λy:

F(λx, λy) = ((λy)² + (λx)(λy)) / (λx)²
= (λ²y² + λ²xy) / λ²x²
= (y² + xy) / x²
= λ⁰ · F(x, y)

∴ F(x, y) is a homogeneous function of degree 0.

∴ The given differential equation is homogeneous. [1]

─────────────────────────────────────────────
(ii) Reducing to separable form and finding the general solution:

Put y = vx ⟹ dy/dx = v + x · dv/dx

Substituting into dy/dx = (y² + xy) / x²,

v + x · dv/dx = (v²x² + x · vx) / x²
= v² + v

⟹ x · dv/dx = v² + v − v

⟹ x · dv/dx = v²

Separating variables (v ≠ 0):

dv / v² = dx / x

Integrating both sides,

∫ v⁻² dv = ∫ dx/x

−1/v = log|x| + C₁ (where C₁ is an arbitrary constant)

Back-substituting v = y/x:

−x/y = log|x| + C₁

∴ Required general solution is −x/y = log|x| + C, where C is an arbitrary constant. [2]

─────────────────────────────────────────────
(iii) Finding the particular solution and value at x = 2:

Applying the initial condition: y = 1 when x = 1,

−1/1 = log|1| + C
−1 = 0 + C
∴ C = −1

∴ Required particular solution is

−x/y = log|x| − 1

⟹ x/y = 1 − log x (for x > 0)

⟹ y = x / (1 − log x)

At x = 2:

y = 2 / (1 − log 2)

∴ The number of people (in hundreds) who have heard the rumour after 2 hours is 2/(1 − log 2) hundred. [1]
Q7Case-based4 marks

A biologist is studying the growth of bacteria in a culture. She observes that the rate of increase of the number of bacteria at any instant is proportional to the number of bacteria present at that instant. Let N be the number of bacteria at time t (in hours).

A biologist is studying the growth of bacteria in a culture. She observes that the rate of increase of the number of bacteria at any instant is proportional to the number of bacteria present at that instant. Let N be the number of bacteria at time t (in hours).

Based on the above information, answer the following questions:

(i) Write the differential equation representing the model described above. Also state its order and degree.

(ii) If the number of bacteria doubles in 2 hours, find the value of the proportionality constant k (in terms of log).

(iii) If the initial count of bacteria is N₀ = 500, find the number of bacteria after 6 hours. (Use the value of k found in part (ii).)

OR

(iii) If N = 1000 when t = 0, find the time t (in hours) at which the number of bacteria becomes 8000.

Show answer
(i) The rate of increase of bacteria is proportional to the number present, so:

dN/dt = kN, where k > 0 is the proportionality constant.

∴ The required differential equation is dN/dt = kN.

Order = 1 (highest derivative is dN/dt), Degree = 1 (power of dN/dt is 1).

(ii) Separating variables in dN/dt = kN:

dN/N = k dt

Integrating both sides,

∫ dN/N = ∫ k dt

⟹ log|N| = kt + C ... (i)

At t = 0, let N = N₀ (initial count). Substituting in (i),

log|N₀| = 0 + C ⟹ C = log|N₀|

∴ log|N| = kt + log|N₀|

⟹ log(N/N₀) = kt ... (ii)

Given: bacteria doubles in 2 hours, i.e., N = 2N₀ when t = 2.

Substituting in (ii),

log(2N₀/N₀) = k × 2

⟹ 2k = log 2

∴ k = (log 2)/2

(iii) From (ii): log(N/N₀) = kt, with N₀ = 500 and k = (log 2)/2.

At t = 6:

log(N/500) = ((log 2)/2) × 6 = 3 log 2 = log(2³) = log 8

⟹ N/500 = 8

∴ N = 4000

∴ The number of bacteria after 6 hours is 4000.

OR

(iii) From (ii): log(N/N₀) = kt, with N₀ = 1000 and k = (log 2)/2.

We need N = 8000:

log(8000/1000) = ((log 2)/2) × t

⟹ log 8 = ((log 2)/2) × t

⟹ 3 log 2 = ((log 2)/2) × t

⟹ t = 3 log 2 × (2/log 2) = 6

∴ The number of bacteria becomes 8000 after t = 6 hours.
Q8Case-based4 marks

A water tank is being filled by a pipe. At time t (in minutes), the rate of change of the volume V (in litres) of water in the tank satisfies the differential equation:

dV/dt + (2/(t+1))·V = 4(t+1)

Initially (at t = 0), the tank contains 1 litre of water.

A water tank is being filled by a pipe. At time t (in minutes), the rate of change of the volume V (in litres) of water in the tank satisfies the differential equation:

dV/dt + (2/(t+1))·V = 4(t+1)

Initially (at t = 0), the tank contains 1 litre of water.

(i) Identify the type of the differential equation and find the Integrating Factor (I.F.).
(ii) Find the general solution of the differential equation.
(iii) Find the particular solution satisfying the initial condition V(0) = 1. Hence find the volume of water in the tank at t = 1 minute.

Show answer
(i) This is a linear differential equation of the form dV/dt + P·V = Q, where P = 2/(t+1) and Q = 4(t+1).

I.F. = e^(∫P dt) = e^(∫2/(t+1) dt) = e^(2·log|t+1|) = e^(log(t+1)²) = (t+1)²

∴ I.F. = (t+1)²

[1 mark]

(ii) Multiplying both sides by I.F. = (t+1)²,

The general solution is given by:

V·(t+1)² = ∫4(t+1)·(t+1)² dt

⟹ V·(t+1)² = ∫4(t+1)³ dt

⟹ V·(t+1)² = 4·(t+1)⁴/4 + C

⟹ V·(t+1)² = (t+1)⁴ + C

∴ Required general solution is V = (t+1)² + C/(t+1)²

[1 mark]

(iii) Applying the initial condition V(0) = 1:

1 = (0+1)² + C/(0+1)²

⟹ 1 = 1 + C

⟹ C = 0

∴ Required particular solution is V = (t+1)²

At t = 1:

V = (1+1)² = (2)² = 4

∴ The volume of water in the tank at t = 1 minute is 4 litres.

[2 marks]
Q9MCQ1 mark

The general solution of the differential equation dy/dx = e^(x+y) is:

Show answer
Option (A) is correct.

Explanation: The given DE is dy/dx = e^(x+y) = e^x · e^y.

Separating variables: e^(−y) dy = e^x dx.

Integrating both sides, ∫e^(−y) dy = ∫e^x dx ⟹ −e^(−y) = e^x − C₁ ⟹ e^x + e^(−y) = C.

∴ The general solution is e^x + e^(−y) = C.
Q10MCQ1 mark

The order of the differential equation d²y/dx² + sin(dy/dx) = x is:

Show answer
Option (C) is correct.

Explanation: The order of a differential equation is the order of the highest-order derivative present in the equation. Here, the highest-order derivative is d²y/dx², which is of order 2. (Note: The degree is not defined since the equation is not a polynomial in its derivatives, but the order is still 2.)

∴ Order = 2.
Q11MCQ1 mark

The degree of the differential equation is:

Show answer
Option (B) is correct.

Explanation: The degree of a differential equation is the power of the highest-order derivative, provided the equation is a polynomial in its derivatives. The highest-order derivative here is d²y/dx², and its power is 2. ∴ Degree = 2.
Q12MCQ1 mark

The degree of the differential equation is:

Show answer
Option (D) is correct.

Explanation: The degree of a differential equation is defined only when it can be expressed as a polynomial in its derivatives. Here, the term sin(dy/dx) is a transcendental (non-polynomial) function of the derivative dy/dx. Since the equation cannot be written as a polynomial in its derivatives, its degree is not defined.
Q13MCQ1 mark

Find the degree of the differential equation d²y/dx² + (dy/dx)³ + 6y = 0.

Show answer
Option (A) is correct.

Explanation: The degree of a differential equation is the power of the highest-order derivative, provided the equation is a polynomial in all its derivatives.

Here, the highest-order derivative is d²y/dx², and its power is 1.

∴ Degree = 1.
Q14Short Answer2 marks

Find the general solution of the differential equation: dy/dx = e^(x+y) + x²e^y

Show answer
dy/dx = e<super>x+y</super> + x<super>2</super>e<super>y</super>

⟹ dy/dx = e<super>y</super>(e<super>x</super> + x<super>2</super>)

Separating the variables,

e<super>−y</super> dy = (e<super>x</super> + x<super>2</super>) dx

Integrating both sides,

∫e<super>−y</super> dy = ∫(e<super>x</sup> + x<super>2</super>) dx

−e<super>−y</super> = e<super>x</super> + x<super>3</super>/3 + C

∴ Required general solution is −e<super>−y</super> = e<super>x</super> + x<super>3</super>/3 + C
Q15Short Answer2 marks

Find the general solution of the differential equation: cos x · dy + y sin x · dx = 0.

Show answer
The given differential equation is:

cos x · dy + y sin x · dx = 0

Separating the variables,

dy/y = −(sin x / cos x) dx = −tan x · dx

Integrating both sides,

∫ dy/y = −∫ tan x · dx

log|y| = −log|sec x| + log C

log|y| = log|C / sec x|

∴ y sec x = C

or equivalently, y = C cos x

∴ Required general solution is y sec x = C (or y = C cos x).
Q16Short Answer2 marks

Find the sum of the order and the degree of the differential equation:

Show answer
The given differential equation is:

(d²y/dx²)<super>3</super> + (dy/dx)<super>2</super> + sin(dy/dx) + 1 = 0

The term sin(dy/dx) involves a transcendental (trigonometric) function of the derivative dy/dx. ∵ the equation is not a polynomial in its derivatives, its degree is not defined.

∴ Order = 2 (the highest order derivative present is d²y/dx²)

∴ Degree = not defined

∴ Sum of order and degree = not defined.
Q17Short Answer2 marks

Find the general solution of the differential equation: dy/dx = e^(x+y).

Show answer
The given differential equation is dy/dx = e<super>x+y</super>.

Separating the variables,

e<super>−y</super> dy = e<super>x</super> dx

Integrating both sides,

∫e<super>−y</super> dy = ∫e<super>x</super> dx

⟹ −e<super>−y</super> = e<super>x</super> + C

∴ Required general solution is e<super>x</super> + e<super>−y</super> = C, where C is an arbitrary constant.
Q18Short Answer3 marks

Find the general solution of the differential equation: (1 + y²) dx = (tan⁻¹y − x) dy.

Show answer
Rewriting the equation:

(1 + y²) dx = (tan⁻¹y − x) dy

⟹ dx/dy = (tan⁻¹y − x)/(1 + y²)

⟹ dx/dy + x/(1 + y²) = tan⁻¹y/(1 + y²)

This is a linear DE of the form dx/dy + Px = Q, where
P = 1/(1 + y²), Q = tan⁻¹y/(1 + y²).

I.F. = e^(∫P dy) = e^(∫dy/(1 + y²)) = e^(tan⁻¹y)

Multiplying both sides by I.F.:

d/dy [x · e^(tan⁻¹y)] = (tan⁻¹y/(1 + y²)) · e^(tan⁻¹y)

Integrating both sides,

x · e^(tan⁻¹y) = ∫ (tan⁻¹y/(1 + y²)) · e^(tan⁻¹y) dy ...(i)

For the integral on R.H.S., put t = tan⁻¹y ⟹ dt = dy/(1 + y²).

∴ ∫ (tan⁻¹y/(1 + y²)) · e^(tan⁻¹y) dy = ∫ t · eᵗ dt

Using integration by parts (ILATE: u = t, dv = eᵗ dt):

= t · eᵗ − ∫ eᵗ dt = t · eᵗ − eᵗ = eᵗ(t − 1)

Back-substituting t = tan⁻¹y:

= e^(tan⁻¹y)(tan⁻¹y − 1)

Substituting in (i):

x · e^(tan⁻¹y) = e^(tan⁻¹y)(tan⁻¹y − 1) + C

∴ Required general solution is x = (tan⁻¹y − 1) + C · e^(−tan⁻¹y)
Q19Short Answer3 marks

Find the particular solution of the differential equation (1 + x²) dy/dx + 2xy = x, given that y = 1 when x = 0.

Show answer
This is a linear DE of the form dy/dx + Py = Q, where P = 2x/(1 + x²) and Q = x/(1 + x²).

Dividing throughout by (1 + x²),

dy/dx + [2x/(1 + x²)]y = x/(1 + x²) ...(i)

Integrating factor:

I.F. = e^(∫P dx) = e^(∫2x/(1 + x²) dx)

Put 1 + x² = t ⟹ 2x dx = dt,

∴ ∫2x/(1 + x²) dx = ∫dt/t = log|t| = log(1 + x²)

∴ I.F. = e^(log(1 + x²)) = (1 + x²)

Multiplying both sides of (i) by I.F. = (1 + x²),

d/dx [y·(1 + x²)] = x/(1 + x²) · (1 + x²) = x

Integrating both sides,

y·(1 + x²) = ∫x dx

y·(1 + x²) = x²/2 + C ...(ii)

Applying the initial condition y = 1 when x = 0:

1·(1 + 0) = 0/2 + C ⟹ C = 1

Substituting in (ii),

∴ Required particular solution is y(1 + x²) = x²/2 + 1
Q20Short Answer3 marks

Find the particular solution of the differential equation x dy = (2y + x³ cos x) dx, given that y = 0 when x = π.

Show answer
Rewriting the equation:

x dy = (2y + x³ cos x) dx

⟹ dy/dx = 2y/x + x² cos x

⟹ dy/dx − (2/x)y = x² cos x

This is a linear DE of the form dy/dx + Py = Q, where P = −2/x and Q = x² cos x.

I.F. = e^(∫P dx) = e^(∫−2/x dx) = e^(−2 log|x|) = e^(log x⁻²) = x⁻² = 1/x²

Multiplying both sides by I.F.:

d/dx [y · (1/x²)] = x² cos x · (1/x²)

⟹ d/dx [y/x²] = cos x

Integrating both sides,

y/x² = ∫cos x dx

y/x² = sin x + C ...(i)

Applying the initial condition y = 0 when x = π:

0/π² = sin π + C

⟹ 0 = 0 + C

⟹ C = 0

Substituting C = 0 in (i):

y/x² = sin x

∴ Required particular solution is y = x² sin x.
Q21Short Answer3 marks

Solve the differential equation: (2y³ + x) dy = y dx, given that x = 1 when y = 1.

Show answer
The given differential equation is:
(2y³ + x) dy = y dx

Rewriting as dx/dy:
⟹ y · dx/dy = 2y³ + x
⟹ dx/dy − x/y = 2y²

This is a linear DE of the form dx/dy + Px = Q, where P = −1/y, Q = 2y².

I.F. = e^(∫P dy) = e^(∫−1/y dy) = e^(−log|y|) = e^(log|1/y|) = 1/y

Multiplying both sides by I.F. = 1/y,

x · (1/y) = ∫2y² · (1/y) dy + C

⟹ x/y = ∫2y dy + C

⟹ x/y = y² + C ...(i)

Applying the initial condition x = 1 when y = 1:

1/1 = (1)² + C
⟹ 1 = 1 + C
⟹ C = 0

Substituting C = 0 in (i):

x/y = y²

∴ Required particular solution is x = y³.
Q22Short Answer3 marks

Solve the following differential equation: x dy − y dx = √(x² + y²) dx

Show answer
Rewriting the equation:

x dy − y dx = √(x² + y²) dx

⟹ x dy = [y + √(x² + y²)] dx

⟹ dy/dx = [y + √(x² + y²)] / x …(i)

This is a homogeneous differential equation (each term is of degree 1).

Put y = vx ⟹ dy/dx = v + x · dv/dx.

Substituting in (i),

v + x · dv/dx = [vx + √(x² + v²x²)] / x

⟹ v + x · dv/dx = [vx + x√(1 + v²)] / x

⟹ v + x · dv/dx = v + √(1 + v²)

⟹ x · dv/dx = √(1 + v²)

Separating variables,

dv / √(1 + v²) = dx / x

Integrating both sides,

∫ dv / √(1 + v²) = ∫ dx / x

log|v + √(1 + v²)| = log|x| + log|C|

⟹ log|v + √(1 + v²)| = log|Cx|

⟹ v + √(1 + v²) = Cx

Back-substituting v = y/x,

y/x + √(1 + y²/x²) = Cx

⟹ y/x + √(x² + y²)/x = Cx

⟹ y + √(x² + y²) = Cx²

∴ Required general solution is y + √(x² + y²) = Cx²
Q23Short Answer3 marks

Solve the differential equation: (x² + 1) dy/dx + 2xy = x² + 1, given that y = 1 when x = 0.

Show answer
This is a linear DE of the form dy/dx + Py = Q, where P = 2x/(x² + 1) and Q = 1.

I.F. = e^(∫P dx) = e^(∫2x/(x² + 1) dx) = e^(log|x² + 1|) = x² + 1

Multiplying both sides by I.F.,

d/dx [y · (x² + 1)] = 1 · (x² + 1)

Integrating both sides,

y(x² + 1) = ∫(x² + 1) dx

y(x² + 1) = x³/3 + x + C ...(i)

Applying initial condition y = 1 when x = 0 in (i),

1(0 + 1) = 0 + 0 + C ⟹ C = 1

Substituting C = 1 in (i),

∴ Required particular solution is y(x² + 1) = x³/3 + x + 1
Q24Case-based4 marks

A tank initially contains 100 litres of pure water. Brine (salt solution) containing 2 grams of salt per litre enters the tank at a rate of 3 litres per minute. The well-mixed solution leaves the tank at the same rate of 3 litres per minute. Let x(t) denote the amount of salt (in grams) in the tank at time t (in minutes).

A tank initially contains 100 litres of pure water. Brine (salt solution) containing 2 grams of salt per litre enters the tank at a rate of 3 litres per minute. The well-mixed solution leaves the tank at the same rate of 3 litres per minute.

Let x(t) denote the amount of salt (in grams) in the tank at time t (in minutes).

(i) Write the differential equation governing x(t).
(ii) Identify the type of differential equation and write its Integrating Factor (I.F.).
(iii) Solve the differential equation to find x(t), given that initially the tank contains no salt. Hence find the amount of salt in the tank after 10 minutes. (Use e⁻⁰·³ ≈ 0·741)

Show answer
(i) [1 mark]

Rate of salt entering = 2 × 3 = 6 grams per minute.

Concentration of salt in tank at time t = x/100 grams per litre.

Rate of salt leaving = (x/100) × 3 = 3x/100 grams per minute.

∴ dx/dt = 6 − 3x/100

(ii) [1 mark]

Rewriting: dx/dt + 3x/100 = 6

This is a linear DE of the form dx/dt + Px = Q, where P = 3/100, Q = 6.

∴ I.F. = e<super>∫(3/100)dt</super> = e<super>3t/100</super>

(iii) [2 marks]

Multiplying both sides by I.F.,

d/dt [x · e<super>3t/100</super>] = 6 · e<super>3t/100</super>

Integrating both sides,

x · e<super>3t/100</super> = 6 · (100/3) · e<super>3t/100</super> + C

x · e<super>3t/100</super> = 200 e<super>3t/100</super> + C

∴ x = 200 + C · e<super>−3t/100</super> ...(i)

Applying initial condition: at t = 0, x = 0,

0 = 200 + C · e<super>0</super> ⟹ C = −200

∴ Required particular solution is x(t) = 200(1 − e<super>−3t/100</super>)

At t = 10 minutes,

x(10) = 200(1 − e<super>−30/100</super>) = 200(1 − e<super>−0·3</super>)

= 200(1 − 0·741)

= 200 × 0·259

∴ Amount of salt in the tank after 10 minutes = 51·8 grams
Q25Case-based4 marks

A pharmaceutical company models the concentration C (in mg/L) of a drug in a patient's bloodstream. The rate of change of concentration satisfies the differential equation dC/dt + (2/(t+1))·C = 6/(t+1)², t ≥ 0, where t is the time in hours after the drug is administered intravenously.

A pharmaceutical company models the concentration C (in mg/L) of a drug in a patient's bloodstream using the differential equation:

dC/dt + (2/(t+1))·C = 6/(t+1)², t ≥ 0

where t is the time in hours after administration.

(i) Identify the integrating factor of the given differential equation. [1]
(ii) Find the general solution of the differential equation. [1]
(iii) If the initial concentration at t = 0 is C = 2 mg/L, find the particular solution and determine the concentration of the drug in the bloodstream at t = 2 hours. [2]

Show answer
(i) This is a linear DE of the form dC/dt + P·C = Q, where P = 2/(t+1) and Q = 6/(t+1)².

I.F. = e^(∫P dt) = e^(∫2/(t+1) dt) = e^(2 log|t+1|) = (t+1)²

∴ Integrating Factor = (t+1)²

(ii) Multiplying both sides by I.F. = (t+1)²,

d/dt [C·(t+1)²] = 6/(t+1)² · (t+1)² = 6

Integrating both sides,

C·(t+1)² = ∫6 dt = 6t + C₁

∴ Required general solution is C·(t+1)² = 6t + C₁

(iii) Applying the initial condition: at t = 0, C = 2 mg/L,

2·(0+1)² = 6(0) + C₁

⟹ C₁ = 2

∴ Required particular solution is C·(t+1)² = 6t + 2

⟹ C = (6t + 2)/(t+1)²

At t = 2 hours,

C = (6×2 + 2)/(2+1)² = (12 + 2)/9 = 14/9

∴ The concentration of the drug at t = 2 hours is 14/9 mg/L.
Q26Case-based4 marks

A scientist is studying the spread of a rumour in a small town of fixed population 1000. She models the rate of spread using the differential equation dP/dt = kP(1000 − P), where P is the number of people who have heard the rumour at time t (in hours), and k is a positive constant.

A scientist is studying the spread of a rumour in a small town of fixed population 1000. She models the rate of spread using the differential equation:

dP/dt = kP(1000 − P)

where P is the number of people who have heard the rumour at time t (in hours), and k is a positive constant.

(i) Initially (at t = 0), 100 people have heard the rumour. At t = 2 hours, 250 people have heard it. Using the method of partial fractions, find the general solution of the differential equation. [2 marks]

(ii) Using the initial conditions given, find the value of k (in terms of logarithm), and hence determine the number of people who have heard the rumour after 4 hours. (Give your answer to the nearest integer.) [2 marks]

Show answer
Part (i): General Solution [2 marks]

The differential equation is:

dP/dt = kP(1000 − P)

Separating variables:

dP / [P(1000 − P)] = k dt

Using partial fractions, let:

1/[P(1000 − P)] = A/P + B/(1000 − P)

On comparing coefficients: 1 = A(1000 − P) + BP

Putting P = 0: 1 = 1000A ⟹ A = 1/1000
Putting P = 1000: 1 = 1000B ⟹ B = 1/1000

∴ 1/[P(1000 − P)] = (1/1000)[1/P + 1/(1000 − P)]

Integrating both sides:

(1/1000) ∫ [1/P + 1/(1000 − P)] dP = ∫ k dt

(1/1000) [log|P| − log|1000 − P|] = kt + C

(1/1000) log|P/(1000 − P)| = kt + C

∴ General solution: log|P/(1000 − P)| = 1000kt + C₁ (where C₁ = 1000C)

∴ P/(1000 − P) = Ae<super>1000kt</super> (where A = e<super>C₁</super>)

Part (ii): Finding k and P at t = 4 hours [2 marks]

Applying initial condition t = 0, P = 100:

100/(1000 − 100) = A · e<super>0</super>

100/900 = A ⟹ A = 1/9

∴ P/(1000 − P) = (1/9) e<super>1000kt</super> ...(i)

Applying condition t = 2, P = 250:

250/(1000 − 250) = (1/9) e<super>2000k</super>

250/750 = (1/9) e<super>2000k</super>

1/3 = (1/9) e<super>2000k</super>

e<super>2000k</super> = 3

∴ 2000k = log 3 ⟹ k = (log 3)/2000

∴ 1000k = (log 3)/2

Substituting in (i):

P/(1000 − P) = (1/9) e<super>(log 3)/2 · t</super> = (1/9) · 3<super>t/2</super>

At t = 4:

P/(1000 − P) = (1/9) · 3<super>2</super> = (1/9) · 9 = 1

P = 1000 − P ⟹ 2P = 1000 ⟹ P = 500

∴ The number of people who have heard the rumour after 4 hours = 500.
Q27Case-based4 marks

A city's water reservoir monitors the rate at which water level changes. An engineer observes that the rate of change of water level h (in metres) with respect to time t (in hours) is jointly proportional to the current water level h and the square of time t. At t = 0, the water level is 2 metres, and at t = 1 hour, the water level is 2e metres.

A city's water reservoir monitors the rate at which water level changes. An engineer observes that the rate of change of water level h (in metres) with respect to time t (in hours) is jointly proportional to the current water level h and the square of time t. At t = 0, the water level is 2 metres, and at t = 1 hour, the water level is 2e metres.

Based on the above information, answer the following questions:
(i) Write the differential equation representing this situation. [1 mark]
(ii) Find the general solution of this differential equation. [1 mark]
(iii) Find the particular solution using the given initial conditions. [2 marks]

Show answer
(i) Since the rate of change of h with respect to t is jointly proportional to h and t², the differential equation is:

dh/dt = k·h·t²

where k is a constant of proportionality.

(ii) Separating the variables:

dh/h = k·t² dt

Integrating both sides,

∫ dh/h = ∫ k·t² dt

⟹ log|h| = k·t³/3 + C

⟹ |h| = e^(kt³/3 + C)

∴ h = Ae^(kt³/3), where A = ±e^C is an arbitrary constant.

(iii) Applying the initial condition h = 2 when t = 0:

2 = A·e^(k·0/3)

⟹ 2 = A·e^0 = A

∴ A = 2

So h = 2e^(kt³/3) ...(i)

Applying the second condition h = 2e when t = 1:

2e = 2e^(k·1/3)

⟹ e^1 = e^(k/3)

⟹ k/3 = 1

⟹ k = 3 ...(ii)

Substituting (ii) into (i),

∴ Required particular solution is h = 2e^(t³)
Q28Case-based4 marks

A biologist models the spread of an invasive plant species in a wetland reserve. The fraction y of the reserve covered by the plant satisfies dy/dx = y(1 − y) − y², where x is time in weeks. At x = 0, the plant covers 1/4 of the reserve.

A biologist is studying the spread of an invasive plant species in a rectangular wetland reserve. She models the rate of change of the fraction y of the reserve covered by the plant as follows: the fraction grows at a rate proportional to the product of the fraction covered and the fraction not yet covered, but is also subject to a removal effort proportional to the square of the fraction covered. Mathematically, she writes:

dy/dx = y(1 − y) − y²,

where x denotes time (in weeks) and y = y(x) denotes the fraction covered, 0 < y < 1.

(i) Show that the differential equation can be written as dy/dx = y − 2y².

(ii) Identify the type of this differential equation and rewrite it in a suitable standard form.

(iii) Find the particular solution given that at x = 0, the plant covers 1/4 of the reserve. Express y explicitly in terms of x.

OR

(iii) Using the substitution v = 1/y, reduce the equation dy/dx = y − 2y² to a linear differential equation in v and x, and hence find the general solution.

Show answer
(i) [1 mark]

dy/dx = y(1 − y) − y²
= y − y² − y²
= y − 2y²

∴ dy/dx = y − 2y² Hence shown.

──────────────────────────────────────

(ii) [1 mark]

Dividing both sides by y², the equation dy/dx = y − 2y² is a Bernoulli equation in y.

Rewrite as:
dy/dx − y = −2y² ...(i)

This is a Bernoulli differential equation of the form dy/dx + P(x)y = Q(x)yⁿ with P = −1, Q = −2, n = 2.

──────────────────────────────────────

(iii) [2 marks — Main option]

Put v = 1/y ⟹ dv/dx = −(1/y²)(dy/dx).

Dividing equation (i) by y²:
(1/y²)(dy/dx) − (1/y) = −2

⟹ −dv/dx − v = −2

⟹ dv/dx + v = 2 ...(ii)

This is a linear DE of the form dv/dx + Pv = Q, where P = 1, Q = 2.

I.F. = e^(∫1 dx) = eˣ

Multiplying (ii) by eˣ:
d/dx(v · eˣ) = 2eˣ

Integrating both sides,
v · eˣ = 2eˣ + C

⟹ v = 2 + Ce^(−x)

Back-substituting v = 1/y:
1/y = 2 + Ce^(−x) ...(iii)

Applying initial condition: at x = 0, y = 1/4, so 1/y = 4.
4 = 2 + C · e⁰
⟹ C = 2

Substituting in (iii):
1/y = 2 + 2e^(−x)

∴ Required particular solution is y = 1/(2 + 2e^(−x)) = 1/(2(1 + e^(−x)))

──────────────────────────────────────

(iii) OR [2 marks — Alternative option]

Put v = 1/y ⟹ dv/dx = −(1/y²)(dy/dx).

Dividing dy/dx = y − 2y² throughout by y²:
(1/y²)(dy/dx) = (1/y) − 2

⟹ −dv/dx = v − 2

⟹ dv/dx + (−1)v = −2 + 2

More carefully:
−dv/dx = v − 2 ⟹ dv/dx = 2 − v ⟹ dv/dx + v = 2

This is a linear DE of the form dv/dx + Pv = Q, where P = 1, Q = 2.

I.F. = e^(∫P dx) = e^(∫1 dx) = eˣ

Multiplying throughout by eˣ:
d/dx(v · eˣ) = 2eˣ

Integrating both sides,
v · eˣ = 2eˣ + C

⟹ 1/y = 2 + Ce^(−x)

∴ Required general solution is 1/y = 2 + Ce^(−x), i.e., y = 1/(2 + Ce^(−x)).
Q29Case-based4 marks

A scientist is modelling the cooling of a liquid in a laboratory. The rate of decrease of temperature T (in °C) of the liquid at any time t (in minutes) is proportional to the difference between T and the surrounding room temperature, which is 25°C. Initially (at t = 0), the temperature of the liquid is 85°C, and after 5 minutes the temperature falls to 65°C.

A scientist is modelling the cooling of a liquid in a laboratory. The rate of decrease of temperature T (in °C) of the liquid at any time t (in minutes) is proportional to the difference between T and the surrounding room temperature, which is 25°C. Initially (at t = 0), the temperature of the liquid is 85°C, and after 5 minutes the temperature falls to 65°C.

Based on this information, answer the following:
(i) Write the differential equation that models this situation. [1]
(ii) Find the general solution of the differential equation obtained in part (i). [1]
(iii) Find the particular solution using the given initial conditions. Hence, find the temperature of the liquid after another 5 minutes (i.e., at t = 10 minutes). [2]

Show answer
(i) The rate of decrease of temperature T with respect to time t is proportional to (T − 25).

∴ The required differential equation is:

dT/dt = −k(T − 25), where k > 0 is a constant of proportionality.

(ii) Separating the variables:

dT/(T − 25) = −k dt

Integrating both sides,

∫ dT/(T − 25) = ∫ −k dt

⟹ log|T − 25| = −kt + C

∴ General solution: log|T − 25| = −kt + C

(iii) Applying initial condition: at t = 0, T = 85°C,

log|85 − 25| = −k(0) + C

⟹ log 60 = C ...(i)

Applying the second condition: at t = 5, T = 65°C,

log|65 − 25| = −5k + C

⟹ log 40 = −5k + log 60 ...(ii)

From (i) and (ii),

−5k = log 40 − log 60 = log(40/60) = log(2/3)

⟹ k = −(1/5) log(2/3) = (1/5) log(3/2)

Substituting C = log 60 and k = (1/5) log(3/2) in the general solution:

log|T − 25| = −(t/5) log(3/2) + log 60

This is the particular solution.

To find T at t = 10 minutes:

log|T − 25| = −(10/5) log(3/2) + log 60

⟹ log|T − 25| = −2 log(3/2) + log 60

⟹ log|T − 25| = log(2/3)² + log 60

⟹ log|T − 25| = log(4/9) + log 60

⟹ log|T − 25| = log(60 × 4/9)

⟹ log|T − 25| = log(240/9)

⟹ T − 25 = 240/9

⟹ T = 25 + 240/9 = (225 + 240)/9 = 465/9

∴ The temperature of the liquid at t = 10 minutes is 465/9 °C ≈ 51.67°C.
Q30Case-based4 marks

A scientist models the spread of a rumour in a closed community of 1000 people. The rate of spread is jointly proportional to the number of people who have heard the rumour (y) and the number who have not yet heard it (1000 − y), where t is time in hours.

A scientist is modelling the spread of a rumour in a closed community of 1000 people. She observes that the rate at which the rumour spreads is jointly proportional to the number of people who have heard it and the number who have not yet heard it. Let y denote the number of people who have heard the rumour at time t (in hours).

(i) Write the differential equation that models this situation.

(ii) Identify the type of this differential equation (variable separable / linear / homogeneous) and separate the variables.

(iii) Solve the differential equation to find y as a function of t, given that at t = 0 exactly 100 people have heard the rumour. Express your answer in the form y = f(t), keeping the proportionality constant as k.

Show answer
(i) Setting up the differential equation

The rate of change of y with respect to t is jointly proportional to y and (1000 − y).

∴ dy/dt = k · y(1000 − y), where k > 0 is the proportionality constant.

(ii) Type and separation of variables

This is a variable separable differential equation.

Separating the variables:

dy / [y(1000 − y)] = k dt ...(i)

(iii) Solving with the initial condition

Integrating both sides of (i),

∫ dy / [y(1000 − y)] = ∫ k dt

Resolving the left side using partial fractions:

1 / [y(1000 − y)] = A/y + B/(1000 − y)

Multiplying both sides by y(1000 − y):

1 = A(1000 − y) + By

Putting y = 0: 1 = 1000A ⟹ A = 1/1000
Putting y = 1000: 1 = 1000B ⟹ B = 1/1000

∴ 1 / [y(1000 − y)] = (1/1000)[1/y + 1/(1000 − y)]

Substituting back:

(1/1000) ∫ [1/y + 1/(1000 − y)] dy = ∫ k dt

Integrating both sides,

(1/1000) [log|y| − log|1000 − y|] = kt + C

(1/1000) log|y / (1000 − y)| = kt + C ...(ii)

Applying the initial condition: at t = 0, y = 100,

(1/1000) log|100 / (1000 − 100)| = k(0) + C

(1/1000) log(100/900) = C

C = (1/1000) log(1/9)

Substituting C back into (ii):

(1/1000) log|y / (1000 − y)| = kt + (1/1000) log(1/9)

(1/1000) log|y / (1000 − y)| − (1/1000) log(1/9) = kt

(1/1000) log[9y / (1000 − y)] = kt

log[9y / (1000 − y)] = 1000kt

9y / (1000 − y) = e<super>1000kt</super>

9y = (1000 − y) e<super>1000kt</super>

9y = 1000 e<super>1000kt</super> − y e<super>1000kt</super>

y(9 + e<super>1000kt</super>) = 1000 e<super>1000kt</super>

∴ y = 1000 e<super>1000kt</super> / (9 + e<super>1000kt</super>)

This is the required particular solution. It can equivalently be written as:

∴ y = 1000 / (1 + 9e<super>−1000kt</super>)

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