Ordinary Differential Equations Questions and Answers – Formation of Ordinary Differential Equations by Elimination of Arbitrary Constants

«
»

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= x2+8?
a) 2x
b) 0
c) 8
d) x
View Answer

Answer: a
Explanation: The slope of the given equation, y= x2+8, is given by,
Slope= \(\frac{dy}{dx}=2x \)
advertisement

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

Answer: d
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

Answer: a
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

Answer: c
Explanation: The equation is, y2=4ax+4a2……………………………………. (1)
Differentiating (1) with respect to x, we get,
2yy’=4a ………………………………………………………………………………………….. (2)
Therefore, substituting the value of 4a in (1), we get,
y2=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

Answer: b
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.
advertisement

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

Answer: b
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

Answer: d
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?
x6y6 dy + (x7y5 +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

Answer: a
Explanation: Given: (x6y6 + 1) dy + x7y5dx = 0, is an example of non-exact differential equation.
Dividing the equation by x we get,
x5y6 dy + x6y5dx + \(\frac{dx}{x} = 0\)
x5y5 (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

Answer: b
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 – x2))/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,
(d2 (a))/(dx2)= -2 < 0 …………………. (i)
We know that the condition for maxima is (d2 (f(x)))/(dx2)<0, which is satisfied by (i), therefore, x = 60 cm maximizes the area of the rectangle.
advertisement

10. In the equation, y = x2+c,c is known as the parameter and x and y are known as the main variables.
a) True
b) False
View Answer

Answer: a
Explanation: Given: y = x2+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.

To practice all areas of Ordinary Differential Equations for Entrance exams, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
advertisement

Leave a Comment

Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn