Partial Differential Equations Questions and Answers – Non-Homogeneous Linear PDE with Constant Coefficient

This set of Partial Differential Equations Questions and Answers for Experienced people focuses on “Non-Homogeneous Linear PDE with Constant Coefficient”.

1. Non-homogeneous which may contain terms which only depend on the independent variable.
a) True
b) False
View Answer

Answer: a
Explanation: Linear partial differential equations can further be classified as:

  • Homogeneous for which the dependent variable (and its derivatives) appear in terms with degree
    exactly one, and
  • Non-homogeneous which may contain terms which only depend on the independent variable

2. Which of the following is a non-homogeneous equation?
a) \(\frac{∂^2 u}{∂t^2}-c^2\frac{∂^2 u}{∂x^2}=0\)
b) \(\frac{∂^2 u}{∂x^2}+\frac{∂^2 u}{∂y^2}=0\)
c) \(\frac{∂^2 u}{∂x^2}+(\frac{∂^2 u}{∂x∂y})^2+\frac{∂^2 u}{∂y^2}=x^2+y^2\)
d) \(\frac{∂u}{∂t}-T \frac{∂^2 u}{∂x^2}=0\)
View Answer

Answer: c
Explanation: As we know that homogeneous equations are those in which the dependent variable (and its derivatives) appear in terms with degree exactly one, hence the equation,
\(\frac{∂^2 u}{∂x^2}+(\frac{∂^2 u}{∂x∂y})^2+\frac{∂^2 u}{∂y^2}=x^2+y^2\) is not a homogenous equation (since its degree is 2).

3. What is the general form of the general solution of a non-homogeneous DE (uh(t)= general solution of the homogeneous equation, up(t)= any particular solution of the non-homogeneous equation)?
a) u(t)=uh (t)/up (t)
b) u(t)=uh (t)*up (t)
c) u(t)=uh (t)+up (t)
d) u(t)=uh (t)-up (t)
View Answer

Answer: c
Explanation: The general solution of an inhomogeneous ODE has the general form: u(t)=uh (t)+up (t), where uh (t) is the general solution of the homogeneous equation, up (t) is any particular solution of the non-homogeneous equation.
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4. While an ODE of order m has m linearly independent solutions, a PDE has infinitely many.
a) False
b) True
View Answer

Answer: b
Explanation: A differential equation is an equation involving an unknown function y of one or more independent variables x, t, …… and its derivatives. These are divided into two types, ordinary or partial differential equations.
An ordinary differential equation is a differential equation in which a dependent variable (say ‘y’) is a function of only one independent variable (say ‘x’).
A partial differential equation is one in which a dependent variable depends on one or more independent variables.

5. Which of the following methods is not used in solving non-homogeneous equations?
a) Exponential Response Formula
b) Method of Undetermined Coefficients
c) Orthogonal Method
d) Variation of Constants
View Answer

Answer: c
Explanation: There are several methods in solving a non-homogeneous equation. Some of them are:

  • Exponential Response Formula
  • Method of Undetermined Coefficients
  • Variation of Constants
  • annihilator method
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6. What is the order of the non-homogeneous partial differential equation,
\(\frac{∂^2 u}{∂x^2}+(\frac{∂^2 u}{∂x∂y})^2+\frac{∂^2 u}{∂y^2}=x^2+y^2\)?
a) Order-3
b) Order-2
c) Order-0
d) Order-1
View Answer

Answer: b
Explanation: The order of an equation is defined as the highest order derivative present in the equation and hence from the equation, \(\frac{∂^2 u}{∂x^2}+(\frac{∂^2 u}{∂x∂y})^2+\frac{∂^2 u}{∂y^2}=x^2+y^2\), it is clear that it of 2nd order.

7. What is the degree of the non-homogeneous partial differential equation,
\((\frac{∂^2 u}{∂x∂y})^5+\frac{∂^2 u}{∂y^2}+\frac{∂u}{∂x}=x^2-y^3\)?
a) Degree-2
b) Degree-1
c) Degree-0
d) Degree-5
View Answer

Answer: d
Explanation: Degree of an equation is defined as the power of the highest derivative present in the equation. Hence from the equation,\((\frac{∂^2 u}{∂x∂y})^5+\frac{∂^2 u}{∂y^2}+\frac{∂u}{∂x}=x^2-y^3\), the degree is 5.
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8. The Integrating factor of a differential equation is also called the primitive.
a) True
b) False
View Answer

Answer: b
Explanation: The general solution of a differential equation is also called the primitive. The solution of a partial differential equation obtained by eliminating the arbitrary constants is called a general solution.

9. A particular solution for an equation is derived by substituting particular values to the arbitrary constants in the complete solution.
a) True
b) False
View Answer

Answer: a
Explanation: A solution which does not contain any arbitrary constants is called a general solution whereas a particular solution is derived by substituting particular values to the arbitrary constants in this solution.
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10. What is the complete solution of the equation, \(q= e^\frac{-p}{α}\)?
a) \(z=ae^\frac{-a}{α} y\)
b) \(z=x+e^\frac{-a}{α} y\)
c) \(z=ax+e^\frac{-a}{α} y+c\)
d) \(z=e^\frac{-a}{α} y\)
View Answer

Answer: c
Explanation: Given: \(q= e^\frac{-p}{α}\)
The given equation does not contain x, y and z explicitly.
Setting p = a and q = b in the equation, we get \(b= e^\frac{-p}{α}.\)
Hence, a complete solution of the given equation is,
z=ax+by+c, with \(b= e^\frac{-a}{α}\)
\(z=ax+e^\frac{-a}{α} y+c.\)

11. In recurrence relation, each further term of a sequence or array is defined as a function of its succeeding terms.
a) True
b) False
View Answer

Answer: b
Explanation: An equation that gives a sequence such that each next term of the sequence or array is defined as a function of its preceding terms, is called a recurrence relation.

12. What is the degree of the differential equation, x3-6x3 y3+2xy=0?
a) 3
b) 5
c) 6
d) 8
View Answer

Answer: c
Explanation: The degree of an equation that has not more than one variable in each term is the exponent of the highest power to which that variable is raised in the equation. But when more than one variable appears in a term, it is necessary to add the exponents of the variables within a term to get the degree of the equation. Hence, the degree of the equation, x3-6x3 y3+2xy=0, is 3+3 = 6.

Sanfoundry Global Education & Learning Series – Fourier Analysis and Partial Differential Equations.

To practice all areas of Partial Differential Equations for Experienced people, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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