For two functions y1,y2y1,y2, the Wronskian W(y1,y2)W(y1,y2) is
A y1+y2y1+y2
B A 2×2 determinant
C y1y2y1y2
D y1/y2y1/y2
The Wronskian of two functions is the determinant formed using y1,y2y1,y2 and their first derivatives. It helps test whether two solutions of a second-order linear ODE are independent.
If W(y1,y2)≠0W(y1,y2)=0 on an interval for a linear ODE, then y1,y2y1,y2 are
A Always equal
B Not differentiable
C Linearly independent
D Linearly dependent
For solutions of a linear homogeneous second-order ODE, a nonzero Wronskian on an interval implies linear independence. Then the two solutions can form a fundamental set for the general solution.
If two solutions are linearly dependent, their Wronskian is
A Always one
B Always positive
C Always increasing
D Always zero
If y2=Cy1y2=Cy1, the rows of the Wronskian determinant become proportional. This makes the determinant zero on the interval where both solutions are defined and differentiable.
A fundamental set of solutions (2nd order) contains
A Two dependent solutions
B Two independent solutions
C One solution only
D Three solutions
A second-order linear homogeneous ODE has a two-dimensional solution space. Two linearly independent solutions are enough to generate every solution by taking linear combinations.
The superposition principle applies directly to
A Linear homogeneous ODE
B Nonlinear ODE only
C Exact equations only
D Clairaut equation only
In linear homogeneous equations, any linear combination of solutions is also a solution. This property allows building the general solution using a basis of independent solutions.
In first-order notation, pp usually denotes
A ∫y dx∫ydx
B y/xy/x
C dy/dxdy/dx
D d2y/dx2d2y/dx2
Many first-order equation forms use pp as shorthand for the derivative dy/dxdy/dx. This is common in equations solvable for xx, yy, or pp.
An equation “solvable for pp” can be written as
A x=f(y)x=f(y)
B p=f(x,y)p=f(x,y)
C y=f(x)y=f(x)
D p=f(x)p=f(x) only
“Solvable for pp” means the derivative p=dy/dxp=dy/dx can be isolated in terms of xx and yy. Then the slope field and solution methods become easier to apply.
A Clairaut equation is commonly written as
A y′+Py=Qy′+Py=Q
B y′′+ay=0y′′+ay=0
C y′=f(x)y′=f(x)
D y=px+f(p)y=px+f(p)
Clairaut’s form is y=px+f(p)y=px+f(p) with p=dy/dxp=dy/dx. Its general solution is a family of straight lines, and it may also have a singular solution as an envelope.
General solution of Clairaut equation has form
A y=Cexy=Cex
B y=Csinxy=Csinx
C y=Cx+f(C)y=Cx+f(C)
D y=C/xy=C/x
In Clairaut’s equation, taking p=Cp=C (constant slope) yields the family y=Cx+f(C)y=Cx+f(C). Each value of CC gives one line; the envelope gives the singular curve.
The singular solution in Clairaut’s equation represents the
A Maximum slope line
B Envelope of lines
C Intersection of axes
D Periodic curve
Clairaut’s general solution gives a one-parameter family of lines. The singular solution is the curve tangent to all these lines, so it is the envelope of the family.
For constant-coefficient linear ODEs, we use the
A Auxiliary equation
B Exactness test
C Bernoulli trick
D Integrating factor
Assume y=emxy=emx and substitute into the ODE. This produces a polynomial equation in mm, called the characteristic or auxiliary equation, which determines the complementary function.
Distinct real roots m1,m2m1,m2 give CF terms
A xm1,xm2xm1,xm2
B em1x,em2xem1x,em2x
C sinm1x,cosm2xsinm1x,cosm2x
D lnxlnx terms
For each distinct real root mm, emxemx is a solution. With two distinct roots, the complementary function is C1em1x+C2em2xC1em1x+C2em2x.
Repeated root mm (double) gives CF terms
A emx,e−mxemx,e−mx
B cosmx,sinmxcosmx,sinmx
C emx,xemxemx,xemx
D 1,x1,x only
A double root needs a second independent solution. Multiplying by xx gives xemxxemx, which is independent of emxemx, ensuring two linearly independent solutions.
Complex roots α±iβα±iβ produce solutions
A eβxcosαxeβxcosαx
B cosαxcosαx only
C sinβxsinβx only
D eαxcosβxeαxcosβx
Complex conjugate roots give oscillatory solutions. The real complementary function uses eαxcosβxeαxcosβx and eαxsinβxeαxsinβx as independent real solutions.
In a linear nonhomogeneous ODE, the complete solution is
A CF × PI
B CF + PI
C CF − PI always
D PI only
The complementary function solves the homogeneous equation, while one particular integral accounts for the forcing term. Adding them gives the general solution of the full nonhomogeneous equation.
“Resonance” in undetermined coefficients happens when RHS
A Becomes zero
B Is discontinuous
C Matches CF term
D Is implicit only
If the trial particular solution duplicates a term in the complementary function, it fails. Multiplying by xx (or higher powers) shifts the form to achieve independence and makes solving possible.
For RHS eaxeax, a basic PI trial is usually
A AeaxAeax
B AsinaxAsinax
C AxaAxa
D AlnxAlnx
Derivatives of eaxeax stay proportional to eaxeax, so a trial AeaxAeax fits. If resonance occurs, multiply the trial by xkxk as needed.
For RHS sinaxsinax, a basic PI trial uses
A AeaxAeax only
B Asinax+BcosaxAsinax+Bcosax
C Ax2Ax2 only
D AlnxAlnx only
Because differentiating sine produces cosine and vice-versa, both terms must appear in the trial. Substituting into the ODE allows solving for coefficients AA and BB.
For polynomial RHS, PI trial is generally
A Only exponential
B Only trig
C Only logarithmic
D Polynomial form
A polynomial forcing suggests a polynomial trial of similar degree. If any term overlaps with the complementary function, multiply the trial by suitable powers of xx to avoid duplication.
The operator DD in ODE methods means
A dx/dydx/dy
B ∫dx∫dx
C d/dxd/dx
D Constant CC
In operator notation, DD is the differentiation operator. For constant-coefficient linear ODEs, writing polynomials in DD can simplify forming CF and handling standard RHS forms.
A Cauchy–Euler equation typically contains terms like
A y′′+ay′y′′+ay′ only
B x2y′′x2y′′ and xy′xy′
C exy′′exy′′ terms
D sinx y′sinxy′
Cauchy–Euler equations have coefficients as powers of xx. The pattern x2y′′+axy′+by=⋯x2y′′+axy′+by=⋯ is typical and can be solved using y=xmy=xm or x=etx=et.
For Cauchy–Euler, a common trial solution is
A y=emxy=emx
B y=sinmxy=sinmx
C y=ln(mx)y=ln(mx)
D y=xmy=xm
Substituting y=xmy=xm makes derivatives produce powers of xx that match the equation’s structure. After simplification, it reduces to an algebraic equation in mm.
Substitution x=etx=et in Cauchy–Euler helps to
A Make equation exact
B Get constant coefficients
C Remove derivatives
D Force separable form
With x=etx=et, powers of xx turn into exponentials in tt, and derivative relations convert the variable-coefficient equation into a constant-coefficient ODE in tt.
Reduction of order is used when you already know
A One solution y1y1
B Two solutions y1,y2y1,y2
C Only the RHS
D Only initial value
If a second-order linear ODE has one known solution y1y1, reduction of order helps find another independent solution y2y2 so the full complementary function can be built.
Variation of parameters primarily finds the
A Characteristic roots
B Wronskian equals zero
C Particular solution
D Exact integrating factor
Variation of parameters constructs a particular solution for nonhomogeneous linear ODEs by letting constants in the homogeneous solution vary as functions, typically involving the Wronskian.
If ypyp and yp∗yp∗ are both particular solutions, then yp−yp∗yp−yp∗ is
A A singular solution
B A homogeneous solution
C Always constant
D Not differentiable
Subtracting two particular solutions cancels the RHS, so the difference satisfies the homogeneous equation. This explains why the particular solution is not unique.
In second-order linear ODEs, if W(x0)≠0W(x0)=0, then W(x)W(x) is
A Always zero nearby
B Always constant
C Always negative
D Never zero nearby
For linear second-order ODEs, the Wronskian cannot change from nonzero to zero within an interval where coefficients are continuous. This supports independence on that interval.
The Wronskian test is mainly used for checking
A Exactness condition
B Separation of variables
C Linear independence
D Laplace inversion
The Wronskian helps determine whether two candidate solutions are linearly independent. Independent solutions are required to form the full general solution of a second-order linear ODE.
For the family y=Cx+f(C)y=Cx+f(C), the parameter CC represents
A x-intercept only
B Slope of line
C y-intercept only
D Curve radius
In Clairaut’s general solution, each CC sets the slope of a straight line. The intercept depends on f(C)f(C). The singular solution is the envelope of these lines.
The envelope condition for y=Cx+f(C)y=Cx+f(C) typically uses
A ∂y/∂C=0∂y/∂C=0
B ∂y/∂x=0∂y/∂x=0
C ∂y/∂y=0∂y/∂y=0
D ∂x/∂y=0∂x/∂y=0
The envelope is found by eliminating CC between the family equation and ∂y/∂C=0∂y/∂C=0. This gives the curve tangent to all lines in the family.
A solution of a homogeneous linear ODE forms a
A Random set only
B Closed curve set
C Discrete set only
D Vector space set
Homogeneous linear ODE solutions are closed under addition and scalar multiplication. This vector-space property is why two independent solutions can generate all solutions for second order.
For constant-coefficient ODEs, “CF” stands for
A Constant factor
B Curvature form
C Complementary function
D Change function
CF is the general solution of the associated homogeneous equation. It depends on the characteristic roots and is combined with a particular integral to form the complete solution.
In undetermined coefficients, if RHS duplicates a CF term, multiply trial by
A lnxlnx factor
B xx factor
C sinxsinx factor
D Constant factor
Multiplying by xx shifts the trial to a new independent form. If duplication persists (higher multiplicity), multiply by higher powers of xx until the trial is independent of the CF.
A boundary value problem usually specifies conditions at
A Two different points
B One point only
C Infinity only
D Origin only
Boundary value problems typically set values at two points, such as y(a)y(a) and y(b)y(b). This differs from initial value problems, which specify conditions at the same point.
A “particular integral” is a solution that satisfies
A Only homogeneous ODE
B Only Clairaut form
C Only Cauchy–Euler
D Only the full ODE
A particular integral (particular solution) satisfies the complete nonhomogeneous equation. It is added to the complementary function to produce the full family of solutions.
In the D-operator method, (D−a)y=0(D−a)y=0 has solution
A CxaCxa
B CeaxCeax
C CsinaxCsinax
D ClnxClnx
(D−a)y=0(D−a)y=0 means dy/dx=aydy/dx=ay. Solving gives y=Ceaxy=Ceax. This is a basic building block for higher-order constant-coefficient equations.
For root m=0m=0 in characteristic equation, a CF term is
A exex
B x−1x−1
C 11
D sinxsinx
If m=0m=0 is a root, then e0x=1e0x=1 is a solution. Repeated zero roots produce terms 1,x,x2,…1,x,x2,… as independent solutions.
If characteristic roots are 00 (double), CF includes
A ex,xexex,xex
B 1,x1,x
C sinx,cosxsinx,cosx
D x2,x3x2,x3
A double root at zero means solutions e0x=1e0x=1 and xe0x=xxe0x=x. These are independent and form the complementary function part for that repeated root.
“Stability” in ODEs is about
A Finding exact solutions
B Response to small changes
C Finding Wronskian only
D Using Laplace tables
Stability asks whether small changes in initial conditions cause only small changes in the solution behavior. It is a qualitative idea used to understand robustness of solutions.
For second-order ODE, two independent solutions are needed because order is
A One
B Three
C Two
D Zero
The order equals the number of integration constants in the general solution. A second-order linear homogeneous ODE needs two constants, so two independent solutions are required.
For Clairaut y=px+f(p)y=px+f(p), if dp/dx≠0dp/dx=0, then it must satisfy
A x+f′(p)=0x+f′(p)=0
B x−f′(p)=0x−f′(p)=0
C x+f(p)=0x+f(p)=0
D x−p=0x−p=0
Differentiating gives (x+f′(p))dp/dx=0(x+f′(p))dp/dx=0. If dp/dx≠0dp/dx=0, then x+f′(p)=0x+f′(p)=0. Eliminating pp from this and original yields the singular solution.
The singular solution of Clairaut’s equation is obtained by
A Separating variables
B Eliminating pp
C Exactness method
D Integrating factor
Use the envelope condition from differentiation and eliminate the parameter pp. This produces a curve not included in the general line family, representing the singular solution.
A linear ODE is called “linear” because yy and derivatives appear
A Only as squares
B Only inside sine
C Only in products
D To first power
In a linear ODE, y,y′,y′′y,y′,y′′ appear only to power one and are not multiplied together. Coefficients may depend on xx, but the dependent variable stays linear.
For Cauchy–Euler homogeneous equation, if roots are equal, solutions involve
A sinxsinx factor
B exex factor
C lnxlnx factor
D Constant only
For repeated root mm in Cauchy–Euler, solutions are xmxm and xmlnxxmlnx. The logarithm factor creates a second independent solution, similar to the xx factor in constant coefficients.
A quick indicator of “constant coefficients” is coefficients are
A Powers of x
B Numbers only
C Trig functions
D Exponential in x
Constant-coefficient linear ODEs have fixed numerical coefficients, like y′′+3y′+2y=⋯y′′+3y′+2y=⋯. This allows the characteristic equation approach using exponential trial solutions.
A quick indicator of “variable coefficients” is coefficients
A Depend on x
B Are all zero
C Are all one
D Depend on y only
Variable-coefficient ODEs have coefficients that change with the independent variable, such as x2y′′+xy′+y=0x2y′′+xy′+y=0. These often need special substitutions or known standard forms.
The method “variation of parameters” typically uses
A Only separation
B Wronskian in formulas
C Only characteristic roots
D Only graphs
Variation of parameters builds a particular solution using integrals involving the Wronskian and the known homogeneous solutions. The Wronskian ensures independence and appears in the denominator.
The “annihilator method” works best with RHS that are
A Random complicated forms
B Only implicit forms
C Standard simple functions
D Only discontinuous
An annihilator is a differential operator that turns the RHS into zero, such as for exponentials or trig functions. This helps convert a nonhomogeneous equation into a higher-order homogeneous one.
Laplace transform is mainly helpful for ODEs with
A Only boundary conditions
B Only Clairaut form
C Only Cauchy–Euler
D Initial conditions
Laplace transform converts derivatives into algebraic expressions and naturally incorporates initial values like y(0)y(0) and y′(0)y′(0). It is especially useful with piecewise or forcing inputs.
In a second-order linear ODE, if one solution is known, the second is often found using
A Separation method
B Reduction of order
C Bernoulli method
D Clairaut method
Reduction of order assumes a second solution of the form y2=v(x)y1y2=v(x)y1. Substitution reduces the problem to a simpler equation for v(x)v(x), yielding an independent second solution.