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Class 12 Maths Chapter 9 Differential Equations MCQs

Class 12 Maths Chapter 9 Differential Equations MCQs are provided here with answers. Students can practice the multiple-choice questions based on the latest exam pattern for Class 12 Maths and score good marks in the board exam. All the MCQ questions for differential equations are prepared according to the CBSE Class 12 syllabus (2022-2023) and the NCERT curriculum. Students can revise chapter 9 Differential equations by solving the MCQs available here for free.

Check: Chapter-wise Class 12 Maths MCQs

Differential Equations Class 12 MCQs with Solutions

Find the differential equations mcqs for class 12 with complete solutions here.

Download PDF – Chapter 9 Differential Equations MCQs

Q.1: What is the order of differential equation y’’ + 5y’ + 6 = 0?

A. 0

B. 1

C. 2

D. 3

Answer: C. 2

Explanation: Given, differential equation y’’ + 5y’ + 6 = 0.

The highest order derivative present in the differential equation is y’’. Hence, the order is 2.

Q.2: What is the degree of differential equation (y’’’)2 + (y’’)3 + (y’)4 + y5 = 0?

A. 2

B. 3

C. 4

D. 5

Answer: A. 2

Explanation: The degree is the power raised to the highest order derivative. Therefore, in the given differential equation, (y’’’)2 + (y’’)3 + (y’)4 + y5 = 0, the degree will be power raised to y’’’.

So, the answer is 2.

\(\begin{array}{l}\text{Q.3: Find the order of differential equations: }2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0\end{array} \)

A. 2

B. 1

C. 0

D. Undefined

Answer: A. 2

Explanation: Given, the differential equation is:

\(\begin{array}{l}2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0\end{array} \)

Or we can write:

2x2 y’’ – 3y’ + y = 0

Order is the highest derivative in the differential equation. Therefore, the order is 2.

\(\begin{array}{l}\text{Q.4: Find the degree of the differential equation: }\left(1+\frac{d y}{d x}\right)^{3}=\left(\frac{d y}{d x}\right)^{2}\end{array} \)

A. 0

B. 1

C. 2

D. 3

Answer: C. 3

Explanation: Given, the differential equation is:

\(\begin{array}{l}\left(1+\frac{d y}{d x}\right)^{3}=\left(\frac{d y}{d x}\right)^{2}\end{array} \)

We can expand it and get:

\(\begin{array}{l}1+3 \frac{d y}{d x}+3\left(\frac{d y}{d x}\right)^{2}+\left(\frac{d y}{d x}\right)^{3}=\left(\frac{d y}{d x}\right)^{2}\end{array} \)

The exponent of highest derivative is the degree. Therefore, the degree is 3.

Q.5: The number of arbitrary constants in the particular solution of a differential equation of third order is:

A. 3

B. 2

C. 1

D. 0

Answer: D. 0

Explanation: The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation.

Q.6: What is the differential equation of the family of circles touching the y-axis at the origin?

A. 2xyy’ + x2 = y2

B. 2xyy’’ + x’ = y2

C. 2xyy’ – x2 = y2

D. xyy’ + x2 = y2

Answer: A. 2xyy’ + x2 = y2

Explanation: Let the center of the circle touch the y- axis at origin lies on the x-axis.

Say, (k, 0) be the center of the circle.

Hence, it touches the y – axis at origin, its radius is p.

Now, the equation of the circle with center (p, 0) and radius (p) is

⇒ (x – k)2 + y2 = k2

⇒ x2 + k2 – 2xk + y2 = p2

Shifting k and – 2xk to RHS then it becomes: k2 and 2xk

⇒ x2 + y2 = k2 – k2 + 2kx

⇒ x2 + y2 = 2kx ….(i)

Differentiating equation on both sides, we have,

⇒ 2x + 2yy’ = 2k

⇒ x + yy’ = k

Now, on substituting the value of ‘k’ in the equation (i), we get,

⇒ x2 + y2 = 2(x + yy’)x

⇒ 2xyy’ + x2 = y2

Therefore, 2xyy’ + x2 = y2 is the required differential equation.

Q.7: Solution of differential equation x.dy – y.dx = Q represents:

A. a rectangular hyperbola

B. parabola whose vertex is at the origin

C. straight line passing through the origin

D. a circle whose centre is at the origin

Answer: C. straight line passing through the origin

\(\begin{array}{l}\text{Q.8: Find the general solution of: }\frac{d y}{d x}=\sqrt{4-y^{2}}(-2<y<2)\end{array} \)

A. sin-1 y = x + c

B. sin-1 y/2 = x + c

C. sin-1 y2 = x + c

D. None of the above

Answer: B. sin-1 y/2 = x + c

Explanation: Given,

\(\begin{array}{l}\frac{d y}{d x}=\sqrt{4-y^{2}}\end{array} \)

Rearranging we get;

dy/√(4 – y2) = dx

Integrating both the sides, we get;

∫dy/√(4 – y2) = ∫dx

We know that, by the formula;

∫1/√(a2 – x2) = sin-1(x/a)

Therefore,

sin-1y/2 = x+c

Q.9: Which of the following is a second-order differential equation?

A. (y’)² + x = y²

B. y’y” + y = sin x

C. y”’ + (y”)² + y = 0

D. y’ = y²

Answer: B. y’y” + y = sin x

Explanation: The order of y’y” + y = sin x is 2. Thus, it is a second-order differential equation.

Q.10: The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is:

A. an ellipse

B. parabola

C. circle

D. hyperbola

Answer: D. hyperbola

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