# Differential Equations Class 12 Mathematics Exam Questions

Please refer to Differential Equations Class 12 Mathematics Exam Questions provided below. These questions and answers for Class 12 Mathematics have been designed based on the past trend of questions and important topics in your Class 12 Mathematics books. You should go through all Class 12 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

## Class 12 MathematicsExam Questions Differential Equations

Class 12 Mathematics students should read and understand the important questions and answers provided below for Differential Equations which will help them to understand all important and difficult topics.

Question. Find the order and the degree of the differential equation

Answer. The given differential equation is

Question. Find the integrating factor of the differential equation

Question. Write the degree of differential equation

Answer. Degree of the given differential equation is 2.

Question. Write the solution of the differential equation

Taking log on both sides to the base 2, we get
log2 2y = log2 [(C + x) log2]
⇒ y = log2 [(C + x) log2],
which is the required solution.

Question. Find the differential equation whose solution is

where A and B are arbitrary constants.

Question. Find the differential equation whose solution is y = mx, where m is an arbitrary constant.
Answer. Here, y = mx                                                 …(i)
Differentiating (i) w.r.t. x, we get

Question. Find the solution of the differential equation

Question. Find the integrating factor of the differential equation

Question. Write the sum of the order and degree of the differential equation

Answer. Order = 2, Degree = 2.
∴ Required Sum = 2 + 2 = 4

Question. Write the integrating factor of the differential equation

Answer. The given differential equation is

Question. Find the differential equation whose solution is y = aebx + 5, where a and b are arbitrary constants
Answer. We have, y = a ebx + 5                                  …(i)
Differentiating (i) w.r.t. x, we get

Question. Find the general solution of the differential equation xey/x dy = (yey/x + x2)dx, x ≠ 0
Answer. xey/xdy = (yey/x + x2 )dx

Question. Find the differential equation whose solution is y = e2x (a + bx), where ‘a’ and ‘b’ are arbitrary constants.
Answer. We have y = e2x(a + bx)                                  ..(i)
On differentiating (i) w.r.t. x, we get

Question. Find the particular solution of the differential equation (1 – y2)(1 + logx)dx + 2xy dy = 0, given that y = 0 when x = 1.
Answer. We have, (1 – y2)(1 + log x) dx + 2xy dy = 0
⇒ (1 – y2)(1 + log x) dx = – 2xy dy

⇒ (1 + log x)2 = 2 log |1 – y2| + 1 is the required particular solution

Question. Find the particular solution of the

Integrating both sides, we get
–log (2 –ey) = log (x + 1) + C                                …(ii)
When x = 0, y = 0
∴ – log (2 – 1) = log (0 + 1) + C ⇒ C = 0
∴ (ii) becomes
– log (2 – ey) = log (x + 1)
⇒ log (x + 1) + log (2 – ey) = 0
⇒ log[(x + 1) (2 – ey)] = 0
⇒ (x + 1) (2 – ey) = 1 is the required particular solution

Question. Find the particular solution of the

∴ y siny = x2 logx + π/2 is the required particular solution.

Question. Solve the differential equation

Question. Find the particular solution of the differential equation x (1 + y2) dx – y (1 + x2)dy = 0, given that y = 1 when x = 0.
Answer. We have, x(1 + y2) dx – y(1 + x2) dy = 0

Integrating both sides, we get log(1 + y2) = log(1+ x2) + log C
⇒ 1 + y2 = C (1+ x2)
When x = 0, y = 1
∴ 1 + 1 = C(1 + 0) ⇒ C = 2
∴ 1 + y2 = 2(1 + x2) is the required particular solution.

Question. Solve the differential equation dy/dx = 1 + x2+ y2 + x2y2, given that y = 1 when x = 0.

Question. Find the particular solution of the

Integrating both sides, we get
y – 2log(y + 2) = x + 2log x + C
when x = 1, y = – 1
So, – 1 – 2log (– 1 + 2) = 1 + 2 log 1 + C
⇒ C = – 1 – 1 = – 2
So, we have y – 2log(y + 2) = x + 2log x – 2
⇒ y – x + 2 = 2log (x(y + 2)).

Question. Find the differential equation whose solution is x2 = 4ay, where a is constant.
Answer. We have x2 = 4ay, where a is the constant.                     …(i)
Differentiating (i) w.r.t. x, we get

Question. Find the general solution of the differential equation xex dy = (x3 + 2y ex)dx.
Answer. We have, xexdy = (x3 + 2yex)dx

Question. Find the particular solution of the

Question. Find the differential equation whose solution is (x + a)2 + (y – a)2 = a2, where a is constant.
Answer. We have (x + a)2 + (y – a)2 = a2 .                                      ..(i)
which has only one arbitrary constant a.
Differentiating (i) w.r.t. x, we get

Question. Solve the differential equation

Question. Find the particular solution of the differential equation ex tan y dx + (2 – ex) sec2y dy = 0, given that y = π/4 when x = 0.
Answer. The given differential equation is,
ex tan y dx + (2 – ex) sec2y dy = 0
⇒ (2 – ex)sec2y dy = –ex tan y dx

∴ Particular solution is log tan y = log (2 – ex)
i.e., ex + tan y – 2 = 0

Question. Find the particular solution of the differential equation

given that y = 1 when x = 0.

Question. Solve the differential equation

Question. Solve the following differential equation y2dx + (x2 – xy + y2)dy = 0
Answer. We have, y2 dx + (x2 – xy + y2) dy = 0

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

Question. Solve the differential equation dy/dx + ycot x = 2cosx, given that y = 0 when x = π/2.

Question. Solve the differential equation

Question. Find the particular solution of the following differential equation

given that when x = 2, y = p.

Question. If y(x) is a solution of the differential

Question. Show that the differential equation

is homogeneous and also solve it.

Question. Find the particular solution of the differential equation (tan–1x–y) dx = (1+x2)dy, given that y = 1 when x = 0.
Answer. We have, (tan–1x – y)dx = (1 + x2)dy

Question. Solve the following differential equation

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

Question. Prove that x2 – y2 = C(x2 + y2)2 is the general solution of the differential equation (x3 – 3xy2)dx = (y3 – 3x2y) dy, where C is a parameter.
Answer. We have, (x3 – 3xy2)dx = (y3 – 3x2y)dy

⇒ v3 – 3v = (Av + B)(1 + v2) + (Cv + D)(1 – v2)
Comparing coeff. of like powers, we get
A – C = 1, A + C = –3, B – D = 0 and B + D = 0
Solving these equations, we get A = –1, B = 0, C = –2, D = 0       ……….(iv)
From (ii), (iii) and (iv), we have

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

CASE STUDY:

A Veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 pm which was 94.6°F. He took the temperature again after one hour; the temperature was lower than the first observation. It was 93.4°F. The room in which the cat was put is always at 70°F. The normal temperature of the cat is taken as 98.6°F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed by the differential equation:
𝑑𝑇/𝑑𝑡 ∝ (𝑇 − 70), where 70°F is the room temperature and T is the temperature of the object at time t.
Substituting the two different observations of T and t made, in the solution of the differential equation 𝑑𝑇/𝑑𝑡 = 𝑘(𝑇 − 70) where k is a constant of proportion, time of death is calculated.

Question. State the degree of the above given differential equation.
Degree is 1

Question. If the temperature was measured 2 hours after 11.30pm, will the time of death change? (Yes/No)
No

Question. Which method of solving a differential equation helped in calculation of the time of death?
a. Variable separable method
b. Solving Homogeneous differential equation
c. Solving Linear differential equation
d. all of the above

A

Question. The solution of the differential equation 𝑑𝑇/𝑑𝑡=𝑘(𝑇−70)is given by,
a. log | T – 70| = kt + C
b. log | T – 70| = log |kt |+ C
c. T – 70 = kt + C
d. T – 70 = kt C