The highest derivative is y′′′y′′′, so order is 3. It appears to power 1, and the equation is polynomial in derivatives, so degree is 1.
Degree offrac1y′+y=0frac1y′+y=0 is
A Not defined
B 2
C 1
D 0
Degree is defined only when the equation is a polynomial in derivatives. Here y′y′ is in the denominator, so it is not polynomial in derivatives, hence degree is not defined.
Order ofleft(y′′right)2+y′′′=0left(y′′right)2+y′′′=0
A 2
B 1
C 3
D 4
The highest derivative present is y′′′y′′′, which is third order. Therefore, order is 3, even though y′′y′′ is squared.
The highest derivative is y′′′y′′′ and it appears to power 1. The equation is polynomial in derivatives, so degree equals 1.
Which equation has undefined degree
A sin(y′)+y=0sin(y′)+y=0
B y′′+y′=0y′′+y′=0
C y′+y=0y′+y=0
D (y′)2+y=0(y′)2+y=0
Degree requires a polynomial in derivatives. Since y′y′ appears inside a sine function, the equation is not polynomial in derivatives, so degree is not defined.
How many constants in general solution of y′′=0y′′=0
A Three constants
B One constant
C Zero constants
D Two constants
y′′=0y′′=0 is second order. Integrating once gives y′=C1y′=C1, integrating again gives y=C1x+C2y=C1x+C2. Hence two arbitrary constants appear.
General solution of dy/dx=0dy/dx=0 is
A y=Cxy=Cx
B y=x+Cy=x+C
C y=Cy=C
D y=Cx2y=Cx2
If dy/dx=0dy/dx=0, the function has zero slope everywhere, so it must be constant. Integrating gives y=Cy=C, which represents a family of horizontal lines.
If y=C1x+C2y=C1x+C2, the differential equation is
A y′′=0y′′=0
B y′+y=0y′+y=0
C y′=0y′=0
D y′′′=0y′′′=0
Differentiate: y′=C1y′=C1 and y′′=0y′′=0. Therefore, the simplest differential equation whose general solution is y=C1x+C2y=C1x+C2 is y′′=0y′′=0.
For first-order IVP, constants are found using
A Two conditions
B One condition
C Three conditions
D No conditions
A first-order ODE general solution typically contains one constant. One initial condition like y(x0)=y0y(x0)=y0 is enough to determine that constant and get a unique particular solution.
Which pair best matches boundary conditions
A y(0),y′(0)y(0),y′(0)
B y′(1),y′′(1)y′(1),y′′(1)
C y(0),y(2)y(0),y(2)
D y(1)y(1) only
Boundary conditions are specified at two different points, such as values at x=0x=0 and x=2x=2. Conditions at the same point are typical for initial value problems.
A differential equation is linear if it has no
A Products yy′yy′
B yy term
C xx term
D constant term
Linearity requires yy and its derivatives to appear only to the first power and not multiplied together. A product like yy′yy′ breaks linearity, making the equation nonlinear.
Which is linear but nonhomogeneous
A y′+2y=0y′+2y=0
B y′+y2=xy′+y2=x
C yy′+y=0yy′+y=0
D y′+2y=xy′+2y=x
It matches y′+P(x)y=Q(x)y′+P(x)y=Q(x) with Q(x)=x≠0Q(x)=x=0. That makes it linear but nonhomogeneous. Option B is homogeneous; C and D are nonlinear.
Which linear ODE has integrating factor e2xe2x
A y′+y=Q(x)y′+y=Q(x)
B y′+2y=Q(x)y′+2y=Q(x)
C y′−2y=Q(x)y′−2y=Q(x)
D 2y′+y=Q(x)2y′+y=Q(x)
For y′+P(x)y=Q(x)y′+P(x)y=Q(x), integrating factor is e∫Pdxe∫Pdx. To get e2xe2x, we need P=2P=2. That occurs in y′+2y=Q(x)y′+2y=Q(x).
Integrating factor for y′+frac1xy=xy′+frac1xy=x
A 1/x1/x
B xx
C exex
D e1/xe1/x
Here P(x)=1/xP(x)=1/x. Integrating factor is IF=e∫(1/x) dx=elnx=xIF=e∫(1/x)dx=elnx=x. Multiplying by xx turns LHS into ddx(xy)dxd(xy).
After multiplying by IF, equation becomes
A ddx(yIF)=QIFdxd(yIF)=QIF
B ddx(IF)=Qdxd(IF)=Q
C y′=QIFy′=QIF
D y=Q/IFy=Q/IF
For y′+Py=Qy′+Py=Q, multiplying by IFIF gives (yIF)′=QIF(yIF)′=QIF. This is the key step because it allows direct integration to find yIFyIF, then divide by IF.
Which equation is homogeneous type
A dy/dx=x+2ydy/dx=x+2y
B dy/dx=x2+ydy/dx=x2+y
C dy/dx=(x+2y)/(x−y)dy/dx=(x+2y)/(x−y)
D dy/dx=sin(x+y)dy/dx=sin(x+y)
(x+2y)/(x−y)(x+2y)/(x−y) is a ratio of first-degree homogeneous expressions. Dividing numerator and denominator by xx gives a function of y/xy/x, so it is homogeneous.
To reduce homogeneous DE, first write RHS as
A F(x+y)F(x+y)
B F(xy)F(xy)
C constant only
D F(y/x)F(y/x)
A first-order homogeneous DE is written as dy/dx=F(y/x)dy/dx=F(y/x). This form signals the substitution y=vxy=vx, which converts the equation into one involving vv and xx only.
If y=vxy=vx, then y′=v+xv′y′=v+xv′. This uses
A Product rule
B Quotient rule
C Chain rule
D L’Hospital rule
Since y=vxy=vx is a product of v(x)v(x) and xx, differentiating gives y′=v⋅1+x⋅v′y′=v⋅1+x⋅v′. This is exactly the product rule.
Exact equation represents differential of
A Only derivative
B Only constant
C Potential function
D Slope field
If Mdx+NdyMdx+Ndy is exact, it equals dϕ(x,y)dϕ(x,y) for some function ϕϕ. Then dϕ=0dϕ=0 implies ϕ(x,y)=Cϕ(x,y)=C, the implicit solution.
Which is exact: Mdx+Ndy=0Mdx+Ndy=0 with M=3x2M=3×2, N=4y3N=4y3
A Not exact
B Exact
C Homogeneous only
D Linear only
Compute My=∂(3×2)/∂y=0My=∂(3×2)/∂y=0 and Nx=∂(4y3)/∂x=0Nx=∂(4y3)/∂x=0. Since My=NxMy=Nx, the equation is exact.
Solution of 3x2dx+4y3dy=03x2dx+4y3dy=0 is
A x2+y3=Cx2+y3=C
B x3−y4=Cx3−y4=C
C x3+y4=Cx3+y4=C
D x+y=Cx+y=C
Integrate 3x2dx3x2dx to get x3x3 and 4y3dy4y3dy to get y4y4. Hence d(x3+y4)=0d(x3+y4)=0, giving implicit solution x3+y4=Cx3+y4=C.
If My−NxMy−Nx is nonzero, we may use
A Integrating factor
B Order reduction
C Laplace transform
D Partial fractions
When My≠NxMy=Nx, the equation is not exact. A suitable integrating factor can sometimes make it exact by adjusting MM and NN so the exactness condition holds.
Separable method requires arranging into
A y′+Py=Qy′+Py=Q
B My=NxMy=Nx
C f(y)dy=g(x)dxf(y)dy=g(x)dx
D y=vxy=vx always
A separable equation allows separation of variables so all yy terms go with dydy and all xx terms go with dxdx. Then integrate both sides to solve.
Solve: dy/dx=2ydy/dx=2y gives
A y=C+2xy=C+2x
B y=Ce2xy=Ce2x
C y=2Cexy=2Cex
D y=Cx2y=Cx2
Separate: dy/y=2dxdy/y=2dx. Integrate: ln∣y∣=2x+Cln∣y∣=2x+C. Exponentiate: y=Ce2xy=Ce2x. This is standard exponential growth with rate 2.
If dy/dx=kydy/dx=ky, doubling time depends on
A Constant kk
B Initial yy only
C xx only
D Degree only
Solution is y=Cekxy=Cekx. Doubling occurs when ekx=2ekx=2, giving x=ln2/kx=ln2/k. So the doubling time depends on kk, not on the initial amount CC.
Newton cooling solution tends to
A Infinite temperature
B Zero always
C Initial temperature
D Surrounding temperature
In dT/dt=−k(T−Ts)dT/dt=−k(T−Ts), the difference T−TsT−Ts decays exponentially to zero. Therefore the object’s temperature approaches the surrounding temperature TsTs over time.
In dT/dt=−k(T−Ts)dT/dt=−k(T−Ts), if T>TsT>Ts then dT/dtdT/dt is
A Positive
B Zero
C Undefined
D Negative
If T>TsT>Ts, then T−Ts>0T−Ts>0. With negative sign, dT/dt=−k(T−Ts)<0dT/dt=−k(T−Ts)<0. So temperature decreases, moving toward TsTs.
Meaning of orthogonal trajectories in DE context
A Curves cut at 90°
B Curves never meet
C Curves are parallel
D Curves coincide
Orthogonal trajectories are families of curves that intersect each other at right angles. In slope terms, if one family has slope mm, the orthogonal family has slope −1/m−1/m.
If a family has slope mm, orthogonal slope is
A 1/m1/m
B −1/m−1/m
C −m−m
D m2m2
Two curves intersect orthogonally when their tangents are perpendicular. The slopes must satisfy m1m2=−1m1m2=−1, so the orthogonal slope to mm is −1/m−1/m.
Which shows “particular integral” idea in linear ODE
A Only homogeneous part
B Only constants part
C A specific forcing response
D Only slope field
In a nonhomogeneous linear ODE, the total solution is complementary (homogeneous) plus a particular solution due to the forcing term Q(x)Q(x). That forced part is the particular integral idea.
Correct statement about homogeneous linear solutions
A Always two constants
B No constants
C Infinite constants
D Only one arbitrary constant
A first-order homogeneous linear ODE y′+P(x)y=0y′+P(x)y=0 yields one integration constant. Higher order would have more constants, but first-order typically gives exactly one.
Which DE is reducible to linear by substitution
A Bernoulli type
B Exact type
C Only separable
D Only constant slope
Bernoulli equation y′+Py=Qyny′+Py=Qyn becomes linear after substitution v=y1−nv=y1−n. This converts the nonlinear power term into a linear equation in vv.
For Bernoulli y′+Py=Qyny′+Py=Qyn, substitution is
A v=ynv=yn
B v=y1−nv=y1−n
C v=xyv=xy
D v=y/xv=y/x
Dividing by ynyn produces y1−ny1−n. Let v=y1−nv=y1−n; then dv/dx=(1−n)y−ny′dv/dx=(1−n)y−ny′, converting the Bernoulli equation into a linear first-order ODE in vv.
Degree of (y′′′)2+y=0(y′′′)2+y=0
A 3
B 1
C 2
D Not defined
The highest derivative is y′′′y′′′. It appears squared as (y′′′)2(y′′′)2. Since the equation is polynomial in derivatives, degree equals 2.
Order of (y′′′)2+y=0(y′′′)2+y=0
A 3
B 2
C 1
D 6
Order is the highest derivative order present, which is y′′′y′′′. The power does not affect order. So order is 3.
Which is an ODE, not PDE
A Uses partial derivative
B Uses two variables
C Uses ordinary derivative
D Uses gradient symbol
ODEs involve ordinary derivatives like dy/dxdy/dx with one independent variable. PDEs involve partial derivatives like ∂y/∂x∂y/∂x, because the function depends on multiple variables.
A solution curve “passing through (1,2)” means
A y′(1)=2y′(1)=2
B y′′(1)=2y′′(1)=2
C x(2)=1x(2)=1
D y(1)=2y(1)=2
The point (1,2) lies on the solution curve means when x=1x=1, the function value is y=2y=2. This is an initial condition type statement used to find constants.
If y=Cekxy=Cekx, then y′/yy′/y equals
A kk
B CC
C xx
D ekxekx
Differentiate y=Cekxy=Cekx: y′=kCekx=kyy′=kCekx=ky. So y′/y=ky′/y=k, a constant. This property is why exponential functions appear in growth/decay models.
Equation y′+y=0y′+y=0 has solution family
A y=Cexy=Cex
B y=Cxy=Cx
C y=Ce−xy=Ce−x
D y=C+xy=C+x
Solve y′+y=0y′+y=0 by separation: dy/y=−dxdy/y=−dx. Integrate: ln∣y∣=−x+Cln∣y∣=−x+C. Thus y=Ce−xy=Ce−x, a family depending on constant CC.
If y′+y=0y′+y=0 and y(0)=5y(0)=5, then CC is
A 0
B 5
C -5
D 1
General solution is y=Ce−xy=Ce−x. Apply y(0)=5y(0)=5: 5=Ce0=C5=Ce0=C. So C=5C=5, giving the particular solution y=5e−xy=5e−x.
If an equation is homogeneous, numerator and denominator are
A Different degree functions
B Constant functions
C Same degree functions
D Trig functions
A common homogeneous type is dy/dx=F(x,y)dy/dx=F(x,y) where FF is a ratio of homogeneous functions of the same degree. Then it can be rewritten as a function of y/xy/x.
If F(tx,ty)=F(x,y)F(tx,ty)=F(x,y), function is
A Homogeneous degree 0
B Homogeneous degree 1
C Not homogeneous
D Constant only
If scaling both variables by tt does not change the value, the function depends only on ratio y/xy/x. Such functions are homogeneous of degree zero, used in homogeneous first-order ODEs.
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=x/ydy/dx=x/y
Rearrange dy/dx=x/ydy/dx=x/y to y dy=x dxydy=xdx. Variables separate cleanly. Integrating gives y2/2=x2/2+Cy2/2=x2/2+C, so it is separable.
For exact solution, after integrating MM w.r.t xx, add
A Function of xx
B Constant zero only
C Function of yy
D Function of xyxy
When integrating M(x,y)M(x,y) with respect to xx, treat yy constant. Any missing dependence on yy is included as an unknown function g(y)g(y), later determined using NN.
If solution is ϕ(x,y)=Cϕ(x,y)=C, then dϕdϕ equals
A 0
B 1
C CC
D x+yx+y
Differentiating ϕ(x,y)=Cϕ(x,y)=C gives dϕ=0dϕ=0 because CC is constant. Exact equations use this idea: Mdx+Ndy=dϕ=0Mdx+Ndy=dϕ=0, leading to the implicit solution.
A slope field at a point shows
A Local tangent direction
B Exactness condition
C Degree number
D Constant value
At a point (x,y)(x,y), the slope field draws a small segment with slope dy/dxdy/dx. This indicates the tangent direction of the solution curve passing through that point.
Which is a correct “medium” check for order
A Count all powers
B Use constant term
C Ignore derivative powers
D Use RHS only
Order depends only on the highest derivative order, not on its power. For example, (y′′)5(y′′)5 still has order 2. This prevents confusion between order and degree.
Lagrange equation preview often involves
A y′+Py=Qy′+Py=Q
B Mdx+Ndy=0Mdx+Ndy=0
C y=xf(p)+g(p)y=xf(p)+g(p)
D dy/dx=f(x)dy/dx=f(x)
Lagrange’s equation is a generalization of Clairaut form and is often written using p=dy/dxp=dy/dx as y=xf(p)+g(p)y=xf(p)+g(p). It may lead to parametric solutions and special cases.
Clairaut equation family solutions are usually
A Parabolas family
B Straight lines family
C Circles family
D Sine curves family
Clairaut equation y=px+f(p)y=px+f(p) produces a family of straight lines parameterized by pp. The singular solution, when present, forms the envelope of these lines.
Best meaning of “formation of DE” from given family
A Eliminate constants
B Add more constants
C Increase degree
D Replace variables
To form a differential equation from a family of curves, differentiate enough times to introduce derivatives, then eliminate arbitrary constants using the original equation and derivatives. The result is the required DE.