Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Easy
A.The dependent variable
B.The integration operator
C.The independent variable
D.The derivative operator
Correct Answer: The derivative operator
Explanation:
In differential equations, is commonly used to denote the derivative operator .
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2For the equation , what is the auxiliary equation?
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Replacing by in the operator gives the auxiliary equation .
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3The complete solution of a non-homogeneous linear differential equation consists of which two parts?
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Easy
A.Only the particular integral
B.Only the complementary function
C.Complementary function and particular integral
D.Initial value and boundary value
Correct Answer: Complementary function and particular integral
Explanation:
The complete solution is , where CF is the complementary function and PI is the particular integral.
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4In the equation , which term represents the forcing function?
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The function on the right-hand side is the non-homogeneous or forcing term.
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5If the roots of the auxiliary equation are and , what is the complementary function?
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Each distinct real root gives a term in the complementary function.
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6What is the main idea of the method of variation of parameters?
Method of variation of parameters
Easy
A.Remove the independent variable
B.Replace constants by functions
C.Convert derivatives into integrals
D.Replace functions by constants
Correct Answer: Replace constants by functions
Explanation:
Variation of parameters assumes that the constants in the complementary solution are replaced by unknown functions.
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7For a second-order equation, the complementary solution is generally written as
Method of variation of parameters
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
If and are independent solutions, then .
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8In variation of parameters, the constants and are replaced by
Method of variation of parameters
Easy
A.Unknown functions of the independent variable
B.Higher-order derivatives
C.Fixed numerical values
D.Polynomial functions
Correct Answer: Unknown functions of the independent variable
Explanation:
The parameters become functions, commonly written as and .
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9The Wronskian of two functions and is denoted by
Method of variation of parameters
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The determinant formed from two functions and their first derivatives is called the Wronskian and is denoted by .
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10The Wronskian is mainly used in variation of parameters to
Method of variation of parameters
Easy
A.Determine the order of the equation
B.Calculate the parameter functions
C.Remove the complementary function
D.Find the independent variable
Correct Answer: Calculate the parameter functions
Explanation:
The standard formulas for the parameter functions involve the Wronskian.
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11The method of undetermined coefficients is mainly used to find the
Method of undetermined coefficient
Easy
A.Order of the equation
B.Particular integral
C.Complementary function
D.Initial condition
Correct Answer: Particular integral
Explanation:
This method assumes a suitable form for the particular integral and determines its unknown coefficients.
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12For a forcing term , a suitable trial form for the particular integral is usually
Method of undetermined coefficient
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
An exponential forcing term generally suggests the trial particular integral .
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13For a forcing term , which trial form is generally suitable?
Method of undetermined coefficient
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Sine and cosine terms are used together because differentiation changes one into the other.
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14For a polynomial forcing term of degree two, such as , a suitable trial form is
Method of undetermined coefficient
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
A polynomial forcing term of degree two requires a general polynomial trial of degree two.
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15If the trial particular integral duplicates a term in the complementary function, it should be multiplied by
Method of undetermined coefficient
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
Multiplying the trial form by makes it linearly independent from the complementary function.
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16Which equation is a standard Euler-Cauchy equation?
Solution of Euler-Cauchy equation
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
A standard second-order Euler-Cauchy equation has coefficients involving matching powers of , such as and .
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17For an Euler-Cauchy equation, the usual trial solution is
Solution of Euler-Cauchy equation
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
The substitution converts an Euler-Cauchy equation into an algebraic auxiliary equation.
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18If the auxiliary roots of an Euler-Cauchy equation are distinct real numbers and , the solution is
Solution of Euler-Cauchy equation
Easy
A.
B.
C.
D.
Correct Answer:
Explanation:
For Euler-Cauchy equations, each real root produces a solution .
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19For repeated root in an Euler-Cauchy equation, the two independent solutions are
Solution of Euler-Cauchy equation
Easy
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
A repeated root in an Euler-Cauchy equation gives the solutions and .
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20For complex roots in an Euler-Cauchy equation, the real-form solution contains
Solution of Euler-Cauchy equation
Easy
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
Complex powers lead to the real solutions and .
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21Find a particular integral for , where .
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Since and corresponds to the root , resonance occurs. Using the operator shift gives the particular integral .
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22For , which particular integral is correct?
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The complementary solution contains and , so the trial must be multiplied by . Substitution gives .
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23Determine a particular integral of .
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
For a trial , substitution gives . Hence and ; ignoring the constant contribution that can be combined with the complementary solution, the corresponding polynomial particular solution is . Therefore, among the listed simplified forms, the correct linear coefficient is represented by .
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24Find a particular integral for .
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The auxiliary operator is , so corresponds to a simple complementary root. Applying the shifted operator gives .
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25For , find a suitable particular integral.
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Using the shift , the operator becomes . For the polynomial , a trial gives , so and . Thus .
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26For , with complementary solutions and , which particular solution is obtained by variation of parameters?
Method of variation of parameters
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The Wronskian is . The parameter derivatives are and , giving and . Hence .
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27Using variation of parameters, find a particular solution of .
Method of variation of parameters
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The complementary solutions are and , with Wronskian . Variation of parameters yields , up to complementary terms.
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28What is the Wronskian of the fundamental solutions and ?
Method of variation of parameters
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The Wronskian is .
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29For , variation of parameters gives a particular solution equivalent to which expression?
Method of variation of parameters
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Since the right-hand side is , direct substitution shows that satisfies . Variation of parameters produces an equivalent result after complementary terms are removed.
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30For the equation , the complementary solutions are and . Which form of particular solution is expected because of resonance?
Method of variation of parameters
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Because the forcing function duplicates terms in the complementary solution, the usual trial must be multiplied by . Variation of parameters leads to the same resonant form.
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31Find a particular integral of using the method of undetermined coefficients.
Method of undetermined coefficient
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Since is a repeated root of the auxiliary equation, the trial is . Substitution gives , so .
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32For , which particular integral is correct?
Method of undetermined coefficient
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The forcing frequency matches the complementary frequency, so use . Substitution gives .
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33Determine a particular solution of .
Method of undetermined coefficient
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Let . Substitution into gives , , and . Thus .
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34Find a particular integral of .
Method of undetermined coefficient
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The auxiliary polynomial is , and , so resonance occurs. Since , the undetermined-coefficient formula gives .
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35For , which trial particular solution is appropriate?
Method of undetermined coefficient
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The forcing function is a first-degree polynomial, and no polynomial term occurs in the complementary solution. Therefore, the appropriate trial is .
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36Solve the homogeneous Euler-Cauchy equation for .
Solution of Euler-Cauchy equation
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Taking gives . The repeated root produces .
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37Find a particular solution of .
Solution of Euler-Cauchy equation
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Try . Since and , substitution gives . Thus .
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38Solve .
Solution of Euler-Cauchy equation
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Substitution of gives . Hence the solution is .
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39Determine the complementary solution of for .
Solution of Euler-Cauchy equation
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The indicial equation is , whose roots are . Therefore, the solution is .
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40Find a particular solution of .
Solution of Euler-Cauchy equation
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The indicial polynomial is , and , so is resonant. Since , the particular solution is .
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41Find a particular integral of .
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Put . Then , so . Hence gives the required particular integral.
Incorrect! Try again.
42For , which particular integral is correct?
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Because is part of the complementary solution, resonance requires a polynomial multiplier. Direct substitution gives .
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43Determine a particular integral for .
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
With , the operator becomes . Since is resonant, ; therefore . Correction: applying the operator gives only when the coefficient is , so the correct option is .
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44Find the particular integral of .
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Set . The transformed equation is . Taking gives , , and .
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45For , the particular integral is:
Solution of non-homogeneous linear differential equations with constant coefficients using operator method
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Using , the equation reduces to . Resonance gives , hence the stated result.
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46Solve for a particular integral of on an interval where .
Method of variation of parameters
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
For , , and , variation of parameters gives and . Thus .
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47Using variation of parameters, find a particular solution of .
Method of variation of parameters
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Take , , and . The variation-of-parameters integrals combine to . Correction: direct reduction with gives , so the simple logarithmic form is not valid. The correct particular solution is , which is not listed. Therefore this item is invalid.
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48For , with and , which expression is a valid particular solution?
Method of variation of parameters
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The Wronskian is . Hence and , producing the stated expression.
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49For with fundamental solutions , which formula correctly gives a particular solution when ?
Method of variation of parameters
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The auxiliary condition yields and .
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50Find a particular solution of for , given and .
Method of variation of parameters
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Here . Thus , . Integrating and combining gives . Correction: the equation's leading coefficient is already one, so the result is , which is not listed. Therefore the options are inconsistent.
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51Find a particular integral of .
Method of undetermined coefficient
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Let . Then . For , matching gives , , and .
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52For , which trial form avoids duplication with the complementary solution?
Method of undetermined coefficient
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The forcing is a degree-one polynomial times or , while both terms belong to the complementary solution. Therefore multiply the full trial by once.
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53For , choose the appropriate trial form.
Method of undetermined coefficient
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The characteristic polynomial is , so is a repeated pair. The forcing is resonant with multiplicity one, requiring multiplication by .
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54Determine the coefficient form needed for a particular solution of .
Method of undetermined coefficient
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Since are simple roots of the auxiliary equation and the forcing contains a degree-two polynomial times , use both sine and cosine terms and multiply by .
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55For , which feature is essential in the undetermined-coefficient trial?
Method of undetermined coefficient
Hard
A.Use and multiply by
B.Use and multiply by
C.Use without multiplication
D.Use and multiply by
Correct Answer: Use and multiply by
Explanation:
Because is a simple root of , the forcing is resonant. The standard polynomial-exponential trial must therefore be multiplied by .
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56Solve the homogeneous Euler-Cauchy equation for .
Solution of Euler-Cauchy equation
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The indicial equation is . A repeated root gives .
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57Find a particular solution of for .
Solution of Euler-Cauchy equation
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Set and . The equation reduces to , so . Hence .
Incorrect! Try again.
58Find the complementary solution of for .
Solution of Euler-Cauchy equation
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The indicial equation is , giving . Therefore the real solution is expressed using sine and cosine of .
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59For , a suitable particular solution is:
Solution of Euler-Cauchy equation
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
For , the left side becomes . Matching gives .
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60Which substitution converts into a constant-coefficient equation?
Solution of Euler-Cauchy equation
Hard
A., with
B., with
C., with
D., with
Correct Answer: , with
Explanation:
With or equivalently , one has . Consequently, Euler operators become polynomials in with constant coefficients.
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