For y1=sinxy1=sinx and y2=cosxy2=cosx, the Wronskian W(y1,y2)W(y1,y2) equals
A 11
B 00
C −1−1
D sinxsinx
W=y1y2′−y2y1′W=y1y2′−y2y1′. Here y2′=−sinxy2′=−sinx and y1′=cosxy1′=cosx. So W=sinx(−sinx)−cosx(cosx)=−(sin2x+cos2x)=−1W=sinx(−sinx)−cosx(cosx)=−(sin2x+cos2x)=−1.
If a second-order linear ODE has a nonzero Wronskian at one point, then on that interval the solutions are
A Independent everywhere
B Dependent everywhere
C Constant everywhere
D Periodic everywhere
By Abel’s result, the Wronskian is either never zero or identically zero on an interval of continuity. If it is nonzero at one point, it cannot become zero later, so solutions remain independent.
In y′′+P(x)y′+Q(x)y=0y′′+P(x)y′+Q(x)y=0, Abel’s formula gives W(x)W(x) proportional to
A e∫Qdxe∫Qdx
B ∫Pdx∫Pdx
C P(x)Q(x)P(x)Q(x)
D e−∫Pdxe−∫Pdx
Abel’s formula states W(x)=C e−∫P(x) dxW(x)=Ce−∫P(x)dx. So the coefficient of y′y′ controls the growth/decay of the Wronskian across the interval.
For y′′+2y′+y=0y′′+2y′+y=0, the complementary function is
A C1ex+C2e−xC1ex+C2e−x
B (C1+C2x)e−x(C1+C2x)e−x
C C1cosx+C2sinxC1cosx+C2sinx
D (C1+C2x)ex(C1+C2x)ex
Characteristic equation is m2+2m+1=0=(m+1)2m2+2m+1=0=(m+1)2. A repeated root m=−1m=−1 gives CF y=(C1+C2x)e−xy=(C1+C2x)e−x.
For y′′−5y′+6y=0y′′−5y′+6y=0, the roots are m=m=
A −2,−3−2,−3
B 1,61,6
C 2,32,3
D −1,−6−1,−6
The characteristic equation is m2−5m+6=0m2−5m+6=0. Factoring gives (m−2)(m−3)=0(m−2)(m−3)=0, so roots are 2 and 3, giving exponential solutions e2xe2x and e3xe3x.
For y′′+9y=0y′′+9y=0, the general solution is
A C1cos3x+C2sin3xC1cos3x+C2sin3x
B C1e3x+C2e−3xC1e3x+C2e−3x
C (C1+C2x)e−3x(C1+C2x)e−3x
D C1cosx+C2sinxC1cosx+C2sinx
Characteristic equation m2+9=0m2+9=0 gives m=±3im=±3i. Pure imaginary roots produce oscillatory solutions, so CF is C1cos3x+C2sin3xC1cos3x+C2sin3x.
For y′′−9y=0y′′−9y=0, the general solution is
A C1x+C2C1x+C2
B C1cos3x+C2sin3xC1cos3x+C2sin3x
C (C1+C2x)e3x(C1+C2x)e3x
D C1e3x+C2e−3xC1e3x+C2e−3x
Characteristic equation m2−9=0m2−9=0 gives m=±3m=±3. Real distinct roots produce exponential solutions e3xe3x and e−3xe−3x, forming the complementary function.
If RHS is e2xe2x and m=2m=2 is a simple root, the PI trial must include
A Ae2xAe2x
B Ax2e2xAx2e2x
C Axe2xAxe2x
D Acos2xAcos2x
If e2xe2x appears in the complementary function, a direct trial Ae2xAe2x duplicates CF. Multiply by xx once for a simple overlap to get an independent trial Axe2xAxe2x.
If RHS is e2xe2x and m=2m=2 is a double root, the PI trial should be
A Ae2xAe2x
B Ax2e2xAx2e2x
C Axe2xAxe2x
D Asin2xAsin2x
A double overlap means e2xe2x and xe2xxe2x are already in CF. Multiply by x2x2 to shift beyond the repeated root duplication, giving PI trial Ax2e2xAx2e2x.
For y′′+y=cosxy′′+y=cosx, the correct PI trial is
A x(Acosx+Bsinx)x(Acosx+Bsinx)
B Acosx+BsinxAcosx+Bsinx
C AexAex
D AxAx
CF for y′′+y=0y′′+y=0 already contains cosxcosx and sinxsinx. Since RHS cosxcosx overlaps, multiply the usual sin-cos trial by xx to ensure independence.
For y′′+4y=sin2xy′′+4y=sin2x, resonance occurs because
A RHS is exponential
B Coefficients variable
C RHS matches CF
D Wronskian is zero
For y′′+4y=0y′′+4y=0, CF includes cos2xcos2x and sin2xsin2x. Since RHS is sin2xsin2x, it duplicates a CF term, so the PI must be multiplied by xx.
For Cauchy–Euler x2y′′+xy′−y=0x2y′′+xy′−y=0, the trial y=xmy=xm gives the equation
A m2+m−1=0m2+m−1=0
B m2+1=0m2+1=0
C m2−2m=0m2−2m=0
D m2−1=0m2−1=0
Substitute y=xmy=xm. Then y′=mxm−1y′=mxm−1, y′′=m(m−1)xm−2y′′=m(m−1)xm−2. Plugging gives [m(m−1)+m−1]xm=(m2−1)xm=0[m(m−1)+m−1]xm=(m2−1)xm=0.
For x2y′′+xy′−y=0x2y′′+xy′−y=0, the roots mm are
A 1,−11,−1
B 0,10,1
C 2,−22,−2
D 1,11,1
From the indicial equation m2−1=0m2−1=0, roots are m=1m=1 and m=−1m=−1. So solutions are xx and x−1x−1, giving the complementary function C1x+C2/xC1x+C2/x.
For Cauchy–Euler repeated root mm, the second solution typically is
A xemxxemx
B emxlnxemxlnx
C xmlnxxmlnx
D sin(mx)sin(mx)
In equidimensional equations, a repeated root in the mm-equation produces solutions xmxm and xmlnxxmlnx. The logarithm creates linear independence.
Variation of parameters is preferred when RHS is
A Always polynomial
B Not simple type
C Always exponential
D Always zero
Undetermined coefficients works for specific RHS forms (polynomial, exponential, trig). For more general RHS, variation of parameters is reliable because it constructs a PI using integrals with homogeneous solutions.
In variation of parameters for y′′+Py′+Qy=Ry′′+Py′+Qy=R, a key requirement is
A P=0P=0
B Q=0Q=0
C R=0R=0
D W≠0W=0
The formulas involve dividing by the Wronskian WW. If W=0W=0, the chosen homogeneous solutions are dependent, and the method cannot uniquely determine the varying parameters.
Reduction of order is most useful when
A RHS is zero
B Coefficients constant
C One solution known
D Order is first
If one solution of a second-order linear ODE is known, reduction of order provides a systematic way to find a second independent solution, completing the fundamental set.
In first-order higher-degree equations, the “solvable for xx” form is commonly
A x=f(y,p)x=f(y,p)
B x=f(x,y)x=f(x,y)
C x=f(y)x=f(y) only
D x=f(p)x=f(p) only
Solvable for xx means isolate xx in terms of yy and p=dy/dxp=dy/dx. Then parametric elimination of pp is performed to obtain the solution relation between xx and yy.
Lagrange-type equation is commonly expressed as
A y=px+f(p)y=px+f(p)
B y=xϕ(p)+ψ(p)y=xϕ(p)+ψ(p)
C y′+Py=Qy′+Py=Q
D y′′+ay=0y′′+ay=0
Lagrange’s form uses p=dy/dxp=dy/dx and writes yy as xϕ(p)+ψ(p)xϕ(p)+ψ(p). It generalizes Clairaut’s form and is treated using parameter methods and elimination.
Clairaut’s equation is a special case of Lagrange when
A ϕ(p)=xϕ(p)=x
B ψ(p)=xψ(p)=x
C ϕ(p)=pϕ(p)=p
D ψ(p)=yψ(p)=y
Lagrange form y=xϕ(p)+ψ(p)y=xϕ(p)+ψ(p). If ϕ(p)=pϕ(p)=p and ψ(p)=f(p)ψ(p)=f(p), it becomes Clairaut y=px+f(p)y=px+f(p), giving a family of straight lines and an envelope.
A singular solution is not obtained by
A Choosing parameter value
B Envelope elimination
C Differentiation step
D Parameter removal
A singular solution does not come from plugging a constant value into the general solution family. It arises from the envelope condition, typically by differentiating with respect to the parameter and eliminating it.
For a nonhomogeneous linear ODE, any two particular solutions differ by
A A constant only
B A singular curve
C A Wronskian value
D A homogeneous solution
If yp1yp1 and yp2yp2 satisfy the full equation, then subtracting gives a solution of the homogeneous equation. Hence yp1−yp2yp1−yp2 is always a homogeneous solution.
In constant-coefficient ODEs, the “auxiliary equation” is also called
A Indicial equation
B Separation equation
C Characteristic equation
D Laplace equation
Substituting y=emxy=emx produces a polynomial equation in mm. This is known as the characteristic (or auxiliary) equation and determines the complementary function structure.
If the characteristic polynomial factors (D−1)(D−3)(D−1)(D−3), then roots are
A −1,−3−1,−3
B 1,31,3
C 0,20,2
D 2,42,4
DD acts like mm in the characteristic equation. (D−1)(D−3)=0(D−1)(D−3)=0 corresponds to roots m=1m=1 and m=3m=3. Hence CF uses exex and e3xe3x.
The annihilator for RHS sinaxsinax is
A D2+a2D2+a2
B D−aD−a
C D2−a2D2−a2
D D+aD+a
Applying D2+a2D2+a2 to sinaxsinax gives zero, since d2dx2sinax=−a2sinaxdx2d2sinax=−a2sinax. This operator “annihilates” sine and cosine terms.
The annihilator for RHS eaxeax is
A D+aD+a
B D2+a2D2+a2
C D−aD−a
D D2−a2D2−a2
(D−a)eax=0(D−a)eax=0 because Deax=aeaxDeax=aeax. So D−aD−a is the annihilator for eaxeax. This idea helps convert nonhomogeneous equations into homogeneous ones.
In operator method, “PI” is a short form for
A Product integral
B Partial iteration
C Period index
D Particular integral
PI stands for particular integral (particular solution) of a nonhomogeneous linear ODE. The complete solution is CF + PI, where CF solves the corresponding homogeneous equation.
A second-order ODE needs two independent solutions because the solution space dimension is
A One-dimensional
B Two-dimensional
C Three-dimensional
D Zero-dimensional
The set of solutions of a second-order linear homogeneous ODE forms a two-dimensional vector space. Therefore, two linearly independent solutions are required to represent every solution.
For y1=exy1=ex and y2=xexy2=xex, the Wronskian is
A e2xe2x
B 00
C −e2x−e2x
D xe2xxe2x
y1′=exy1′=ex, y2′=ex+xexy2′=ex+xex. So W=y1y2′−y2y1′=ex(ex+xex)−xex⋅ex=e2x≠0W=y1y2′−y2y1′=ex(ex+xex)−xex⋅ex=e2x=0, hence independent.
A key use of Wronskian in reduction of order is to
A Confirm independence
B Remove initial data
C Confirm independence
D Find boundary points
Reduction of order aims to build a second solution independent of the first. The Wronskian helps verify independence and also appears in related methods like variation of parameters.
A linear ODE is homogeneous when the RHS is
A Zero function
B Constant only
C Polynomial only
D Zero function
In a linear ODE, “homogeneous” means the forcing term is zero. Then solutions satisfy the superposition principle and form a vector space, enabling systematic construction of general solutions.
For y′′+y′=0y′′+y′=0, the characteristic equation is
A m2+m=0m2+m=0
B m2−m=0m2−m=0
C m2+1=0m2+1=0
D m2=0m2=0
Replace yy by emxemx. Then y′=memxy′=memx, y′′=m2emxy′′=m2emx. Substitute to get m2emx+memx=0⇒m2+m=0m2emx+memx=0⇒m2+m=0.
For y′′+y′=0y′′+y′=0, the complementary function is
A C1ex+C2e−xC1ex+C2e−x
B C1cosx+C2sinxC1cosx+C2sinx
C C1+C2e−xC1+C2e−x
D C1x+C2C1x+C2
From m2+m=0⇒m(m+1)=0m2+m=0⇒m(m+1)=0, roots are 00 and −1−1. Thus solutions are 11 and e−xe−x, giving CF C1+C2e−xC1+C2e−x.
For y′′+y=xy′′+y=x, a valid PI trial type is
A Exponential type
B Polynomial type
C Logarithmic type
D Trig type
The RHS is a polynomial xx. For constant coefficients, choose a polynomial PI of suitable degree, typically Ax+BAx+B. Then substitute and match coefficients to solve.
The “standard linear form” of first-order ODE is
A dy/dx=P(x)Q(x)dy/dx=P(x)Q(x)
B dy/dx=y2dy/dx=y2
C dy/dx=xydy/dx=xy only
D dy/dx+P(x)y=Q(x)dy/dx+P(x)y=Q(x)
A first-order linear ODE is written as dy/dx+P(x)y=Q(x)dy/dx+P(x)y=Q(x). It is solved using an integrating factor e∫Pdxe∫Pdx, producing a direct formula for yy.
For first-order linear ODE, integrating factor is
A e∫Qdxe∫Qdx
B ∫Pdx∫Pdx
C e∫Pdxe∫Pdx
D ∫Qdx∫Qdx
Multiply the equation by e∫Pdxe∫Pdx to make the left side a derivative of (y⋅IF)(y⋅IF). This converts the ODE into an exact derivative, allowing direct integration.
A typical “medium” check for Wronskian is it is computed using
A Determinant of derivatives
B Product of solutions
C Sum of derivatives
D Ratio of solutions
The Wronskian is a determinant built from the solutions and their derivatives. For two functions, it uses y1,y2y1,y2 and y1′,y2′y1′,y2′, showing dependence or independence.
For y=px+f(p)y=px+f(p), general family is obtained by setting
A x=Cx=C constant
B p=Cp=C constant
C y=Cy=C constant
D f(p)=0f(p)=0
In Clairaut’s equation, the general solution comes by taking pp as a constant slope CC. Substituting gives y=Cx+f(C)y=Cx+f(C), a family of straight lines.
The singular solution is found by eliminating pp between
A y=px+f(p)y=px+f(p) and y=0y=0
B y=0y=0 and p=0p=0
C x=px=p and y=py=p
D y=px+f(p)y=px+f(p) and x+f′(p)=0x+f′(p)=0
The envelope condition comes from differentiation and gives x+f′(p)=0x+f′(p)=0. Eliminating the parameter pp between this and the original family yields the singular curve.
In Laplace method, the transform variable is usually
A mm variable
B pp variable
C ss variable
D tt variable
Laplace transform converts a function of time (often tt) into a function of ss. This turns differential equations into algebraic equations in ss, incorporating initial conditions naturally.
For constant coefficients, “damping” is linked to roots having
A Negative real part
B Zero real part
C Positive real part
D No real part
In physical models, if complex roots have a negative real part α<0α<0, the factor eαxeαx causes amplitude to decay. This corresponds to damping in oscillatory solutions.
For constant coefficients, “undamped oscillation” corresponds to roots
A Distinct real
B Pure imaginary
C Repeated real
D Zero roots only
Pure imaginary roots ±iβ±iβ give solutions cosβxcosβx and sinβxsinβx with no exponential decay or growth. This matches undamped oscillations in many standard models.
For y′′+4y′+5y=0y′′+4y′+5y=0, roots are
A 2±i2±i
B −1±2i−1±2i
C −2±i−2±i
D 1±2i1±2i
Characteristic equation m2+4m+5=0m2+4m+5=0. Discriminant is 16−20=−416−20=−4. Roots are −4±2i2=−2±i2−4±2i=−2±i, giving damped oscillatory solutions.
For y′′+4y′+5y=0y′′+4y′+5y=0, the CF is
A e−2x(C1cosx+C2sinx)e−2x(C1cosx+C2sinx)
B e2x(C1cosx+C2sinx)e2x(C1cosx+C2sinx)
C C1e−2x+C2e−xC1e−2x+C2e−x
D C1cosx+C2sinxC1cosx+C2sinx
Roots −2±i−2±i give e−2xcosxe−2xcosx and e−2xsinxe−2xsinx. Combining gives y=e−2x(C1cosx+C2sinx)y=e−2x(C1cosx+C2sinx), representing damped oscillations.
A “fundamental set” for a second-order ODE means
A Two dependent solutions
B One particular solution
C One constant solution
D Two independent solutions
A fundamental set consists of two linearly independent solutions of the homogeneous second-order ODE. Any homogeneous solution can be written as a linear combination of these two.
If W=0W=0 for two solutions, then variation of parameters cannot be used because
A No integration needed
B Division by zero
C RHS becomes zero
D Order becomes first
The method uses formulas containing 1/W1/W. If W=0W=0, the homogeneous solutions are not independent, making the system for parameter derivatives unsolvable and causing division by zero.
In equation solvable for pp, “degree” refers to power of
A xx in equation
B yy in equation
C pp in equation
D constants only
First-order higher-degree equations may involve pp nonlinearly, like p2p2 or higher. The “degree” usually means the highest power of p=dy/dxp=dy/dx after removing radicals and fractions.
In Clairaut’s equation, the singular solution exists when the envelope condition gives
A Real elimination possible
B Only constant p
C Only zero slope
D Real elimination possible
The envelope requires eliminating pp between equations. A real-valued relation must exist after elimination. If elimination yields no real curve, then no real singular solution appears.
A good reason to prefer undetermined coefficients over variation of parameters is
A Faster for standard RHS
B Works for any RHS
C Needs Wronskian zero
D Avoids CF finding
For standard RHS like polynomials, exponentials, and trig functions, undetermined coefficients is quick and direct. Variation of parameters is more general but involves integrals and more algebra.
The main practical role of Abel’s formula in MCQs is to decide if Wronskian can
A Become constant always
B Become zero inside
C Become polynomial always
D Become negative always
Abel’s formula shows W(x)=Ce−∫PdxW(x)=Ce−∫Pdx. If C≠0C=0, the exponential never becomes zero, so the Wronskian cannot vanish within the interval—confirming independence throughout.