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Edition: Complete Solutions Manual latest = = = = =
2026/2027 =
Complete Q&A with Rationales for Differential = = = = = =
Equations Success = =
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SECTION A: INTRODUCTION TO DIFFERENTIAL EQUATIONS (Ch
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apter 1) = =
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Question 1 = =
The order of a differential equation is determined by:
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A) The highest power of the dependent variable
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B) The highest derivative present in the equation
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C) The number of independent variables
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D) The number of terms in the equation
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Answer: B) The highest derivative present in the equation
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Rationale: The order of a differential equation is defined as the ord
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er of the highest derivative appearing in the equation . This is a funda
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mental concept that distinguishes first-
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,order, secondorder, and higher-
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order differential equations. The degree of a differential equation r
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efers to the power of the highest derivative.
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Question 2 = =
A differential equation is considered linear if:
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A) All derivatives are of first order
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B) The dependent variable and its derivatives appear to the first po
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wer only = =
C) The equation has constant coefficients
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D) There are no products of the dependent variable with its derivati
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ves =
Answer: B) The dependent variable and its derivatives appear to th
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e first power only
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Rationale: A linear differential equation is one in which the depende
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nt variable and all its derivatives occur only to the first power and ar
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e not multiplied together . This is a key distinction between linear and
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nonlinear equations. The standard form of a linear ODE is an(x)y(n)+
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...+a1(x)y′+a0(x)y=g(x)an(x)y(n)+...+a1 (x)y′+a0(x)y=g(x). = =
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, Question 3 = =
Which of the following is a solution to the differential equation y′
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=2xy′=2x? =
A) y=x2+1y=x2+1 =
B) y=x2y=x2 =
C) y=2xy=2x =
D) y=x2+Cy=x2+C =
Answer: D) y=x2+Cy=x2+C
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Rationale: The general solution to y′=2xy′=2x is found by integratin
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g both sides: y=x2+Cy=x2+C. This represents a family of solutions.
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A particular solution is obtained by specifying the constant CC using
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an initial condition.
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Question 4 = =
The differential equation y′=3x2y′=3x2 with initial condition y
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(1)=4y(1)=4 has the particular solution: = = = = =
A) y=x3+4y=x3+4 =
B) y=x3+3y=x3+3 =
C) y=x3+1y=x3+1 =
D) y=x3−3y=x3−3 =