The highest derivative is y′′y′′, so order is 2. It appears as (y′′)3(y′′)3 and the equation is polynomial in derivatives, so degree equals 3.
Degree ofsqrty′+y=0sqrty′+y=0 is
A 1
B Not defined
C 2
D 1/2
Degree is defined only when the equation is polynomial in derivatives. Here y′y′ is inside a square root, so it is not polynomial in derivatives, hence degree is not defined.
Order ofleft(fracd2ydx2right)2+fracdydx=0left(fracd2ydx2right)2+fracdydx=0
A 2
B 1
C 3
D 4
The highest derivative present is d2y/dx2d2y/dx2, so the order is 2. Powers do not affect order, only the derivative order matters.
The equation is polynomial in derivatives and the highest derivative is y′′y′′. The highest power of y′′y′′ is 2, so degree is 2.
How many conditions needed for unique solution of 2nd order ODE
A One condition
B Two conditions
C Three conditions
D No conditions
A second-order ODE typically gives two arbitrary constants in its general solution. Two independent conditions, like y(x0)y(x0) and y′(x0)y′(x0), are needed to determine them uniquely.
Which pair is valid for a second-order IVP
A y(0),y(2)y(0),y(2)
B y(1),y(3)y(1),y(3)
C y′(1),y′(4)y′(1),y′(4)
D y(0),y′(0)y(0),y′(0)
An initial value problem provides conditions at the same initial point. For second order, giving y(x0)y(x0) and y′(x0)y′(x0) at one point is standard for determining both constants.
Which indicates a family of curves in solution
A No constant present
B Only one point used
C Arbitrary constant present
D Only derivative used
A general solution contains arbitrary constants and represents many possible solution curves. Each choice of constant produces a different curve, and conditions select one curve from the family.
Which statement best fits “singular solution” idea
A Not from general family
B Always constant solution
C Always linear solution
D Needs no condition
A singular solution is a special solution that cannot be obtained by choosing constants in the general solution. It often acts as an envelope of the family of solutions, introduced at basic level.
In linear ODE, integrating factor is chosen to make
A RHS equal zero
B LHS exact derivative
C Order increase
D Degree decrease
Multiplying by integrating factor converts y′+P(x)yy′+P(x)y into ddx(y IF)dxd(yIF). This makes the left side a single derivative, allowing direct integration to solve the equation.
Integrating factor for y′−frac2xy=xy′−frac2xy=x
A x2x2
B exex
C x−2x−2
D e−xe−x
With P(x)=−2/xP(x)=−2/x, integrating factor is e∫−2/x dx=e−2lnx=x−2e∫−2/xdx=e−2lnx=x−2. Multiplying by it makes the LHS ddx(yx−2)dxd(yx−2).
Which equation is homogeneous in x,yx,y
A dy/dx=x+ydy/dx=x+y
B dy/dx=x2+ydy/dx=x2+y
C dy/dx=x−y2dy/dx=x−y2
D dy/dx=(x+y)/xdy/dx=(x+y)/x
(x+y)/x=1+y/x(x+y)/x=1+y/x, which depends only on y/xy/x. This matches the form dy/dx=F(y/x)dy/dx=F(y/x), so it is homogeneous and solvable via y=vxy=vx.
After substitution y=vxy=vx, equation becomes in
A yy and xx
B vv and xx
C vv and yy
D Only constants
Substitution y=vxy=vx gives dy/dx=v+x dv/dxdy/dx=v+xdv/dx. This replaces yy by vv and reduces the equation to one involving vv and xx, often separable.
Which step makes separable after y=vxy=vx
A Replace x/yx/y by vv
B Replace xx by yy
C Replace y/xy/x by vv
D Replace yy by constant
In homogeneous equations, expressions like y/xy/x appear. Since y=vxy=vx, we get y/x=vy/x=v. That single change simplifies the function and helps separate variables in vv and xx.
Identify linear ODE among these
A y′+y2=xy′+y2=x
B y′+(sinx)y=exy′+(sinx)y=ex
C yy′+x=0yy′+x=0
D (y′)2+y=0(y′)2+y=0
A first-order linear ODE has y′y′ and yy only to first power with coefficient depending on xx. Option A fits y′+P(x)y=Q(x)y′+P(x)y=Q(x) with P=sinxP=sinx, Q=exQ=ex.
Exact equation must satisfy condition
A My=NxMy=Nx
B Mx=NyMx=Ny
C M=NM=N
D MN=1MN=1
For Mdx+Ndy=0Mdx+Ndy=0 to be exact, mixed partial derivatives must match: ∂M/∂y=∂N/∂x∂M/∂y=∂N/∂x. Then a potential function ϕϕ exists with dϕ=Mdx+Ndydϕ=Mdx+Ndy.
For equation Mdx+Ndy=0Mdx+Ndy=0, MM and NN depend on
A xx only
B yy only
C xx and yy
D constants only
In exact differential equations, MM and NN are functions of both xx and yy. Their partial derivatives are compared to test exactness before finding the potential function.
If MyneqNxMyneqNx, equation is
A Always separable
B Not exact
C Always linear
D Always homogeneous
Exactness requires My=NxMy=Nx. If this condition fails, the equation is not exact in current form. Sometimes an integrating factor can be used to convert it into an exact equation.
Integrating factor depends only on xx when
A M=NM=N
B My=NxMy=Nx already
C Equation is separable
D Expression becomes function of xx
A common test uses My−NxNNMy−Nx. If this ratio depends only on xx, then an integrating factor μ(x)μ(x) exists. Multiplying by μ(x)μ(x) can make the equation exact.
Integrating factor depends only on yy when
A M=NM=N
B Order is 2
C Expression becomes function of yy
D Degree is 1
If Nx−MyMMNx−My simplifies to a function of yy only, then integrating factor μ(y)μ(y) can be used. After multiplying, the equation often becomes exact and solvable.
Which is a separable form directly
A dy/dx=x2dy/dx=x2
B y′+xy=1y′+xy=1
C yy′+x=0yy′+x=0
D y′+y2=0y′+y2=0
When dy/dxdy/dx depends only on xx, integrate directly: y=∫x2dx+Cy=∫x2dx+C. This is a simplest separable-type case with no yy terms to rearrange.
Solution of dy/dx=3x2dy/dx=3×2 is
A y=3×2+Cy=3×2+C
B y=x3+Cy=x3+C
C y=x2+Cy=x2+C
D y=3x+Cy=3x+C
Integrate both sides: y=∫3x2dx=x3+Cy=∫3x2dx=x3+C. Differentiating x3+Cx3+C gives 3x23x2, confirming the solution.
A differential equation from motion often relates
A Area and volume
B Position and velocity
C Angles and sides
D Matrix and vector
Many physical laws connect a quantity to its rate of change. In motion, velocity is the derivative of position with respect to time, so differential equations naturally model such relationships.
Which indicates uniqueness idea in IVP
A Many solutions always
B One solution for given data
C No solution possible
D Only constant solutions
Uniqueness means that for a given differential equation and initial condition, there is only one solution passing through that initial point. Basic conditions on functions often guarantee this behavior.
Which indicates existence idea in IVP
A At least one solution
B Exactly two solutions
C No solutions ever
D Only polynomial solutions
Existence means a solution actually exists for the given equation and initial condition in some interval. Even if unique, existence must hold first, otherwise no solution is possible.
Which is a Bernoulli equation
A y′+y=xy′+y=x
B dy/dx=x2dy/dx=x2
C y′+y=xy2y′+y=xy2
D Mdx+Ndy=0Mdx+Ndy=0
Bernoulli equation has form y′+Py=Qyny′+Py=Qyn with n≠0,1n=0,1. Here P=1P=1, Q=xQ=x, and n=2n=2, so it is Bernoulli type.
In Bernoulli, dividing by ynyn helps to
A Remove xx terms
B Increase order
C Make exact always
D Make linear in y1−ny1−n
Dividing by ynyn produces terms like y1−ny1−n. Substitution v=y1−nv=y1−n converts the Bernoulli equation into a linear first-order equation in vv.
Orthogonal trajectories require slopes satisfy
A m1+m2=0m1+m2=0
B m1m2=−1m1m2=−1
C m1=m2m1=m2
D m1/m2=1m1/m2=1
For two curves to intersect at right angles, their tangent slopes at intersection satisfy m1m2=−1m1m2=−1 (negative reciprocal). This is the key condition used in orthogonal trajectory problems.
“Envelope” of a curve family means
A Curve cutting all
B Curve always parallel
C Curve tangent to family
D Curve never meets
The envelope is a curve that touches (is tangent to) every member of a family of curves at some point. Singular solutions in certain ODEs may represent such envelopes.
Which equation is NOT exact
A (2x)dx+(2y)dy=0(2x)dx+(2y)dy=0
B y dx+x dy=0ydx+xdy=0
C (x+y)dx+(x−y)dy=0(x+y)dx+(x−y)dy=0
D (xy)dx+(x+y)dy=0(xy)dx+(x+y)dy=0
For exactness, My=NxMy=Nx. Here M=xy⇒My=xM=xy⇒My=x, N=x+y⇒Nx=1N=x+y⇒Nx=1. Since x≠1x=1 in general, the condition fails, so it’s not exact.
Solution of (2x)dx+(2y)dy=0(2x)dx+(2y)dy=0 is
A x+y=Cx+y=C
B x2+y2=Cx2+y2=C
C xy=Cxy=C
D x2−y2=Cx2−y2=C
Integrate: 2xdx2xdx gives x2x2 and 2ydy2ydy gives y2y2. So d(x2+y2)=0d(x2+y2)=0 implies x2+y2=Cx2+y2=C, the implicit solution.
Which ODE is homogeneous linear
A y′+(1/x)y=0y′+(1/x)y=0
B y′+(1/x)y=1y′+(1/x)y=1
C y′+y2=0y′+y2=0
D yy′+y=0yy′+y=0
It matches y′+P(x)y=0y′+P(x)y=0 with P(x)=1/xP(x)=1/x. Since RHS is zero, it is homogeneous linear. Others are nonhomogeneous or nonlinear.
Solve quickly: y′+0⋅y=sinxy′+0⋅y=sinx
A y=cosx+Cy=cosx+C
B y=sinx+Cy=sinx+C
C y=−cosx+Cy=−cosx+C
D y=tanx+Cy=tanx+C
The equation is y′=sinxy′=sinx. Integrate: y=∫sinxdx=−cosx+Cy=∫sinxdx=−cosx+C. This is a direct integration case of dy/dx=f(x)dy/dx=f(x).
Which form is reducible to separable using substitution
A Second-order linear
B Homogeneous DE
C Constant coefficient only
D Pure exact only
Homogeneous first-order DEs dy/dx=F(y/x)dy/dx=F(y/x) become separable after substitution y=vxy=vx. The substitution reduces the equation to a separable form in vv and xx.
In y′+Py=Qy′+Py=Q, after solution formula, yy equals
A (∫QIFdx+C)/IF(∫QIFdx+C)/IF
B (∫PIFdx+C)/IF(∫PIFdx+C)/IF
C (∫IFdx+C)/Q(∫IFdx+C)/Q
D IF/(∫Qdx)IF/(∫Qdx)
Multiply by IFIF: ddx(yIF)=QIFdxd(yIF)=QIF. Integrate: yIF=∫QIFdx+CyIF=∫QIFdx+C. Divide by IFIF to get the final expression for yy.
Which indicates “polynomial in derivatives”
A Derivatives inside sin
B Derivatives in denominator
C Derivatives only with powers
D Derivatives under root
Polynomial in derivatives means derivatives appear as algebraic terms with nonnegative integer powers, without being inside trig/exponential, denominators, or radicals. Only then degree can be defined.
Order of y′′′+(y′′)2=0y′′′+(y′′)2=0
A 2
B 1
C 4
D 3
The highest derivative is y′′′y′′′, which is third order. Even though y′′y′′ is squared, the presence of y′′′y′′′ makes the order 3.
Degree of y′′′+(y′′)2=0y′′′+(y′′)2=0
A 2
B 1
C 3
D Not defined
Highest derivative is y′′′y′′′, and it appears to power 1. The equation is polynomial in derivatives, so degree equals 1, regardless of lower derivative powers.
Which equation is not polynomial in derivatives
A (y′)2+y=0(y′)2+y=0
B ey′+y=0ey′+y=0
C y′′+y′=0y′′+y′=0
D y′+y=0y′+y=0
Because derivative y′y′ appears inside an exponential, the equation is not polynomial in derivatives. Therefore, concepts like degree are not applicable to such equations.
A solution satisfying DE but not initial data is
A Particular solution
B Not IVP solution
C Unique solution
D Boundary solution
It may solve the differential equation, but an initial value problem requires satisfying both the DE and the given initial condition. If it fails the condition, it is not the required IVP solution.
For exact equation, cross-check is done by
A Comparing mixed partials
B Comparing degrees
C Comparing constants
D Comparing slopes only
In Mdx+Ndy=0Mdx+Ndy=0, compute MyMy and NxNx. If they are equal, the equation is exact. This is the main diagnostic check before integrating.
If M=yM=y and N=xN=x, equation Mdx+Ndy=0Mdx+Ndy=0 is
A Not exact
B Separable
C Exact
D Second order
Here M=yM=y so My=1My=1. Also N=xN=x so Nx=1Nx=1. Since My=NxMy=Nx, the equation is exact and has potential function ϕ=xyϕ=xy.
Solution of y dx+x dy=0ydx+xdy=0 is
A x+y=Cx+y=C
B x/y=Cx/y=C
C x2+y2=Cx2+y2=C
D xy=Cxy=C
Since it is exact with M=yM=y, integrate MM w.r.t. xx: ϕ=xy+g(y)ϕ=xy+g(y). Differentiating gives N=xN=x, so g′(y)=0g′(y)=0. Hence xy=Cxy=C.
Equation y=xy′y=xy′ suggests substitution
A y=v+xy=v+x
B y=vxy=vx
C y=v/xy=v/x
D y=v2y=v2
When expressions involve yy and xy′xy′ together, it often indicates homogeneity. Substitution y=vxy=vx converts y′y′ into v+xv′v+xv′, simplifying the relation.
In growth model dy/dt=kydy/dt=ky, solution shape is
A Straight line
B Exponential curve
C Circular arc
D Parabola always
Solving dy/dt=kydy/dt=ky gives y=Cekty=Cekt. This is exponential growth or decay depending on sign of kk. The curve changes rapidly compared to linear models.
If k=0k=0 in dy/dt=kydy/dt=ky, solution is
A Constant yy
B Linear yy
C Quadratic yy
D Periodic yy
If k=0k=0, the equation becomes dy/dt=0dy/dt=0. This means yy does not change with time, so y=Cy=C, a constant solution.
“Nonhomogeneous” linear means
A Order is two
B RHS not zero
C Degree not defined
D Uses substitution
For linear equation y′+P(x)y=Q(x)y′+P(x)y=Q(x), it is nonhomogeneous when Q(x)≠0Q(x)=0. That nonzero forcing term creates a particular part in the solution.
A quick test for linearity checks presence of
A Only xx terms
B Only constants
C Powers of yy
D Only dxdx
In a linear ODE, yy and its derivatives appear only to the first power and are not multiplied together. If y2y2, (y′)2(y′)2, or yy′yy′ appears, it’s nonlinear.
Equation dy/dx=(y/x)2dy/dx=(y/x)2 is solved using
A Exact method
B Constant coefficient
C y=vxy=vx method
D Laplace method
The RHS depends only on y/xy/x, so it is homogeneous. Put y=vxy=vx, giving dy/dx=v+xv′dy/dx=v+xv′. Substitute and separate variables to integrate and get solution.
What does “solution verification” mean
A Integrate twice always
B Differentiate then substitute
C Compare constants only
D Change variable always
To verify, compute derivatives required by the DE and substitute them back into the equation. If the equality holds for all values in the domain, the function is a correct solution.
Which statement about degree is correct
A Uses power of highest derivative
B Uses order of highest derivative
C Always equals order
D Always undefined
Degree is the power of the highest order derivative after expressing the equation as a polynomial in derivatives. It is different from order, which only depends on derivative order present.