For y1=eaxy1=eax and y2=ebxy2=ebx with a≠ba=b, the Wronskian equals
A (a+b)e(a−b)x(a+b)e(a−b)x
B 00 always
C (b−a)e(a+b)x(b−a)e(a+b)x
D (a−b)e(a+b)x(a−b)e(a+b)x
W=y1y2′−y2y1′W=y1y2′−y2y1′. Here y2′=bebxy2′=bebx, y1′=aeaxy1′=aeax. So W=eaxbebx−ebxaeax=(b−a)e(a+b)x=(a−b)(−1)e(a+b)xW=eaxbebx−ebxaeax=(b−a)e(a+b)x=(a−b)(−1)e(a+b)x, giving (b−a)e(a+b)x(b−a)e(a+b)x; written as (a−b)e(a+b)x(a−b)e(a+b)x with sign depends order.
If y1=e2xy1=e2x, y2=e5xy2=e5x, then W(y1,y2)W(y1,y2) is
A 3e7x3e7x
B −3e7x−3e7x
C 7e3x7e3x
D 00
W=y1y2′−y2y1′W=y1y2′−y2y1′. Here y2′=5e5xy2′=5e5x, y1′=2e2xy1′=2e2x. So W=e2x(5e5x)−e5x(2e2x)=(5−2)e7x=3e7xW=e2x(5e5x)−e5x(2e2x)=(5−2)e7x=3e7x.
In y′′+P(x)y′+Q(x)y=0y′′+P(x)y′+Q(x)y=0, if P(x)=2xP(x)=x2, then Abel’s formula gives W(x)∝W(x)∝
A x2x2
B x−2x−2
C e2xe2x
D e−2xe−2x
Abel gives W=Ce−∫PdxW=Ce−∫Pdx. Here ∫Pdx=∫2/x dx=2lnx∫Pdx=∫2/xdx=2lnx. So W=Ce−2lnx=Cx−2W=Ce−2lnx=Cx−2, valid on intervals not crossing x=0x=0.
For a linear second-order ODE, if W(x0)≠0W(x0)=0, then W(x)W(x)
A Never zero there
B Becomes zero later
C Must be constant
D Must be negative
By Abel’s formula, W(x)=Ce−∫PdxW(x)=Ce−∫Pdx. If W(x0)≠0W(x0)=0, then C≠0C=0. An exponential term never becomes zero, so W(x)W(x) cannot vanish on that interval.
Consider y′′+y=sinxy′′+y=sinx. Correct particular trial is
A x(Asinx)x(Asinx)
B AsinxAsinx
C AcosxAcosx
D AxAx
CF of y′′+y=0y′′+y=0 contains sinxsinx and cosxcosx, so sinxsinx overlaps. Multiply the usual trial by xx to avoid duplication; a full trial is x(Asinx+Bcosx)x(Asinx+Bcosx).
For y′′+y=cosxy′′+y=cosx, the minimal correct PI trial must include
A exex term
B lnxlnx term
C xsinxxsinx term
D x2x2 term
Since cosxcosx is part of CF, PI must be multiplied by xx. A correct trial is x(Acosx+Bsinx)x(Acosx+Bsinx). This necessarily includes an xsinxxsinx term.
For y′′−2y′+y=exy′′−2y′+y=ex, the PI trial should be
A Ax2exAx2ex
B AxexAxex
C AexAex
D Ax3exAx3ex
Characteristic equation (m−1)2=0(m−1)2=0 has m=1m=1 double, so CF contains exex and xexxex. RHS exex overlaps with multiplicity 2, so multiply by x2x2: Ax2exAx2ex.
For y′′−2y′+y=exy′′−2y′+y=ex, the CF is
A C1ex+C2e−xC1ex+C2e−x
B C1cosx+C2sinxC1cosx+C2sinx
C (C1+C2x)e−x(C1+C2x)e−x
D (C1+C2x)ex(C1+C2x)ex
The auxiliary 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)exy=(C1+C2x)ex, standard for repeated roots.
For y′′+4y=sin2xy′′+4y=sin2x, the correct PI trial form is
A Acos2x+Bsin2xAcos2x+Bsin2x
B Ae2xAe2x
C x(Acos2x+Bsin2x)x(Acos2x+Bsin2x)
D AxAx
CF of y′′+4y=0y′′+4y=0 contains cos2xcos2x and sin2xsin2x. Since RHS is sin2xsin2x, resonance occurs. Multiply by xx to get an independent PI trial.
For Cauchy–Euler x2y′′−xy′+y=0x2y′′−xy′+y=0, the indicial equation becomes
A (m−1)2=0(m−1)2=0
B m2+m+1=0m2+m+1=0
C m2−1=0m2−1=0
D m2−2m=0m2−2m=0
With y=xmy=xm: y′=mxm−1y′=mxm−1, y′′=m(m−1)xm−2y′′=m(m−1)xm−2. Substitute: x2m(m−1)xm−2−x(mxm−1)+xm=[m(m−1)−m+1]xm=(m−1)2xmx2m(m−1)xm−2−x(mxm−1)+xm=[m(m−1)−m+1]xm=(m−1)2xm.
For x2y′′−xy′+y=0x2y′′−xy′+y=0, the two independent solutions are
A x,x2x,x2
B x,xlnxx,xlnx
C 1,lnx1,lnx
D x−1,xx−1,x
The indicial equation gives a repeated root m=1m=1. For Cauchy–Euler with repeated root, solutions are xmxm and xmlnxxmlnx. Hence y1=xy1=x, y2=xlnxy2=xlnx.
In variation of parameters for y′′+Py′+Qy=Ry′′+Py′+Qy=R, the condition u1′y1+u2′y2=0u1′y1+u2′y2=0 is used to
A Force homogeneity
B Make PI zero
C Simplify derivatives
D Change order
The extra condition u1′y1+u2′y2=0u1′y1+u2′y2=0 removes second derivatives of u1,u2u1,u2 after substitution. This reduces the algebra to a solvable system for u1′u1′ and u2′u2′.
For second-order ODE, variation of parameters yields u1′u1′ proportional to
A −y2R/W−y2R/W
B y2R/Wy2R/W
C −y1R/W−y1R/W
D y1R/Wy1R/W
Standard formulas: u1′=−y2RWu1′=−Wy2R, u2′=y1RWu2′=Wy1R, where WW is the Wronskian of y1,y2y1,y2. Signs come from solving the linear system.
In the same setup, u2′u2′ is proportional to
A −y1R/W−y1R/W
B y2R/Wy2R/W
C −y2R/W−y2R/W
D y1R/Wy1R/W
With yp=u1y1+u2y2yp=u1y1+u2y2 and the simplifying condition, solving gives u2′=y1RWu2′=Wy1R. This is why W≠0W=0 is necessary for the method to work.
For y′′+P(x)y′+Q(x)y=0y′′+P(x)y′+Q(x)y=0, if two solutions are independent, then their Wronskian is
A Zero at some point
B Never zero on interval
C Always constant only
D Always equals one
Independence implies W(x0)≠0W(x0)=0 at some point. Abel’s formula then forces W(x)≠0W(x)=0 everywhere on that interval of continuity, so the Wronskian never vanishes there.
A first-order equation “solvable for pp” may have a singular solution found using
A Integrating factor
B Separation only
C Envelope condition
D Laplace method
Such equations often define a one-parameter family through pp. A singular solution arises as an envelope of that family, obtained by differentiating with respect to the parameter and eliminating it.
For Clairaut y=px+p2y=px+p2, the general solution family is
A y=Cx+C2y=Cx+C2
B y=Cx−C2y=Cx−C2
C y=C2x+Cy=C2x+C
D y=x+C2y=x+C2
Clairaut form is y=px+f(p)y=px+f(p) with f(p)=p2f(p)=p2. General solution sets p=Cp=C, giving y=Cx+C2y=Cx+C2, a family of straight lines parameterized by slope CC.
For y=px+p2y=px+p2, the singular solution is
A y=x2/4y=x2/4
B y=−x2y=−x2
C y=−x2/4y=−x2/4
D y=x2y=x2
Envelope condition: x+f′(p)=0⇒x+2p=0⇒p=−x/2x+f′(p)=0⇒x+2p=0⇒p=−x/2. Substitute into y=px+p2y=px+p2: y=(−x/2)x+(x2/4)=−x2/2+x2/4=−x2/4y=(−x/2)x+(x2/4)=−x2/2+x2/4=−x2/4.
For Clairaut y=px+1py=px+p1, envelope condition gives
A x+1p2=0x+p21=0
B x+1p=0x+p1=0
C x−1p=0x−p1=0
D x−1p2=0x−p21=0
Here f(p)=1/pf(p)=1/p, so f′(p)=−1/p2f′(p)=−1/p2. Envelope condition is x+f′(p)=0⇒x−1/p2=0x+f′(p)=0⇒x−1/p2=0. Then eliminate pp to get the singular curve.
For y=px+1py=px+p1, the singular solution in real form is
A y=±2xy=±2x
B y=2/xy=2/x
C y=x/2y=x/2
D y=x2y=x2
From x=1/p2x=1/p2, take p=±1/xp=±1/x for x>0x>0. Substitute: y=px+1/p=±x±xy=px+1/p=±x±x. The envelope gives y=2xy=2x and also y=−2xy=−2x.
For constant coefficients, if RHS is xeaxxeax and aa is a root of multiplicity 1, PI trial should be
A xeaxxeax poly
B x2eaxx2eax poly
C eaxeax only
D sinaxsinax only
For xeaxxeax, trial normally is (Ax+B)eax(Ax+B)eax. If eaxeax overlaps with CF (simple root), multiply by xx. So PI becomes x(Ax+B)eaxx(Ax+B)eax, i.e., x2eaxx2eax times a linear polynomial.
If roots are m=am=a (double) and RHS is eaxeax, PI needs factor
A xx
B x2x2
C No factor
D lnxlnx
A double root means CF includes eaxeax and xeaxxeax. To avoid duplication, multiply the trial by x2x2. This ensures the PI is linearly independent from the CF terms.
The Wronskian of y1=xy1=x and y2=xlnxy2=xlnx equals
A xx
B x2x2
C 11
D 00
Compute W=y1y2′−y2y1′W=y1y2′−y2y1′. Here y1=x⇒y1′=1y1=x⇒y1′=1. y2=xlnx⇒y2′=lnx+1y2=xlnx⇒y2′=lnx+1. So W=x(lnx+1)−(xlnx)(1)=xW=x(lnx+1)−(xlnx)(1)=x.
For y1=xy1=x and y2=xlnxy2=xlnx, the correct Wronskian is
A 11
B xx
C x2x2
D 00
y1=xy1=x, y1′=1y1′=1; y2=xlnxy2=xlnx, y2′=lnx+1y2′=lnx+1. Then W=x(lnx+1)−(xlnx)⋅1=xW=x(lnx+1)−(xlnx)⋅1=x. Nonzero for x>0x>0, so independent.
In Cauchy–Euler, the substitution x=etx=et mainly converts x ddxxdxd into
A d2dt2dt2d2
B tddttdtd
C etddtetdtd
D ddtdtd
With x=etx=et, we have dx/dt=xdx/dt=x, so d/dx=(1/x)d/dtd/dx=(1/x)d/dt. Thus x(d/dx)=d/dtx(d/dx)=d/dt, which is the key simplification for equidimensional equations.
For x=etx=et, the relation between derivatives is
A dydx=xdydtdxdy=xdtdy
B dydx=1xdydtdxdy=x1dtdy
C dydx=dydtdxdy=dtdy
D dydx=e−tdydxdxdy=e−tdxdy
Chain rule gives dydx=dy/dtdx/dtdxdy=dx/dtdy/dt. Since dx/dt=xdx/dt=x, we get dydx=1xdydtdxdy=x1dtdy, used to transform Cauchy–Euler equations.
For linear ODEs, the solution set of the homogeneous equation forms a
A Vector space
B Closed interval
C Discrete set
D Random set
Homogeneous linear ODE solutions are closed under addition and scalar multiplication. This makes them a vector space, allowing basis solutions, dimension arguments, and systematic construction of general solutions.
For y′′+P(x)y′+Q(x)y=0y′′+P(x)y′+Q(x)y=0, if y1y1 is known, reduction of order often sets
A y2=y12y2=y12
B y2=vy1y2=vy1
C y2=1/y1y2=1/y1
D y2=lny1y2=lny1
The standard approach assumes y2=v(x)y1y2=v(x)y1. Substituting into the ODE reduces to a first-order equation in v′v′ (or vv), producing a second independent solution.
In reduction of order, the unknown function is usually
A P(x)P(x)
B Q(x)Q(x)
C v(x)v(x)
D W(x)W(x)
Since y2=v(x)y1y2=v(x)y1, the task is to determine v(x)v(x). After substitution and simplification, one solves for v′v′ and integrates to obtain vv, hence y2y2.
If the Wronskian of two solutions is identically zero, then the solutions are
A Dependent solutions
B Independent solutions
C Orthogonal always
D Complex always
An identically zero Wronskian indicates linear dependence for solutions of second-order linear homogeneous ODEs with continuous coefficients. One solution can be written as a constant multiple of the other.
For constant coefficients, the operator identity (D−a)2y=0(D−a)2y=0 implies solutions include
A eax,e−axeax,e−ax
B cosax,sinaxcosax,sinax
C 1,x1,x
D eax,xeaxeax,xeax
(D−a)2y=0(D−a)2y=0 corresponds to a repeated root m=am=a. Repeated roots generate solutions eaxeax and xeaxxeax. Higher powers give x2eaxx2eax, etc.
For (D2+1)y=0(D2+1)y=0, solutions are
A sinx,cosxsinx,cosx
B ex,e−xex,e−x
C x,1x,1
D lnx,1lnx,1
D2y=−yD2y=−y gives y′′+y=0y′′+y=0. Its characteristic equation m2+1=0m2+1=0 gives m=±im=±i, producing cosxcosx and sinxsinx as real independent solutions.
If RHS is a polynomial and m=0m=0 is a root of multiplicity 2, then PI trial for constant RHS should include
A AA constant
B AxAx
C Ax2Ax2
D A/xA/x
Constant RHS suggests PI as constant, but if m=0m=0 double, CF includes 11 and xx. Multiply by x2x2 to avoid overlap, giving PI trial Ax2Ax2.
For y′′=1y′′=1, the general solution contains which term?
A exex
B x222x2
C sinxsinx
D lnxlnx
Integrate twice: y′=x+C1y′=x+C1, y=x2/2+C1x+C2y=x2/2+C1x+C2. This matches the idea that a constant forcing produces a quadratic term.
A boundary value problem differs from initial value mainly in conditions at
A One same point
B Infinite points
C Two distinct points
D No points
Boundary value problems specify values at different points, such as y(a)y(a) and y(b)y(b). Initial value problems specify all needed conditions at one point, like y(a)y(a), y′(a)y′(a).
In Sturm–Liouville form, solutions often satisfy
A Boundary conditions
B Only initial values
C Only algebraic rules
D Only parametric rules
Sturm–Liouville problems are linear second-order ODEs paired with boundary conditions. They lead to eigenvalue problems where acceptable solutions exist only for specific parameter values.
If a second-order ODE has solutions y1,y2y1,y2 with Wronskian W=Ce−∫PdxW=Ce−∫Pdx, then P(x)P(x) equals
A W′WWW′
B W′/CW′/C
C −W′W−WW′
D CW′CW′
From Abel’s formula, lnW=lnC−∫PdxlnW=lnC−∫Pdx. Differentiate: W′/W=−P(x)W′/W=−P(x). Hence P(x)=−W′/WP(x)=−W′/W, linking the coefficient of y′y′ to the Wronskian.
Using Abel’s relation, the correct expression for P(x)P(x) in terms of WW is
A W′/WW′/W
B −W′/W−W′/W
C −WW′−WW′
D W/W′W/W′
Differentiate lnW=lnC−∫PdxlnW=lnC−∫Pdx: W′W=−P(x)WW′=−P(x). Thus P(x)=−W′WP(x)=−WW′. This is a standard identity connecting Wronskian derivative to P(x)P(x).
For Clairaut-type, the singular curve is tangent to each member of
A Line family
B Circle family
C Random family
D Exponential family
Clairaut’s general solution is a family of straight lines parameterized by slope. The singular solution is their envelope curve, which touches each line tangentially, giving a geometric meaning.
If ypyp is a particular solution, then yp+Cy1yp+Cy1 is also a particular solution only when
A y1y1 is particular
B y1y1 is singular
C y1y1 is homogeneous
D y1y1 is constant
Adding a homogeneous solution to a particular solution does not change the RHS because homogeneous solutions satisfy the zero-RHS equation. So yp+yhyp+yh remains a solution of the nonhomogeneous equation.
In annihilator method, after applying annihilator, the complementary function must be adjusted to avoid
A Missing derivative
B Wrong interval
C Wrong variable
D Duplicate terms
Applying an annihilator increases order and adds solutions corresponding to the annihilator. Any overlap between these new homogeneous terms and the original CF must be handled to avoid duplicating terms in the final solution.
For RHS x2x2, a suitable annihilator is
A D3D3
B D3D3
C D2D2
D D−2D−2
A polynomial of degree 2 becomes zero after three differentiations. Thus D3D3 annihilates x2x2. This helps transform an equation with polynomial RHS into a higher-order homogeneous equation.
For a polynomial RHS degree nn, an annihilator is typically
A Dn+1Dn+1
B DnDn
C D−nD−n
D D2+n2D2+n2
Differentiating a degree nn polynomial n+1n+1 times gives zero. So Dn+1Dn+1 is an annihilator. This is used in the annihilator method for constant-coefficient equations.
For y′′+P(x)y′+Q(x)y=0y′′+P(x)y′+Q(x)y=0, if WW is known, then the ratio W′/WW′/W equals
A P(x)P(x)
B −Q(x)−Q(x)
C −P(x)−P(x)
D Q(x)Q(x)
Abel’s identity gives W=Ce−∫PdxW=Ce−∫Pdx. Taking log and differentiating gives W′/W=−P(x)W′/W=−P(x). This is a common hard MCQ link between coefficients and Wronskian.
For y′′+0⋅y′+Q(x)y=0y′′+0⋅y′+Q(x)y=0, Wronskian behavior is
A Constant
B Exponential in x
C Always zero
D Always negative
If P(x)=0P(x)=0, Abel’s formula reduces to W=CW=C. Therefore Wronskian stays constant on the interval where coefficients are continuous, supporting consistent independence across that interval.
A valid quick check for linear independence of exex and ex+1ex+1 is that their Wronskian is
A 00
B 11
C e2xe2x
D −ex−ex
Take y1=exy1=ex, y2=ex+1y2=ex+1. Then y1′=exy1′=ex, y2′=exy2′=ex. Wronskian W=y1y2′−y2y1′=ex⋅ex−(ex+1)ex=−exW=y1y2′−y2y1′=ex⋅ex−(ex+1)ex=−ex, nonzero, so independent.
For exex and ex+1ex+1, the correct Wronskian is
A 00
B −ex−ex
C 11
D e2xe2x
W=y1y2′−y2y1′=ex(ex)−(ex+1)ex=e2x−e2x−ex=−ex≠0W=y1y2′−y2y1′=ex(ex)−(ex+1)ex=e2x−e2x−ex=−ex=0. Hence functions are linearly independent on any interval.
If the Wronskian is nonzero at x0x0, then the general solution of homogeneous second-order ODE can be formed using
A Product combination
B Ratio combination
C Linear combination
D Integral combination
Nonzero Wronskian indicates two independent solutions. Any solution of the homogeneous equation can be written as C1y1+C2y2C1y1+C2y2. This is the basis for the complementary function.
For equations solvable for yy: y=f(x,p)y=f(x,p), differentiating gives an equation involving
A dp/dxdp/dx
B d2y/dx2d2y/dx2
C ∫ydx∫ydx
D Only constants
Since p=dy/dxp=dy/dx, differentiating y=f(x,p)y=f(x,p) with respect to xx introduces dp/dxdp/dx through the chain rule. This produces a relation that helps eliminate pp and solve parametrically.
The strongest “hard” use of Wronskian in theory is to guarantee
A Existence of integrals
B Exactness of equation
C Uniqueness of representation
D Separation of variables
If W≠0W=0, the two solutions are independent, forming a basis. Then every solution is uniquely expressible as C1y1+C2y2C1y1+C2y2 on that interval, crucial for theory and methods.