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 Mathematics** **Exam 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.

**Very Short Answer Type Questions**

** 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**

**Answer.**

**Question**. Write the degree of differential equation

**Answer.** Degree of the given differential equation is 2.

**Question**. Write the solution of the differential equation

**Answer.**

Taking log on both sides to the base 2, we get

log_{2} 2^{y} = log_{2} [(C + x) log_{2}]

⇒ y = log_{2} [(C + x) log_{2}],

which is the required solution.

**Question**. Find the differential equation whose solution**is**

**where A and B are arbitrary constants.****Answer.**

**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

**Answer.**

**Question**. Find the integrating factor of the differential**equation**

**Answer.**

**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

**Short Answer Type Questions -I**

**Question**. Find the differential equation whose**solution is y = ae ^{bx + 5}, where a and b are arbitrary**

**constants**

**Answer.**We have, y = a e

^{bx}+ 5

**…(i)**

Differentiating (i) w.r.t. x, we get

**Question**. Find the general solution of the differential**equation xe ^{y/x} dy = (ye^{y/x} + x^{2})dx, x ≠ 0**

**Answer.**xe

^{y/x}dy = (ye

^{y/x}+ x

^{2})dx

**Question**. Find the differential equation whose solution**is y = e ^{2x} (a + bx), where ‘a’ and ‘b’ are arbitrary**

**constants.**

**Answer.**We have y = e

^{2x}(a + bx)

**..(i)**

On differentiating (i) w.r.t. x, we get

**Question**. Find the particular solution of the**differential equation (1 – y ^{2})(1 + logx)dx + 2xy**

**dy = 0, given that y = 0 when x = 1.**

**Answer.**We have, (1 – y

^{2})(1 + log x) dx + 2xy dy = 0

⇒ (1 – y

^{2})(1 + log x) dx = – 2xy dy

⇒ (1 + log x)^{2} = 2 log |1 – y^{2}| + 1 is the required particular solution

**Question**. Find the particular solution of the

**Answer.**

Integrating both sides, we get

–log (2 –e^{y}) = log (x + 1) + C …(ii)

When x = 0, y = 0

∴ – log (2 – 1) = log (0 + 1) + C ⇒ C = 0

∴ (ii) becomes

– log (2 – e^{y}) = log (x + 1)

⇒ log (x + 1) + log (2 – e^{y}) = 0

⇒ log[(x + 1) (2 – e^{y})] = 0

⇒ (x + 1) (2 – e^{y}) = 1 is the required particular solution

**Question**. Find the particular solution of the

**Answer.**

∴ y siny = x^{2} logx + π/2 is the required particular solution.

**Question**. Solve the differential equation

**Answer.**

**Question**. Find the particular solution of the differential equation x (1 + y^{2}) dx – y (1 + x^{2})dy = 0, given that y = 1 when x = 0.**Answer.** We have, x(1 + y^{2}) dx – y(1 + x^{2}) dy = 0

Integrating both sides, we get log(1 + y^{2}) = log(1+ x^{2}) + log C

⇒ 1 + y^{2} = C (1+ x^{2})

When x = 0, y = 1

∴ 1 + 1 = C(1 + 0) ⇒ C = 2

∴ 1 + y^{2} = 2(1 + x^{2}) is the required particular solution.

**Question**. Solve the differential equation**dy/dx = 1 + x ^{2}+ y^{2} + x^{2}y^{2}, given that y = 1 when x = 0.**

**Answer.**

**Question**. Find the particular solution of the

**Answer.**

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)).

**Short Answer Type Questions -II**

**Question**. Find the differential equation whose**solution is x ^{2} = 4ay, where a is constant.**

**Answer.**We have x

^{2}= 4ay, where a is the constant. …(i)

Differentiating (i) w.r.t. x, we get

**Question**. Find the general solution of the differential equation xe^{x} dy = (x^{3} + 2y e^{x})dx.**Answer.** We have, xe^{x}dy = (x^{3} + 2ye^{x})dx

**Question**. Find the particular solution of the

**Answer.**

**Question**. Find the differential equation whose solution is (x + a)^{2} + (y – a)^{2} = a^{2}, where a is constant.**Answer.** We have (x + a)^{2} + (y – a)^{2} = a^{2} . ..(i)

which has only one arbitrary constant a.

Differentiating (i) w.r.t. x, we get

**Question**. Solve the differential equation

**Answer.**

**Question**. Find the particular solution of the**differential equation e ^{x} tan y dx + (2 – e^{x}) sec^{2}y**

**dy = 0, given that y = π/4 when x = 0.**

**Answer.**The given differential equation is,

e

^{x}tan y dx + (2 – e

^{x}) sec

^{2}y dy = 0

⇒ (2 – e

^{x})sec

^{2}y dy = –e

^{x}tan y dx

∴ Particular solution is log tan y = log (2 – e^{x})

i.e., e^{x} + tan y – 2 = 0

**Question**. Find the particular solution of the**differential equation**

**given that y = 1 when x = 0.****Answer.**

**Question**. Solve the differential equation

**Answer.**

**Question**. Solve the following differential equation**y ^{2}dx + (x^{2 }– xy + y^{2})dy = 0**

**Answer.**We have, y

^{2}dx + (x

^{2}– xy + y

^{2}) dy = 0

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

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

**Question**. Solve the differential equation

**Answer.**

**Question**. Find the particular solution of the following**differential equation**

**given that when x = 2, y = p.****Answer.**

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

**Answer.**

**Question**. Show that the differential equation

**is homogeneous and also solve it.****Answer.**

**Long Answer Type Questions**

**Question**. Find the particular solution of the differential equation (tan^{–1}x–y) dx = (1+x^{2})dy, given that**y = 1 when x = 0.****Answer.** We have, (tan^{–1}x – y)dx = (1 + x^{2})dy

**Question**. Solve the following differential equation

**Answer.**

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

**Question**. Prove that x^{2} – y^{2} = C(x^{2} + y^{2})^{2} is the general solution of the differential equation (x^{3} – 3xy^{2})dx = (y^{3} – 3x^{2}y) dy, where C is a parameter.**Answer.** We have, (x^{3} – 3xy^{2})dx = (y^{3} – 3x^{2}y)dy

⇒ v^{3 }– 3v = (Av + B)(1 + v^{2}) + (Cv + D)(1 – v^{2})

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.****Answer.**

**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.

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.Answer :** Degree is 1

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

## Answer

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

## Answer

A

**Question. If t = 0 when T is 72, then the value of c is**a. -2

b. 0

c. 2

d. Log 2

## Answer

D