For solutions of y′′+P(x)y′+Q(x)y=0y′′+P(x)y′+Q(x)y=0, Abel’s formula mainly helps you find
A Particular integral
B Laplace inverse
C Wronskian form
D Series radius
Abel’s formula shows how the Wronskian depends on P(x)P(x). It gives W(x)=C e−∫P(x) dxW(x)=Ce−∫P(x)dx, so you can know whether WW can vanish in the interval.
If P(x)=0P(x)=0 in y′′+P(x)y′+Q(x)y=0y′′+P(x)y′+Q(x)y=0, then Abel’s formula implies the Wronskian is
A Constant on interval
B Always increasing
C Always zero
D Always negative
With P(x)=0P(x)=0, Abel’s formula becomes W(x)=Ce0=CW(x)=Ce0=C. So the Wronskian stays constant wherever the coefficients are continuous, confirming consistent independence or dependence.
If two solutions of a linear ODE satisfy W(x0)=0W(x0)=0 at some point, then on that interval they are
A Independent solutions
B Orthogonal always
C Dependent solutions
D Periodic always
For second-order linear homogeneous ODEs with continuous coefficients, if the Wronskian is zero at one point, Abel’s result forces it to be zero everywhere, meaning solutions are linearly dependent.
For y1=e2xy1=e2x and y2=e−xy2=e−x, the Wronskian is
A 3ex3ex
B exex
C −3ex−3ex
D 00
W=y1y2′−y2y1′W=y1y2′−y2y1′. Here y2′=−e−xy2′=−e−x, y1′=2e2xy1′=2e2x. So W=e2x(−e−x)−e−x(2e2x)=−3ex≠0W=e2x(−e−x)−e−x(2e2x)=−3ex=0.
For a second-order homogeneous linear ODE, the general solution has how many arbitrary constants?
A One constant
B Two constants
C Three constants
D No constants
The order of the ODE equals the number of independent constants in the general solution. A second-order homogeneous linear equation needs two independent solutions, giving two constants.
In the form “solvable for yy” of first-order higher-degree equations, the typical rearrangement is
A p=f(x,y)p=f(x,y)
B x=f(y)x=f(y)
C y=f(x,p)y=f(x,p)
D p=f(x)p=f(x)
“Solvable for yy” means you can explicitly express yy using xx and p=dy/dxp=dy/dx. Then a parametric method is often used to eliminate pp.
In equations solvable for xx, the useful parametric idea is to treat pp as
A Parameter variable
B Constant always
C Random number
D Integral limit
Many solvable-for-xx or solvable-for-yy equations become easier when pp is treated as a parameter. You find relations in xx and yy with pp, then eliminate pp.
The “singular solution” is commonly obtained as the
A Parallel family
B Constant solution
C Closed orbit
D Envelope curve
A singular solution often appears as the envelope of a one-parameter family of solutions. It is not obtained by simply choosing a parameter value from the general solution.
Clairaut’s equation gives a general family of
A Circles only
B Parabolas only
C Straight lines
D Hyperbolas only
The general solution of Clairaut’s equation is y=Cx+f(C)y=Cx+f(C). This represents a family of straight lines with slope CC. The singular solution is their envelope when it exists.
For Clairaut y=px+f(p)y=px+f(p), the envelope condition uses
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 p+f(p)=0p+f(p)=0
Differentiating gives (x+f′(p))dp/dx=0(x+f′(p))dp/dx=0. For the singular solution, take dp/dx≠0dp/dx=0, so x+f′(p)=0x+f′(p)=0. Eliminate pp to get the envelope curve.
A correct statement for constant-coefficient linear ODEs is
A Only separable method
B Always exact form
C CF from roots
D PI from Wronskian
For constant coefficients, the complementary function comes from solving the characteristic equation. The particular integral depends on the RHS form and is found using suitable methods like undetermined coefficients.
If the characteristic equation has roots m=1,2m=1,2, then CF is
A C1sinx+C2cosxC1sinx+C2cosx
B C1ex+C2e2xC1ex+C2e2x
C C1x+C2C1x+C2
D C1e2x+C2xexC1e2x+C2xex
Each distinct real root mm gives a solution emxemx. With roots 1 and 2, the complementary function is C1ex+C2e2xC1ex+C2e2x.
If a repeated root is m=3m=3 (double), CF must include
A e3x,xe3xe3x,xe3x
B e3x,e−3xe3x,e−3x
C 1,x1,x
D cos3x,sin3xcos3x,sin3x
A repeated root needs an extra independent solution. For a double root mm, solutions are emxemx and xemxxemx. This ensures linear independence.
Roots ±iβ±iβ (pure imaginary) produce CF terms
A eβx,e−βxeβx,e−βx
B xβ,x−βxβ,x−β
C 1,x1,x
D cosβx,sinβxcosβx,sinβx
If roots are 0±iβ0±iβ, then α=0α=0. The real solutions become cos(βx)cos(βx) and sin(βx)sin(βx), representing undamped oscillations.
For y′′−y=exy′′−y=ex, a correct PI trial is
A AexAex
B AcosxAcosx
C AxexAxex
D Ax2Ax2
The RHS is exex. Since exex is not part of CF for m=±1m=±1? Actually m=1m=1 is a root, so exex is in CF, so multiply by xx. Hence PI should be AxexAxex.
For y′′−y=exy′′−y=ex, the correct PI form is
A AxexAxex
B AsinxAsinx
C AA constant
D AexAex
The characteristic roots are m=±1m=±1, so exex already appears in the complementary function. To avoid duplication, the particular trial must be multiplied by xx, giving AxexAxex.
For y′′+y=sinxy′′+y=sinx, a good PI trial is
A Asinx+BcosxAsinx+Bcosx
B AexAex
C x(Asinx+Bcosx)x(Asinx+Bcosx)
D AxAx
Derivatives of sinxsinx and cosxcosx cycle between each other, so the trial must include both. Coefficients are found by substituting into the ODE and matching terms.
For y′′+y=sinxy′′+y=sinx, since sinxsinx is in CF, the corrected PI trial should be
A Asinx+BcosxAsinx+Bcosx
B x(Asinx+Bcosx)x(Asinx+Bcosx)
C AexAex
D Ax2Ax2
For y′′+y=0y′′+y=0, CF is C1cosx+C2sinxC1cosx+C2sinx. Since RHS sinxsinx overlaps with CF, multiply the trial by xx to get an independent PI form.
A Cauchy–Euler equation is recognized by coefficients like
A Constant numbers
B Only trig functions
C Only exponentials
D Powers of xx
In Cauchy–Euler equations, the coefficient of y′′y′′ is x2x2, of y′y′ is xx, and so on. This structure allows solution by y=xmy=xm or by x=etx=et.
For Cauchy–Euler x2y′′+axy′+by=0x2y′′+axy′+by=0, the trial y=xmy=xm leads to
A Integral in xx
B Trig identity
C Algebraic in mm
D Laplace table
Substituting y=xmy=xm makes each term proportional to xmxm. Cancelling xmxm yields an algebraic equation in mm, similar to the characteristic equation method.
If Cauchy–Euler gives a repeated root mm, solutions are
A xm,xmlnxxm,xmlnx
B emx,xemxemx,xemx
C cosmx,sinmxcosmx,sinmx
D 1,x1,x only
For Cauchy–Euler, repeated root produces a logarithmic factor. The second solution becomes xmlnxxmlnx, ensuring independence from xmxm in the solution space.
The substitution x=etx=et is used mainly to
A Remove variable powers
B Make equation exact
C Convert to constants
D Remove RHS
Setting x=etx=et changes derivatives with respect to xx into derivatives with respect to tt. This often transforms Cauchy–Euler variable-coefficient equations into constant-coefficient equations in tt.
Variation of parameters requires knowing
A Homogeneous solutions
B Only RHS function
C Only initial values
D Only Wronskian zero
Variation of parameters builds the particular solution using the known fundamental solutions of the homogeneous equation. The Wronskian of these solutions appears in the formulas for the varying coefficients.
In variation of parameters (2nd order), the Wronskian appears typically in the
A Numerator only
B Exponent only
C Denominator
D Constant term
The formulas for the functions u1′(x)u1′(x) and u2′(x)u2′(x) involve the Wronskian in the denominator. A nonzero Wronskian ensures the method is valid on that interval.
Reduction of order often assumes the second solution as
A y2=y12y2=y12
B y2=lny1y2=lny1
C y2=1/y1y2=1/y1
D y2=v(x)y1y2=v(x)y1
With a known solution y1y1, assume y2=v(x)y1y2=v(x)y1. Substituting into the ODE reduces the order and yields an equation for v(x)v(x), producing an independent solution.
In constant coefficients, if RHS is polynomial, a PI trial is usually a
A Sine-cos pair
B Polynomial of degree
C Exponential of x
D Logarithmic form
For polynomial forcing, choose a polynomial trial of the same degree as RHS. If any term duplicates the CF, multiply by xkxk to remove overlap and solve for coefficients.
The annihilator method idea is to apply an operator that makes RHS
A Larger degree
B Periodic always
C Zero function
D Undefined
An annihilator operator is chosen so that applying it to the RHS produces zero. Applying the same operator to both sides creates a higher-order homogeneous equation helpful for finding a PI.
The operator method works smoothly when coefficients are
A Constant coefficients
B Variable coefficients
C Random coefficients
D Nonlinear coefficients
Differential operators like DD behave algebraically for constant coefficients, making factorization and trial methods convenient. With variable coefficients, operator algebra is usually not straightforward.
A boundary value problem typically gives conditions at
A One point
B Zero point
C Two points
D Random points
Boundary conditions specify values at different points, such as y(a)y(a) and y(b)y(b). This contrasts with initial value problems, where conditions like y(0)y(0) and y′(0)y′(0) are at one point.
A Sturm–Liouville problem is usually a special type of
A Initial value
B Separable only
C Exact only
D Boundary value
Sturm–Liouville problems are typically second-order linear differential equations with boundary conditions. They lead to eigenvalues and orthogonal eigenfunctions in many physical and mathematical applications.
For homogeneous linear ODE solutions, if y1,y2y1,y2 are solutions, then y1+y2y1+y2 is
A Never a solution
B Also a solution
C Only sometimes
D Not differentiable
The solution set of a homogeneous linear ODE is closed under addition. This is part of the superposition principle, allowing linear combinations to produce more solutions.
In nonhomogeneous linear ODE, y1+y2y1+y2 is a solution if both are
A Particular solutions
B Solutions of full ODE
C Homogeneous solutions
D Any differentiable
Superposition works directly for the homogeneous part. The sum of two solutions of the homogeneous equation remains a solution. For full nonhomogeneous equations, adding two full solutions changes the RHS.
If ypyp is a particular solution, then y=yp+yhy=yp+yh is
A Complete solution
B Singular solution
C Only CF
D Only PI
Any solution of the nonhomogeneous linear equation can be written as one particular solution plus a homogeneous solution. This forms the complete family of solutions.
If W≠0W=0, variation of parameters is valid because solutions are
A Dependent set
B Periodic set
C Independent set
D Constant set
A nonzero Wronskian indicates the fundamental solutions are linearly independent. This independence is necessary to solve for varying coefficients uniquely in variation of parameters.
In first-order higher-degree equations, “elimination of parameter” means eliminating
A xx
B yy
C Constant CC
D pp
Many such equations are expressed using p=dy/dxp=dy/dx. Solutions are found in parametric form involving pp, then pp is eliminated to obtain a relation between xx and yy.
The “discriminant” idea in singular solution finding is related to
A Initial condition
B Envelope condition
C Cauchy theorem
D Laplace shift
Singular solutions arise where the family has a common tangent, which is an envelope condition. Discriminant-type reasoning appears when eliminating parameters leads to repeated roots representing tangency.
If the auxiliary equation has roots 2±3i2±3i, CF contains
A e3xcos2xe3xcos2x
B e2xcos3xe2xcos3x
C cos2xcos2x only
D e2xe2x only
Complex roots α±iβα±iβ produce eαxcosβxeαxcosβx and eαxsinβxeαxsinβx. Here α=2α=2, β=3β=3, so those terms appear in CF.
For constant coefficients, if RHS is eaxeax and aa is not a root, PI trial is
A AxeaxAxeax
B AcosaxAcosax
C AeaxAeax
D Ax2Ax2
When eaxeax does not overlap with the complementary function, the simplest trial is AeaxAeax. Substitute and solve for AA. Overlap requires multiplying by xx.
For constant coefficients, if RHS is cosaxcosax and ±ia±ia are not roots, PI trial is
A Acosax+BsinaxAcosax+Bsinax
B AeaxAeax
C AxAx
D AlnxAlnx
Differentiation mixes sine and cosine, so both must be included. If ±ia±ia are not roots of the characteristic equation, this trial does not overlap with CF.
In a Cauchy–Euler equation, the change x=etx=et implies t=t=
A x2x2
B 1/x1/x
C exex
D lnxlnx
If x=etx=et, then taking natural logs gives t=lnxt=lnx. This substitution is used to convert equidimensional equations to constant-coefficient equations in the new variable tt.
A correct statement about Wronskian in linear ODE is
A Nonzero at one point ⇒ zero everywhere
B Always equals one
C Zero at one point ⇒ zero everywhere
D Always equals determinant constant
With continuous coefficients in a second-order linear homogeneous ODE, Abel’s formula forces the Wronskian to be either identically zero or never zero on the interval, so zero once means zero always.
In reduction of order, why multiply by v(x)v(x)?
A To ensure independence
B To make it exact
C To remove RHS
D To reduce degree
If y2=v(x)y1y2=v(x)y1, choosing a variable v(x)v(x) allows a new solution not just a constant multiple of y1y1. This is needed to get a truly independent second solution.
The term “particular integral” is mostly used for
A Homogeneous linear
B Nonhomogeneous linear
C Nonlinear only
D Separable only
A particular integral is a specific solution that satisfies the full nonhomogeneous equation. It is added to the complementary function to get the complete general solution.
If y′′+4y=0y′′+4y=0, the nature of solutions is
A Exponential growth
B Polynomial solutions
C Oscillatory solutions
D Logarithmic solutions
The characteristic equation is m2+4=0m2+4=0 giving m=±2im=±2i. Pure imaginary roots produce sinusoidal solutions, so the motion is oscillatory with sine and cosine terms.
If y′′−4y=0y′′−4y=0, solutions are mainly
A Sinusoidal type
B Constant only
C Logarithmic only
D Exponential type
The characteristic equation is m2−4=0m2−4=0 giving m=±2m=±2. Real roots produce exponential solutions e2xe2x and e−2xe−2x, so the solution is exponential in nature.
If y′′+4y′+4y=0y′′+4y′+4y=0, characteristic root type is
A Distinct real roots
B Pure imaginary
C Repeated real root
D Complex nonrepeated
The characteristic equation (m+2)2=0(m+2)2=0 gives a repeated root m=−2m=−2. Hence the complementary function is (C1+C2x)e−2x(C1+C2x)e−2x.
A “medium” check for resonance is whether the RHS form is
A Already in CF
B Always polynomial
C Always exponential
D Always zero
Resonance occurs when the trial PI duplicates a complementary function term. You test by comparing RHS form with CF terms; if overlap exists, multiply the trial by xx to fix it.
For variation of parameters, if Wronskian becomes zero, the method
A Becomes easier
B Gives same result
C Fails there
D Gives constant PI
Variation of parameters requires dividing by the Wronskian. If the Wronskian is zero, the formulas break down, indicating the chosen solutions are not independent at that point.
In first-order higher-degree equations, “solvable for xx” often gives
A x=f(y)x=f(y) only
B x=f(p)x=f(p) only
C x=f(x,y)x=f(x,y)
D x=f(y,p)x=f(y,p)
“Solvable for xx” means the equation can be rearranged to express xx using yy and pp. Then parametric elimination of pp is used to obtain the final relation.
The main purpose of Wronskian in advanced ODE methods is
A Exactness verification
B Stability verification
C Boundary verification
D Independence verification
The Wronskian is primarily a tool to check linear independence of solutions. Independence is essential for forming the correct general solution and for methods like variation of parameters and reduction of order.