Degree is defined only if the DE is polynomial in derivatives. Here the derivative appears under a square root, so it is not polynomial in derivatives, hence degree is not defined.
Order of y′′+sin(y′)=0y′′+sin(y′)=0
A 1
B 3
C 2
D Not defined
The highest derivative present is y′′y′′, which is second order. Even though y′y′ appears inside sine, order depends only on the highest derivative order.
Degree of y′′+sin(y′)=0y′′+sin(y′)=0
A 1
B Not defined
C 2
D 0
Because y′y′ appears inside a sine function, the equation is not polynomial in derivatives. Therefore, degree cannot be defined, even though y′′y′′ appears linearly.
Which is linear in yy but not linear ODE
A y′+y2=xy′+y2=x
B y′+xy=lnxy′+xy=lnx
C y′+y=0y′+y=0
D y′′+y=0y′′+y=0
Linearity requires yy and derivatives to appear only to first power. In y′+y2=xy′+y2=x, the term y2y2 makes it nonlinear, even though it still contains yy.
A first-order ODE has two different particular solutions for same IVP; this violates
A Existence
B Order rule
C Degree rule
D Uniqueness
If the same initial condition leads to two different solutions, uniqueness fails. Existence only means at least one solution exists; uniqueness means exactly one solution fits given initial data.
If an IVP has no solution near x0x0, it violates
A Degree
B Existence
C Homogeneity
D Exactness
Existence means at least one solution should be possible near the initial point. If none exists, the existence condition fails, regardless of whether a solution would be unique.
For dy/dx=F(y/x)dy/dx=F(y/x), correct substitution is
A x=vyx=vy
B y=v+xy=v+x
C y=vxy=vx
D y=v2xy=v2x
Homogeneous first-order equations depend on y/xy/x. With y=vxy=vx, ratio y/x=vy/x=v and derivative becomes y′=v+xv′y′=v+xv′, transforming the ODE into a separable equation in vv.
If dy/dx=(y−x)/(y+x)dy/dx=(y−x)/(y+x), it is homogeneous because
A Contains y′y′ only
B Contains no constants
C Has only powers
D Ratio of same degree
Both numerator and denominator are first-degree homogeneous expressions in xx and yy. Dividing by xx reduces RHS to a function of y/xy/x, confirming homogeneous type.
After y=vxy=vx, which relation is always true
A y/x=vy/x=v
B x/y=vx/y=v
C y+x=vy+x=v
D xy=vxy=v
Since y=vxy=vx, dividing by xx gives y/x=vy/x=v. This replacement is the main reason homogeneous equations reduce neatly after substitution.
For linear ODE y′+P(x)y=Q(x)y′+P(x)y=Q(x), IF is
A eintQdxeintQdx
B intPdxintPdx
C eintPdxeintPdx
D intQdxintQdx
Integrating factor depends only on P(x)P(x). Multiplying by eintPdxeintPdx converts the left side into ddx(yIF)dxd(yIF), which then equals QIFQIF for easy integration.
IF for y′+(2/x)y=x3y′+(2/x)y=x3
A 1/x21/x2
B x2x2
C e2xe2x
D ex2ex2
Here P(x)=2/xP(x)=2/x. So IF=eint2/x dx=e2lnx=x2IF=eint2/xdx=e2lnx=x2. Multiplying by x2x2 makes the LHS ddx(x2y)dxd(x2y).
Which term makes an ODE nonlinear
A P(x)yP(x)y term
B Q(x)Q(x) term
C y′y′ term
D sin(y)sin(y) term
Linearity requires yy and derivatives to appear only to power 1 and not inside nonlinear functions. A term like sin(y)sin(y) violates this, so the ODE becomes nonlinear.
For exact equation Mdx+Ndy=0Mdx+Ndy=0, potential function satisfies
A phix=N,phiy=Mphix=N,phiy=M
B phix=M,phiy=Nphix=M,phiy=N
C phi=MNphi=MN
D phi=M+Nphi=M+N
Exactness means Mdx+Ndy=dphiMdx+Ndy=dphi. Therefore MM equals partial derivative of ϕϕ with respect to xx and NN equals partial derivative with respect to yy.
Check exactness for M=2xy,N=x2M=2xy,N=x2
A Exact
B Not exact
C Homogeneous only
D Linear only
Compute My=partial(2xy)/partialy=2xMy=partial(2xy)/partialy=2x. Compute Nx=partial(x2)/partialx=2xNx=partial(x2)/partialx=2x. Since My=NxMy=Nx, the equation is exact.
Solve exact: 2xy dx+x2 dy=02xydx+x2dy=0 gives
A xy2=Cxy2=C
B x2+y2=Cx2+y2=C
C x+y=Cx+y=C
D x2y=Cx2y=C
From exactness, integrate M=2xyM=2xy w.r.t xx: ϕ=x2y+g(y)ϕ=x2y+g(y). Differentiate w.r.t yy gives x2+g′(y)x2+g′(y). Match with N=x2N=x2, so g′(y)=0g′(y)=0. Hence x2y=Cx2y=C.
Order and degree of (y′′)1/3+y=0(y′′)1/3+y=0
A Order 2, degree 3
B Order 1, degree 3
C Order 2, degree undefined
D Order 3, degree 2
Highest derivative is y′′y′′, so order is 2. But it appears with fractional power 1/31/3, so equation is not polynomial in derivatives, making degree undefined.
Order and degree of (y′′)3+y=0(y′′)3+y=0
A Order 3, Degree 2
B Order 2, Degree 3
C Order 2, Degree 1
D Order 1, Degree 3
Highest derivative is y′′y′′, so order is 2. It appears as (y′′)3(y′′)3 in a polynomial form, hence degree is 3.
Which equation is separable after rearranging
A y′+xy=1y′+xy=1
B y′+y=xy′+y=x
C y′′+y=0y′′+y=0
D dy/dx=(x2+1)ydy/dx=(x2+1)y
Write dy/y=(x2+1)dxdy/y=(x2+1)dx. All yy terms go with dydy and all xx terms with dxdx, so it is separable and solvable by integration.
General solution of dy/dx=(x2+1)ydy/dx=(x2+1)y is
A y=Cex2+1y=Cex2+1
B y=C(x3/3+x)y=C(x3/3+x)
C y=Cex3/3+xy=Cex3/3+x
D y=Cex3+x2y=Cex3+x2
Separate: dy/y=(x2+1)dxdy/y=(x2+1)dx. Integrate: ln∣y∣=x3/3+x+Cln∣y∣=x3/3+x+C. Exponentiate: y=Cex3/3+xy=Cex3/3+x. This is the standard separable solution.
A solution is “singular” when it is
A Always constant
B Envelope type solution
C Always exponential
D Only linear curve
Singular solutions often form the envelope of a one-parameter family of solutions and cannot be obtained by selecting the constant in the general solution. This is a key conceptual distinction.
Clairaut equation standard form is
A y′+Py=Qy′+Py=Q
B dy/dx=F(y/x)dy/dx=F(y/x)
C Mdx+Ndy=0Mdx+Ndy=0
D y=xy′+f(y′)y=xy′+f(y′)
Clairaut equation is written as y=xy′+f(y′)y=xy′+f(y′). Putting p=y′p=y′ gives y=px+f(p)y=px+f(p), producing a family of straight lines and possibly a singular envelope.
In Clairaut, general solution family is
A Straight lines
B Circles only
C Parabolas only
D Random curves
Clairaut general solution is y=cx+f(c)y=cx+f(c) where cc is a constant slope parameter. Each value of cc gives a straight line, forming a one-parameter family.
Euler–Cauchy equation typically looks like
A y′′+y′=0y′′+y′=0
B y′+Py=Qy′+Py=Q
C x2y′′+xy′+y=0x2y′′+xy′+y=0
D Mdx+Ndy=0Mdx+Ndy=0
Euler–Cauchy equations have coefficients as powers of xx, like x2y′′x2y′′ and xy′xy′. This special structure allows trial solutions y=xmy=xm or substitution x=etx=et.
If solution of dy/dx=kydy/dx=ky passes through (0,2)(0,2), then CC is
A 0
B 2
C -2
D 1
General solution is y=Cekxy=Cekx. Apply y(0)=2y(0)=2: 2=Ce0=C2=Ce0=C. So constant C=2C=2, giving particular solution y=2ekxy=2ekx.
In Newton cooling, if T
A Negative
B Zero
C Undefined
D Positive
If T
For exact equation, after integrating MM in xx, add
A g(x)g(x)
B g(xy)g(xy)
C g(y)g(y)
D g(x+y)g(x+y)
While integrating M(x,y)M(x,y) with respect to xx, treat yy as constant. Any missing part depending only on yy is included as g(y)g(y), later fixed using NN.
When forming DE from family y=Cxy=Cx, order of resulting DE is
A 2
B 1
C 0
D 3
Family y=Cxy=Cx has one arbitrary constant. Differentiate once: y′=Cy′=C. Eliminate CC using C=y/xC=y/x or from derivative to get xy′−y=0xy′−y=0, a first-order DE.
Form DE from y=C1x+C2y=C1x+C2 gives order
A 2
B 1
C 3
D 0
The family has two arbitrary constants, so differentiate twice to eliminate them. Differentiating: y′=C1y′=C1, y′′=0y′′=0. Thus the required DE is y′′=0y′′=0, which is second order.
If Mdx+Ndy=0Mdx+Ndy=0 has My=NxMy=Nx, then solution is unique up to
A Variable xx
B Variable yy
C Degree value
D Constant CC
Integrating an exact equation gives ϕ(x,y)=Cϕ(x,y)=C. Different solutions correspond to different constant values, so the solution is determined up to an additive constant in the implicit relation.
A common mistake in degree is to use
A Highest derivative power
B Polynomial condition check
C Power of lower derivative
D Highest derivative order
Degree depends on the power of the highest order derivative, not on powers of lower derivatives. Many errors occur when students pick the largest power anywhere instead of focusing on highest derivative.
Degree of y′′′+(y′′)5=0y′′′+(y′′)5=0 is
A 5
B 1
C 3
D Not defined
Highest derivative is y′′′y′′′. It appears to power 1. Even though y′′y′′ has power 5, it is a lower order derivative, so degree is 1.
If a DE has order 2, max constants in general solution are
A One constant
B Three constants
C Zero constants
D Two constants
Typically, an order-2 differential equation needs two integrations, producing two arbitrary constants. These are then determined by two independent conditions to get a particular solution.
A first-order linear ODE can be written as
A y′+P(x)y=Q(x)y′+P(x)y=Q(x)
B y′+P(y)x=Q(x)y′+P(y)x=Q(x)
C y′2+Py=Qy′2+Py=Q
D yy′+P=0yy′+P=0
In a first-order linear ODE, coefficients depend only on the independent variable xx. yy and y′y′ appear to first power, with no products or nonlinear functions.
If a function F(x,y)F(x,y) satisfies F(tx,ty)=tnF(x,y)F(tx,ty)=tnF(x,y), it is
A Degree undefined
B Exact always
C Homogeneous degree nn
D Linear always
This is the definition of a homogeneous function of degree nn. Such functions are important in identifying homogeneous differential equations and possible integrating factors.
For homogeneous function of degree 0, F(tx,ty)F(tx,ty) equals
A tF(x,y)tF(x,y)
B F(x,y)F(x,y)
C t2F(x,y)t2F(x,y)
D 00
Degree 0 means scaling variables does not change function value. Such functions depend only on ratio y/xy/x, which is why homogeneous first-order ODEs reduce using substitution y=vxy=vx.
Slope field is most useful when
A Degree is zero
B Order is one
C Equation constant
D Exact solution hard
Slope fields help visualize solution behavior without solving explicitly. They are especially useful when an exact analytic solution is difficult or when we want qualitative understanding of growth, decay, or stability.
If y=Ce−xy=Ce−x, then it satisfies
A y′−y=0y′−y=0
B y′′+y=0y′′+y=0
C y′+y=0y′+y=0
D y′+xy=0y′+xy=0
Differentiate y=Ce−xy=Ce−x: y′=−Ce−x=−yy′=−Ce−x=−y. So y′+y=0y′+y=0. This verifies the solution by substitution, which is the standard method.
A DE’s “order” is unaffected by
A Highest derivative order
B Power on derivative
C Presence of derivative
D Type of derivative
Order depends only on the highest derivative order present, not the exponent on it. For example, (y′′)7(y′′)7 still has order 2. Power affects degree, not order.
A DE’s “degree” is unaffected by
A Highest derivative power
B Lower derivative powers
C Polynomial requirement
D Highest derivative
Degree is the power of the highest order derivative (after ensuring polynomial form). Powers of lower derivatives do not decide degree if the highest derivative appears with a different power.
If y=mx+cy=mx+c is family, the formed DE is
A y′′=0y′′=0
B y′=0y′=0
C y′′′=0y′′′=0
D y′+y=0y′+y=0
The family has two arbitrary constants mm and cc. Differentiate: y′=my′=m. Differentiate again: y′′=0y′′=0. This eliminates constants, giving the required DE.
If a solution is given implicitly, best verification is
A Convert to explicit always
B Differentiate then substitute
C Ignore constant
D Multiply by IF
For implicit relations F(x,y)=CF(x,y)=C, differentiate implicitly to find y′y′ (and higher derivatives if needed). Then substitute into the original DE to check it satisfies the equation.
Which indicates equation is not separable
A Only xx on RHS
B Only yy on RHS
C Product xyxy only
D Terms mix as x+yx+y
Separable form requires factors that can be split into pure xx and pure yy parts. A sum like x+yx+y typically prevents separation unless it can be transformed into ratio y/xy/x etc.
Which equation becomes separable after y=vxy=vx
A dy/dx=x+ydy/dx=x+y
B dy/dx=x2+ydy/dx=x2+y
C dy/dx=1+y/xdy/dx=1+y/x
D dy/dx=sin(x+y)dy/dx=sin(x+y)
RHS depends on y/xy/x, so it is homogeneous. Substituting y=vxy=vx gives v+xv′=1+vv+xv′=1+v, leading to xv′=1xv′=1, which is separable and integrable.
From dy/dx=1+y/xdy/dx=1+y/x, after y=vxy=vx, dv/dxdv/dx equals
A xx
B 1/x1/x
C −1/x−1/x
D 1+x1+x
With y=vxy=vx, dy/dx=v+xv′dy/dx=v+xv′. Substitute into equation: v+xv′=1+vv+xv′=1+v. Cancel vv to get xv′=1xv′=1, so dv/dx=1/xdv/dx=1/x.
A nonhomogeneous linear solution is sum of
A Complementary + particular
B Separable + exact
C Homogeneous + separable
D Order + degree
For linear ODE with RHS not zero, the total solution is the solution of the homogeneous part (complementary function) plus one particular solution caused by the forcing term.
Which option best describes “particular solution”
A Contains constants
B Always implicit
C Always singular
D Fits given conditions
A particular solution is the member of the general solution family that satisfies the given initial or boundary conditions. It has no arbitrary constants remaining after applying those conditions.
If an ODE is order 1, its general solution typically forms
A Two-parameter family
B Three-parameter family
C One-parameter family
D No-parameter family
First-order ODEs generally yield one integration constant, giving a one-parameter family of curves. An extra condition selects the unique particular curve from this family.
Which statement about exact equations is true
A Always separable
B Solution is ϕ=Cϕ=C
C Always linear
D Degree always 1
Exact equation means Mdx+Ndy=dϕMdx+Ndy=dϕ. Setting dϕ=0dϕ=0 gives ϕ(x,y)=Cϕ(x,y)=C. Exact equations are not necessarily separable or linear; degree may vary.
To avoid wrong degree, first ensure equation is
A Linear in xx
B Homogeneous in x,yx,y
C Exact in form
D Polynomial in derivatives
Degree exists only when the equation is polynomial in derivatives. Always rewrite the DE first to check derivatives are not inside roots, denominators, trig, or exponentials before attempting degree.