CBSE Maths › Class 12 PYQs › Differential Equations

Differential Equations: 4 marks Questions (CBSE Class 12)

3 different 4 marks questions on Differential Equations from CBSE Class 12 Maths board exams 2024–2026, newest first.

1 mark (50)3 marks (31)4 marks (3)5 marks (7)

Q384 marksCase StudyDifferential Equations CBSE 2025 · 65/4/1

During a heavy gaming session, the temperature of a student’s laptop processor increases significantly. After the session, the processor begins to cool down, and the rate of cooling is proportional to the difference between the processor’s temperature and the room temperature ( $2 5^{\circ}$ C). Initially the processor’s temperature is $8 5^{\circ}$ C. The rate of cooling is defined by the equation $\frac{d}{d t} (T (t)) = - k (T (t) - 25)$ ,
where T(t) represents the temperature of the processor at time t (in minutes) and k is a constant.
Based on the above information, answer the following questions :
(i) Find the expression for temperature of processor, T(t) given that $T (0) = 8 5^{\circ}$ C. (2)
(ii) How long will it take for the processor’s temperature to reach $4 0^{\circ}$ C ? Given that k = 0.03, $lo g_{e} 4 = 1.3863$ . (2)

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Answer: (i)

T (t) = 25 + 60 e^{- k t}

(ii) about 46.21 minutes

(i) $\frac{d T}{T - 25} = - k d t \Rightarrow lo g ∣ T - 25∣ = - k t + C$ .
$T - 25 = A e^{- k t}$ ; $T (0) = 85 \Rightarrow A = 60$ . So $T (t) = 25 + 60 e^{- k t}$ .
(ii) $40 = 25 + 60 e^{- 0.03 t} \Rightarrow e^{- 0.03 t} = \frac{1}{4} \Rightarrow 0.03 t = lo g_{e} 4$ .
$t = \frac{1.3863}{0.03} = 46.21$ minutes (approx.).

Also asked in: 2025 65/4/2, 2025 65/4/3

Q374 marksCase StudyDifferential Equations CBSE 2025 · 65/7/1

Camphor is a waxy, colourless solid with strong aroma that evaporates through the process of sublimation, if left in the open at room temperature.
(Cylindrical-shaped Camphor tablets)
A cylindrical camphor tablet whose height is equal to its radius (r) evaporates when exposed to air such that the rate of reduction of its volume is proportional to its total surface area. Thus, $\frac{d V}{d t} = k S$ is the differential equation, where V is the volume, S is the surface area and t is the time in hours.
Based upon the above information, answer the following questions :
(i) Write the order and degree of the given differential equation. (1)
(ii) Substituting $V = π r^{3}$ and $S = 2 π r^{2}$ , we get the differential equation $\frac{d r}{d t} = \frac{2}{3} k$ . Solve it, given that r(0) = 5 mm. (1)
(iii) (a) If it is given that r = 3 mm when t = 1 hour, find the value of k. Hence, find t for r = 0 mm. (2)
OR
(iii) (b) If it is given that r = 1 mm when t = 1 hour, find the value of k. Hence, find t for r = 0 mm. (2)

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Answer: (i) Order 1, degree 1 (ii)

r = \frac{2}{3} k t + 5

(iii) (a)

k = - 3

;

t = \frac{5}{2}

hours OR (iii) (b)

k = - 6

;

t = \frac{5}{4}

hours

(i) The highest derivative is $\frac{d V}{d t}$ (first order) and it appears to power 1: order 1, degree 1.
(ii) $d r = \frac{2}{3} k d t$ gives $r = \frac{2}{3} k t + C$ ; $r (0) = 5$ gives $C = 5$ , so $r = \frac{2}{3} k t + 5$ .
(iii) (a) $3 = \frac{2}{3} k + 5$ gives $k = - 3$ ; then $r = 5 - 2 t$ , and $r = 0$ when $t = \frac{5}{2}$ hours.
(iii) (b) $1 = \frac{2}{3} k + 5$ gives $k = - 6$ ; then $r = 5 - 4 t$ , and $r = 0$ when $t = \frac{5}{4}$ hours.

Also asked in: 2025 65/7/2, 2025 65/7/3

Q374 marksCase StudyDifferential Equations CBSE 2024 · 65/3/1

A bacteria sample of certain number of bacteria is observed to grow exponentially in a given amount of time. Using exponential growth model, the rate of growth of this sample of bacteria is calculated.
The differential equation representing the growth of bacteria is given as :
$\frac{d P}{d t} = k P$ , where P is the population of bacteria at any time ‘t’.
Based on the above information, answer the following questions :
(i) Obtain the general solution of the given differential equation and express it as an exponential function of ‘t’. (2)
(ii) If population of bacteria is 1000 at t = 0, and 2000 at t = 1, find the value of k. (2)

Diagram for CBSE 2024 Class 12 Maths question 37

Show answer & solution

Answer: (i)

P = C e^{k t}

(ii)

k = lo g 2

(i) Separate variables: $\frac{d P}{P} = k d t$ .
Integrate: $lo g ∣ P ∣ = k t + c_{1}$ , so $P = C e^{k t}$ (C = $e^{c_{1}}$ ).
(ii) t = 0, P = 1000 gives C = 1000, so $P = 1000 e^{k t}$ .
t = 1, P = 2000: $2000 = 1000 e^{k}$ , $e^{k} = 2$ , $k = lo g 2$ (= $lo g_{e} 2$ ).

Also asked in: 2024 65/3/2, 2024 65/3/3

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