Which part decides the order of a differential equation
A Highest derivative order
B Highest power only
C Number of variables
D Constant term only
The order depends on the highest order derivative present, like y′y′, y′′y′′, or y′′′y′′′. Powers of derivatives do not change order; only the derivative’s order matters.
Which situation makes degree not defined
A Only first derivative
B No constant term
C Non-polynomial derivatives
D Linear in yy
Degree is defined only if the equation is a polynomial in derivatives. If derivatives occur inside roots, fractions, exponentials, or trigonometric functions, degree cannot be defined.
Order of y′′+xy′=0y′′+xy′=0 is
A 1
B 3
C 0
D 2
The highest derivative present is y′′y′′, which is the second derivative. Therefore, the order is 2 regardless of other lower derivatives appearing.
Degree of y′′+xy′=0y′′+xy′=0 is
A 1
B 2
C 0
D Not defined
The highest derivative is y′′y′′ and it appears to the first power. The equation is polynomial in derivatives, so degree is 1.
Order of y′+(y′)2=0y′+(y′)2=0
A 2
B 1
C 0
D 3
The highest derivative is y′y′ (first derivative). Squaring it does not change the order, so the equation remains first order.
Degree of y′+(y′)2=0y′+(y′)2=0
A 1
B 0
C 2
D Not defined
The equation is polynomial in y′y′. The highest power of the highest derivative y′y′ is 2, so the degree is 2.
Which is a first-order linear ODE
A y′+2y=xy′+2y=x
B (y′)2+y=0(y′)2+y=0
C yy′+x=0yy′+x=0
D y′+y2=0y′+y2=0
First-order linear ODE has the form y′+P(x)y=Q(x)y′+P(x)y=Q(x). Option A matches this exactly, while others involve products or powers of yy or y′y′, making them nonlinear.
Which equation is nonlinear due to product
A y′+xy=0y′+xy=0
B y′−y=xy′−y=x
C y′+y=1y′+y=1
D yy′=xyy′=x
Linearity requires yy and its derivatives to appear only to power 1 and not multiplied together. In yy′=xyy′=x, yy multiplies y′y′, so it is nonlinear.
General solution usually contains
A Only one point
B No derivatives
C Arbitrary constants
D Fixed numbers
General solution represents a family of solutions and includes arbitrary constants obtained during integration. These constants are later determined using initial or boundary conditions.
Particular solution is obtained by
A Using given conditions
B Removing constants
C Increasing order
D Changing variables
A particular solution is found by applying conditions like y(x0)=y0y(x0)=y0. These conditions help determine the arbitrary constants in the general solution.
How many constants in a 3rd order ODE solution
A One constant
B Three constants
C Two constants
D Zero constants
The number of arbitrary constants in the general solution typically equals the order. A third-order ODE generally requires three integrations, producing three constants.
Which set shows a boundary value problem
A y(0)=1y(0)=1 only
B y′(0)=2y′(0)=2 only
C y(0)=1,y(2)=3y(0)=1,y(2)=3
D No conditions
Boundary value problems specify conditions at two different points of the independent variable. Here values are given at x=0x=0 and x=2x=2, so it is a BVP.
Which set shows an initial value problem
A y(0)=2y(0)=2
B y(1)=3,y(4)=0y(1)=3,y(4)=0
C y(2)=1,y(5)=7y(2)=1,y(5)=7
D y(0)=2,y(3)=4y(0)=2,y(3)=4
An initial value problem specifies the value of the solution (and possibly derivatives) at a single starting point. One point condition like y(0)=2y(0)=2 is typical for first-order IVP.
A solution check is done by
A Comparing degrees
B Changing the equation
C Ignoring derivatives
D Substituting in equation
To verify a solution, compute needed derivatives and substitute them into the original differential equation. If both sides match identically, the function is a correct solution.
Implicit solution is commonly written as
A y=f(x)y=f(x)
B x=g(y)x=g(y) only
C F(x,y)=CF(x,y)=C
D y=mx+cy=mx+c
An implicit solution keeps xx and yy together in one relation like F(x,y)=CF(x,y)=C. Many exact and separable equations naturally lead to such forms.
Explicit solution is commonly written as
A y=f(x)y=f(x)
B F(x,y)=CF(x,y)=C
C Mdx+Ndy=0Mdx+Ndy=0
D y′′=0y′′=0
An explicit solution expresses yy directly as a function of xx. Sometimes an implicit relation can be rearranged to explicit form, but not always.
A first-order homogeneous DE can be written as
A dy/dx=F(x+y)dy/dx=F(x+y)
B dy/dx=F(y/x)dy/dx=F(y/x)
C dy/dx=F(xy)dy/dx=F(xy)
D dy/dx=F(x)dy/dx=F(x)
A first-order homogeneous equation in xx and yy depends only on the ratio y/xy/x. This allows the substitution y=vxy=vx to reduce it to a separable equation.
Best substitution for homogeneous DE is
A y=v+xy=v+x
B y=v/xy=v/x
C y=vxy=vx
D x=vy2x=vy2
For equations depending on y/xy/x, setting y=vxy=vx converts dy/dxdy/dx into v+x dv/dxv+xdv/dx. Then the equation becomes separable in vv and xx.
If y=vxy=vx, then y/xy/x equals
A xx
B yy
C 1/v1/v
D vv
Dividing y=vxy=vx by xx gives y/x=vy/x=v. This is why the substitution works: it replaces the ratio y/xy/x by a single variable vv.
If y=vxy=vx, then dy/dxdy/dx equals
A v+xv′v+xv′
B v−xv′v−xv′
C x+vv′x+vv′
D v′/xv′/x
Differentiate y=vxy=vx using the product rule: dy/dx=v⋅1+x⋅dv/dxdy/dx=v⋅1+x⋅dv/dx. This transforms homogeneous equations into separable form.
A separable DE can be arranged as
A y′+Py=Qy′+Py=Q
B Mdx+Ndy=0Mdx+Ndy=0 always
C f(y)dy=g(x)dxf(y)dy=g(x)dx
D y′′+y=0y′′+y=0
In separable equations, all terms involving yy are moved with dydy and all terms involving xx with dxdx. Then both sides can be integrated easily.
Which equation is separable
A dy/dx=xydy/dx=xy
B y′+xy=1y′+xy=1
C y′′+y=xy′′+y=x
D y′+y=xyy′+y=xy
Rewrite dy/dx=xydy/dx=xy as dy/y=x dxdy/y=xdx. Variables separate clearly, allowing integration to give ln∣y∣=x2/2+Cln∣y∣=x2/2+C.
In linear ODE, P(x)P(x) is coefficient of
A y′y′ term
B yy term
C Constant term
D x2x2 term
In y′+P(x)y=Q(x)y′+P(x)y=Q(x), P(x)P(x) multiplies yy. It determines the integrating factor e∫P(x) dxe∫P(x)dx, making it central to solving linear first-order ODEs.
Integrating factor for y′+P(x)y=Q(x)y′+P(x)y=Q(x)
A ∫Pdx∫Pdx
B ∫Qdx∫Qdx
C e∫Qdxe∫Qdx
D e∫Pdxe∫Pdx
The integrating factor is IF=e∫P(x) dxIF=e∫P(x)dx. Multiplying the equation by IF turns the left side into d(y⋅IF)/dxd(y⋅IF)/dx, which can be integrated directly.
After multiplying by IF, LHS becomes
A Derivative of yIFyIF
B Derivative of IFIF
C Derivative of QQ
D Derivative of PP
For linear ODE, multiplying by the integrating factor makes the left side equal to ddx(y⋅IF)dxd(y⋅IF). This simplifies solving because it reduces the problem to integration.
Which is a homogeneous linear equation
A y′+3y=2y′+3y=2
B y′+y2=0y′+y2=0
C y′+3y=0y′+3y=0
D yy′+y=0yy′+y=0
A linear equation y′+P(x)y=Q(x)y′+P(x)y=Q(x) is homogeneous when Q(x)=0Q(x)=0. Option A matches this, while B has a nonzero RHS and others are nonlinear.
Which is a nonhomogeneous linear equation
A y′+2y=xy′+2y=x
B y′+2y=0y′+2y=0
C y′+y2=1y′+y2=1
D yy′+x=0yy′+x=0
Nonhomogeneous linear first-order ODE has Q(x)≠0Q(x)=0. In y′+2y=xy′+2y=x, the RHS is xx, so it is linear but nonhomogeneous.
Exact differential equation is written as
A y′+Py=Qy′+Py=Q
B dy/dx=F(y/x)dy/dx=F(y/x)
C f(y)dy=g(x)dxf(y)dy=g(x)dx
D Mdx+Ndy=0Mdx+Ndy=0
Exact equations are expressed as M(x,y) dx+N(x,y) dy=0M(x,y)dx+N(x,y)dy=0. If exact, this form represents a total differential dϕ(x,y)=0dϕ(x,y)=0.
Exactness condition is
A Mx=NyMx=Ny
B M=NM=N
C My=NxMy=Nx
D MN=1MN=1
An equation Mdx+Ndy=0Mdx+Ndy=0 is exact when ∂M/∂y=∂N/∂x∂M/∂y=∂N/∂x. Then a potential function ϕ(x,y)ϕ(x,y) exists such that dϕ=Mdx+Ndydϕ=Mdx+Ndy.
If exact, the solution is
A ϕ(x,y)=Cϕ(x,y)=C
B y=Cxy=Cx
C y=mx+cy=mx+c
D M=N=CM=N=C
For exact equations, Mdx+Ndy=dϕ=0Mdx+Ndy=dϕ=0. Integrating gives a potential function ϕ(x,y)ϕ(x,y), and the implicit solution is ϕ(x,y)=Cϕ(x,y)=C.
Meaning of total differential is
A ϕ=ϕx+ϕyϕ=ϕx+ϕy
B dϕ=ϕxdx+ϕydydϕ=ϕxdx+ϕydy
C dϕ=ϕx+ϕydϕ=ϕx+ϕy
D dϕ=ϕxϕydϕ=ϕxϕy
A total differential combines changes in both variables: dϕ=(∂ϕ/∂x) dx+(∂ϕ/∂y) dydϕ=(∂ϕ/∂x)dx+(∂ϕ/∂y)dy. Exact equations match this structure.
An integrating factor is used to make equation
A Exact
B Second order
C Polynomial only
D Constant slope
When My≠NxMy=Nx, the equation is not exact. Multiplying by a suitable integrating factor can make it exact, allowing solution through a potential function ϕ(x,y)ϕ(x,y).
If dy/dx=f(x)dy/dx=f(x), then solution is
A y=f(x)+Cy=f(x)+C
B y=∫fdx+Cy=∫fdx+C
C y=1/f(x)+Cy=1/f(x)+C
D y=∫fdyy=∫fdy
When the derivative depends only on xx, integrate both sides with respect to xx. This directly gives y=∫f(x) dx+Cy=∫f(x)dx+C, one of the simplest ODE forms.
If dy/dx=g(y)dy/dx=g(y), separate as
A dx=g(y)dydx=g(y)dy
B g(y)dy=dxg(y)dy=dx
C dy=dx/g(y)dy=dx/g(y)
D dy/g(y)=dxdy/g(y)=dx
Rewrite dy/dx=g(y)dy/dx=g(y) as dy/g(y)=dxdy/g(y)=dx. Then integrate both sides. This method works because variables are separated into yy-side and xx-side.
Basic growth/decay model is
A dy/dt=kydy/dt=ky
B dy/dt=k+tdy/dt=k+t
C dy/dt=ktdy/dt=kt
D dy/dt=y+kdy/dt=y+k
In exponential growth/decay, the rate of change is proportional to the current amount: dy/dt=kydy/dt=ky. Positive kk gives growth, negative kk gives decay in many applications.
Newton’s cooling law uses form
A dT/dt=kT2dT/dt=kT2
B dT/dt=k+tdT/dt=k+t
C dT/dt=−k(T−Ts)dT/dt=−k(T−Ts)
D dT/dt=T/kdT/dt=T/k
Newton’s law of cooling states temperature changes proportional to difference from surrounding temperature TsTs. The negative sign shows temperature approaches TsTs over time.
In dT/dt=−k(T−Ts)dT/dt=−k(T−Ts), TsTs is
A Surrounding temperature
B Initial temperature
C Final constant kk
D Cooling rate
TsTs represents the ambient or surrounding temperature. The model says the object’s temperature moves toward TsTs, and the rate depends on the difference T−TsT−Ts.
Slope field gives information about
A Degree value
B Solution directions
C Order only
D Constants count
A slope field shows small line segments representing dy/dxdy/dx at many points. These segments guide the shape of solution curves and help visualize behavior without solving.
Orthogonal trajectories mean curves meet at
A Same slope
B Parallel lines
C Random angles
D Right angles
Two families are orthogonal trajectories if every curve from one family intersects each curve of the other family at 90°. This concept links differential equations to geometry of curve families.
In linear ODE, solution uses formula
A yIF=∫QIF dx+CyIF=∫QIFdx+C
B y=∫Pdx+Cy=∫Pdx+C
C y=∫Qdx+Cy=∫Qdx+C
D y=∫IFdx+Cy=∫IFdx+C
For y′+Py=Qy′+Py=Q, multiply by IFIF to get ddx(yIF)=QIFdxd(yIF)=QIF. Integrate both sides: yIF=∫QIF dx+CyIF=∫QIFdx+C, then divide by IFIF.
Bernoulli equation is linear when nn equals
A 2 only
B 3 only
C 0 or 1
D Any real nn
Bernoulli form is y′+Py=Qyny′+Py=Qyn. If n=0n=0, it becomes linear with constant RHS. If n=1n=1, it becomes linear directly. Otherwise, substitution is needed.
Clairaut equation typically has form
A y=px+f(p)y=px+f(p)
B y′+Py=Qy′+Py=Q
C Mdx+Ndy=0Mdx+Ndy=0
D dy/dx=F(y/x)dy/dx=F(y/x)
Clairaut equations are of the form y=x y′+f(y′)y=xy′+f(y′), often written using p=y′p=y′. They produce a family of straight lines and may have a singular solution envelope.
Euler–Cauchy equation has typical coefficients like
A Only constants
B Only trig terms
C Only exponentials
D Powers of xx
In Euler–Cauchy equations, coefficients of derivatives involve powers of xx, like x2y′′+xy′+y=0x2y′′+xy′+y=0. Such structure allows substitution x=etx=et or trial y=xmy=xm.
Second-order linear ODE contains
A y′′′y′′′ term
B y′′y′′ term
C Only y′y′
D No derivatives
A second-order ODE includes the second derivative y′′y′′ as the highest derivative. It may also include y′y′ and yy, but y′′y′′ must appear for order 2.
If a DE models decay, constant kk in dy/dt=kydy/dt=ky is
A Negative
B Positive
C Zero always
D Undefined
In dy/dt=kydy/dt=ky, if k<0k<0 the solution decreases exponentially over time, representing decay. If k>0k>0, it increases, representing growth.
A homogeneous integrating factor case often uses
A Constant only
B Function of x+yx+y
C Function of y/xy/x
D Function of x2+y2x2+y2
For some non-exact equations where MM and NN are homogeneous of same degree, an integrating factor can be a function of y/xy/x. This can turn the equation into an exact one.
Exact equation solution step starts by finding
A Potential function
B Integrating factor always
C Degree of DE
D Order of DE
Once exactness is confirmed, we look for ϕ(x,y)ϕ(x,y) such that ϕx=Mϕx=M and ϕy=Nϕy=N. Integrating and combining terms gives ϕ(x,y)=Cϕ(x,y)=C as the solution.
If ϕx=Mϕx=M, then ϕϕ is found by
A Integrating MM w.r.t yy
B Differentiating MM
C Squaring MM
D Integrating MM w.r.t xx
From ϕx=Mϕx=M, integrate MM with respect to xx treating yy as constant. Then add a function of yy to account for missing yy-dependence.
Equation dy/dx=y/xdy/dx=y/x is best classified as
A Exact second-order
B Linear second-order
C Homogeneous first-order
D Non-ODE type
dy/dx=y/xdy/dx=y/x depends on y/xy/x, which is the hallmark of homogeneous first-order equations. Using y=vxy=vx reduces it to separable form and it can be solved easily.
Meaning of “solution family” in ODE is
A Only one curve
B Many curves from constants
C Only straight lines
D Only closed curves
A solution family arises from the general solution containing arbitrary constants. Different constant values produce different curves. Initial or boundary conditions select one specific curve from the family.