This set of Ordinary Differential Equations Questions and Answers for Entrance exams focuses on “Formation of Ordinary Differential Equations by Elimination of Arbitrary Constants”.

1. What is the slope of the equation, y= x^{2}+8?

a) 2x

b) 0

c) 8

d) x

View Answer

Explanation: The slope of the given equation, y= x

^{2}+8, is given by,

Slope= \(\frac{dy}{dx}=2x \)

2. Which of the following is true with respect to formation of differential equation by elimination of arbitrary constants?

a) The given equation should be differentiated with respect to independent variable

b) Elimination of the arbitrary constant by replacing it using derivative

c) If ‘n’ arbitrary constant is present, the given equation should be differentiated ‘n’ number of times

d) To eliminate the arbitrary constants, the given equation must be integrated with respect to the dependent variable

View Answer

Explanation: Consider a general equation, f(x,y,c)=0 ……………………………………… (1)

To form a differential equation by elimination of arbitrary constant, the following steps need to be followed:

- Differentiate (1) with respect to x
- In case of ‘n’ arbitrary constants, the equation should be differentiated ‘n’ number of times
- Eliminate the arbitrary constant using (1) and the derivatives

3. In the formation of differential equation by elimination of arbitrary constants, after differentiating the equation with respect to independent variable, the arbitrary constant gets eliminated.

a) False

b) True

View Answer

Explanation: In the formation of differential equation by elimination of arbitrary constants, the first step is to differentiate the equation with respect to the dependent variable.

**Sometimes**, the arbitrary constant gets eliminated after differentiation.

4. What is the differential equation of a family of parabolas with the foci at the origin and axis along the X-axis?

a) 2xy’+ 4y(y’)^{2}-y=0

b) xy’+ y(y’)^{2}-y=0

c) 2xy’+ y(y’)^{2}-y=0

d) 2xy’+2y(y’)^{2}-y=0

View Answer

Explanation: The equation is, y

^{2}=4ax+4a

^{2}……………………………………. (1)

Differentiating (1) with respect to x, we get,

2yy’=4a ………………………………………………………………………………………….. (2)

Therefore, substituting the value of 4a in (1), we get,

y

^{2}=2yy’x+(yy’)

^{2}

So, the required differential equation is given by,

2xy’+y(y’)

^{2}-y=0

5. What is the nature of the equation, \(xy^3 (\frac{dy}{dx})^2+yx^2+\frac{dy}{dx}=0\)?

a) Second order, third degree, linear differential equation

b) First order, third degree, non-linear differential equation

c) First order, third degree, linear differential equation

d) Second order, third degree, non-linear differential equation

View Answer

Explanation: Since the equation has only first derivative, i.e. \((\frac{dy}{dx}),\) it is a first order equation.

Degree is defined as the highest power of the highest order derivative involved. Hence it is 2.

The equation has one/more terms having a variable of degree two/higher; hence it is non-linear.

6. Which of the following is a type of Iterative method of solving non-linear equations?

a) Graphical method

b) Interpolation method

c) Trial and Error methods

d) Direct Analytical methods

View Answer

Explanation: There are 2 types of Iterative methods, (i) Interpolation methods (or Bracketing methods) and (ii) Extrapolation methods (or Open-end methods).

7. The half-interval method in numerical analysis is also known as __________

a) Newton-Raphson method

b) Regula Falsi method

c) Taylor’s method

d) Bisection method

View Answer

Explanation: The Bisection method, also known as binary chopping or half-interval method, is a starting method which is used, where applicable, for few iterations, to obtain a good initial value.

8. What is the solution of the given equation?

x^{6}y^{6} dy + (x^{7}y^{5} +1) dx = 0

a) \(\frac{(xy)^6}{6} + lnx = c\)

b) \(\frac{(xy)^5}{6} + lny = c\)

c) \(\frac{(xy)^5}{5} + lnx = c\)

d) \(\frac{(xy)^6}{6} + lny = c\)

View Answer

Explanation: Given: (x

^{6}y

^{6}+ 1) dy + x

^{7}y

^{5}dx = 0, is an example of non-exact differential equation.

Dividing the equation by x we get,

x

^{5}y

^{6}dy + x

^{6}y

^{5}dx + \(\frac{dx}{x} = 0\)

x

^{5}y

^{5}(ydy + xdx) + \(\frac{dx}{x} = 0 \)

(xy)

^{5}(d(xy)) + \(\frac{dx}{x} = 0\)

\(\frac{(xy)^6}{6} + lnx = c\)

9. A rectangular frame is to be made of 240 cm long. Determine the value of the length of the rectangle required to maximize the area.

a) 24 cm

b) 60 cm

c) 240 cm

d) 120 cm

View Answer

Explanation: Let us consider ‘x’ as length and ‘y’ as the breadth of the rectangle.

Given: Perimeter 2(x + y) = 240 cm

x + y = 120

y = 120 – x

Area of the rectangle, a = x*y = x(120-x) = 120x – x2

Finding the derivative, we get, (d(a))/dx = (d(120x – x

^{2}))/dx=120-2x

To find the value of x that maximizes the area, we substitute (d(a))/dx = 0.

Therefore, we get, 120 – 2x =0

2x = 120

x = 60 cm

To check if x = 60 cm is the value that maximizes the area, we find the second derivative of the area,

(d

^{2}(a))/(dx

^{2})= -2 < 0 …………………. (i)

We know that the condition for maxima is (d

^{2}(f(x)))/(dx

^{2})<0, which is satisfied by (i), therefore, x = 60 cm maximizes the area of the rectangle.

10. In the equation, y = x^{2}+c,c is known as the parameter and x and y are known as the main variables.

a) True

b) False

View Answer

Explanation: Given: y = x

^{2}+c, where c is known as an arbitrary constant. It is also referred to as the parameter to differentiate it from the main variables x and y.

**Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.**

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