Ordinary Differential Equations Questions and Answers

Answer: A

The operator method has recipes for an exponential, a sine or cosine, a polynomial and their products. For something like \( \sec x \) or \( \ln x \) there is no such recipe, and variation of parameters builds the particular integral from the complementary function by integration instead. Repeated roots and higher order are handled by either method

Answer: D

A product of a function of each variable separates directly. The linear form takes an integrating factor built from the coefficient of \( y \). Bernoulli's form is neither, but the substitution \( v = y^{1-n} \) turns it into the linear case, after which the integrating factor applies

Answer: C

Distinct real roots give two independent exponentials. A repeated root gives only one, so the second solution carries a factor of \( x \). A conjugate pair combines into a real exponential envelope times an oscillation, the real part fixing the envelope and the imaginary part the frequency

Answer: B

The auxiliary equation comes first because its roots decide both the complementary function and whether the trial particular integral needs a factor of \( x \). The two parts are then added, and only the complete general solution can absorb the initial conditions, since the constants belong to the complementary function

Answer: A

The first two hold. The third overstates the case: a homogeneous equation becomes separable only after the substitution \( y = vx \), and as written it usually mixes the variables. That substitution is precisely what the classification exists to point to

Answer: D

The resonant case does need the extra factor, and variation of parameters does build the particular integral from the complementary solutions, so it cannot be attempted first. A logarithm is outside the operator method's repertoire, which covers exponentials, sinusoids, polynomials and their products

Answer: C

Linearity requires the unknown and its derivatives to appear only to the first power and never multiplied together; the coefficients may depend on \( x \) as freely as they like. The product \( yy' \), the squared derivative and \( \sin y \) each break that, and it is exactly this that puts the pendulum equation beyond elementary treatment

Answer: B

The number of arbitrary constants matches the *order*, not the degree: a second-order equation carries two whatever powers appear. The other three are the standard structure of the solution, which is why the homogeneous and forced parts can be found independently and then added

Answer: A

Each integration introduces one constant, so an order-two equation carries two, and the initial or boundary conditions fix them. This holds whether the coefficients are constant or variable. The degree affects how hard the equation is to solve, not how many constants the solution needs

Answer: D

The complementary function is by construction a solution of the homogeneous equation, so substituting it returns zero rather than the forcing term. The particular integral is what supplies that term, and the sum is what satisfies the full equation; neither part is optional