Which statement best defines a differential equation
A Equation with limits
B Equation with derivatives
C Equation with vectors
D Equation with matrices
A differential equation is an equation that involves an unknown function and one or more of its derivatives. It connects a function to its rate of change with respect to a variable.
In an ODE, the dependent variable usually depends on
A Only constants
B Two independent variables
C No independent variable
D One independent variable
An ordinary differential equation (ODE) involves derivatives with respect to a single independent variable, like xx or tt. If multiple independent variables appear, it becomes a PDE.
What is the order offracd3ydx3+y=0fracd3ydx3+y=0
A 2
B 1
C 3
D 0
The order of a differential equation is the highest order derivative present. Here the highest derivative is d3y/dx3d3y/dx3, so the order is 3.
Order ofleft(fracdydxright)2+y=0left(fracdydxright)2+y=0 is
A 2
B 1
C Not defined
D 0
Even though the first derivative is squared, the highest derivative appearing is still dy/dxdy/dx. Therefore, the order depends only on derivative order, not powers.
Degree is defined only when the equation is
A Polynomial in derivatives
B Separable only
C Linear in xx
D Free of constants
Degree is defined only when the differential equation can be expressed as a polynomial in derivatives (no radicals, fractions, or trig functions of derivatives). Then degree is the highest power of highest derivative.
The highest derivative present is d2y/dx2d2y/dx2. It appears to the first power, and the equation is polynomial in derivatives, so degree is 1.
Order of (y′′)2+y=0(y′′)2+y=0
A 4
B 1
C 2
D 3
The highest derivative appearing is y′′y′′ (second derivative). Order depends on the highest derivative order, so the order is 2 even if it is squared.
Degree of (y′′)2+y=0(y′′)2+y=0
A Not defined
B 1
C 3
D 2
The equation is polynomial in derivatives and the highest derivative is y′′y′′. Its highest power is 2, so the degree is 2.
Which is a first-order differential equation
A y′+y=0y′+y=0
B y′′′+y=0y′′′+y=0
C (y′′)2+y=0(y′′)2+y=0
D y′′+y=0y′′+y=0
A first-order differential equation contains only the first derivative y′y′ and no higher derivatives. Option B has only y′y′, so it is first order.
Which equation is nonlinear
A y′+2y=xy′+2y=x
B y′+xy=0y′+xy=0
C (y′)2+y=0(y′)2+y=0
D y′+y=sinxy′+y=sinx
A differential equation is nonlinear if the dependent variable or its derivatives appear with powers other than 1, products, or inside nonlinear functions. Here (y′)2(y′)2 makes it nonlinear.
Which is a linear first-order ODE form
A (y′)2+P(x)y=0(y′)2+P(x)y=0
B y′+P(x)y=Q(x)y′+P(x)y=Q(x)
C yy′+P(x)=0yy′+P(x)=0
D sin(y′)+y=0sin(y′)+y=0
The standard linear first-order ODE has the form y′+P(x)y=Q(x)y′+P(x)y=Q(x), where coefficients depend only on xx and yy is not multiplied by itself or its derivatives.
A solution containing arbitrary constants is called
A Particular solution
B Singular point
C Exact form
D General solution
A general solution represents a family of solutions and contains arbitrary constants (usually equal to the order). These constants are determined by applying initial or boundary conditions.
A solution satisfying given conditions is called
A Complementary set
B Particular solution
C Homogeneous form
D Degree solution
A particular solution is obtained by using initial or boundary conditions to find values of arbitrary constants from the general solution, giving one specific member of the solution family.
For first-order ODE, general solution has constants
A 1
B 2
C 0
D 3
The number of arbitrary constants in the general solution typically equals the order of the differential equation. A first-order ODE usually yields one constant after integration.
For second-order ODE, general solution has constants
A 1
B 3
C 4
D 2
A second-order differential equation generally needs two integrations, producing two arbitrary constants. Therefore, its general solution usually contains two independent constants.
What does an initial condition usually specify
A Domain limits
B Highest degree
C Value at a point
D Slope field only
Initial conditions provide the value of the solution (and possibly derivatives) at a specific point, such as y(x0)y(x0) or y′(x0)y′(x0). They help determine constants uniquely.
Which is an initial value problem example
A y′+y=0y′+y=0 only
B y′+y=0,y(0)=2y′+y=0,y(0)=2
C y′′+y=0y′′+y=0 only
D y′=xy′=x only
An initial value problem includes a differential equation plus conditions at a specific initial point. Here, y(0)=2y(0)=2 fixes the constant in the general solution.
A boundary value problem gives conditions at
A No points
B One point only
C Random points
D Two points
Boundary value problems specify conditions at two different points, such as y(a)y(a) and y(b)y(b). These are common in physical problems like heat flow or beam bending.
Which indicates an ODE, not a PDE
A Multiple partial derivatives
B partial2y/partialxpartialypartial2y/partialxpartialy
C Only dy/dxdy/dx
D partialy/partialxpartialy/partialx
ODEs use ordinary derivatives like dy/dxdy/dx with one independent variable. PDEs involve partial derivatives (∂) because the function depends on more than one variable.
Homogeneous DE in x,yx,y means RHS is
A Function of xyxy
B Function of y/xy/x
C Function of x+yx+y
D Constant only
A first-order homogeneous differential equation can be written as dy/dx=F(y/x)dy/dx=F(y/x). Such equations become separable after substituting y=vxy=vx.
Standard substitution for homogeneous DE is
A y=v/xy=v/x
B y=v+xy=v+x
C x=vyx=vy
D y=vxy=vx
For homogeneous equations dy/dx=F(y/x)dy/dx=F(y/x), substituting y=vxy=vx gives dy/dx=v+xdv/dxdy/dx=v+xdv/dx, reducing it to a separable equation in vv and xx.
If y=vxy=vx, then dy/dxdy/dx equals
A v+x,dv/dxv+x,dv/dx
B v/x+dv/dxv/x+dv/dx
C x+v,dv/dxx+v,dv/dx
D v−x,dv/dxv−x,dv/dx
Differentiate y=vxy=vx using product rule: dy/dx=vcdot1+xcdotdv/dxdy/dx=vcdot1+xcdotdv/dx. This step is key to transforming homogeneous equations into separable form.
A separable equation can be written as
A y′′+y=0y′′+y=0
B Mdx+Ndy=0Mdx+Ndy=0 always
C f(y)dy=g(x)dxf(y)dy=g(x)dx
D y′+Py=Qy2y′+Py=Qy2 only
A separable differential equation allows variables to be separated, arranging all yy terms with dydy on one side and all xx terms with dxdx on the other for integration.
In y′+P(x)y=Q(x)y′+P(x)y=Q(x), integrating factor is
A intQ(x)dxintQ(x)dx
B eintP(x)dxeintP(x)dx
C eintQ(x)dxeintQ(x)dx
D intP(x)dxintP(x)dx
For a linear first-order ODE, the integrating factor is IF=eintP(x)dxIF=eintP(x)dx. Multiplying the equation by IF converts the left side into the derivative of ycdotIFycdotIF.
Main use of integrating factor is to make LHS
A Constant slope
B Quadratic form
C Separable always
D Exact derivative
Multiplying by the integrating factor makes the left side equal to d(ycdotIF)/dxd(ycdotIF)/dx. This turns the equation into a direct integration problem and helps obtain the solution systematically.
Exact equation has standard form
A dy/dx=F(y/x)dy/dx=F(y/x)
B y′+Py=Qy′+Py=Q
C Mdx+Ndy=0Mdx+Ndy=0
D f(y)dy=g(x)dxf(y)dy=g(x)dx
An exact differential equation is written as M(x,y)dx+N(x,y)dy=0M(x,y)dx+N(x,y)dy=0. If it is exact, it represents a total differential dphi=0dphi=0.
Exactness condition is
A McdotN=1McdotN=1
B partialM/partialy=partialN/partialxpartialM/partialy=partialN/partialx
C partialM/partialx=partialN/partialypartialM/partialx=partialN/partialy
D M=NM=N
A differential equation Mdx+Ndy=0Mdx+Ndy=0 is exact if mixed partial derivatives match: My=NxMy=Nx. Then there exists a potential functionphi(x,y)phi(x,y) with dphi=Mdx+Ndydphi=Mdx+Ndy.
If equation is exact, solution is
A phi(x,y)=Cphi(x,y)=C
B y=mx+cy=mx+c
C M=N=CM=N=C
D y=Cxy=Cx
For an exact equation, Mdx+Ndy=dphi=0Mdx+Ndy=dphi=0. Integrating gives a potential functionphi(x,y)phi(x,y), and the implicit solution isphi(x,y)=Cphi(x,y)=C.
In exact method,phi(x,y)phi(x,y) is called
A Trial solution
B Auxiliary curve
C Degree function
D Potential function
The functionphi(x,y)phi(x,y) whose total differential equals Mdx+NdyMdx+Ndy is the potential function. The solution of the exact equation is obtained by setting this function equal to a constant.
Which indicates equation may need integrating factor
A Degree undefined
B Always separable
C Not exact initially
D Order zero
If MyneNxMyneNx, the equation is not exact. Sometimes multiplying by an integrating factor depending only on xx or only on yy can make it exact and solvable.
Integrating factor for linear ODE depends on
A Both x,yx,y
B P(x)P(x) only
C yy only
D Q(x)Q(x) only
In y′+P(x)y=Q(x)y′+P(x)y=Q(x), the integrating factor is eintP(x)dxeintP(x)dx. It depends only on the coefficient of yy, not on the forcing term Q(x)Q(x).
If dy/dx=f(x)dy/dx=f(x), then yy equals
A f(x)+Cf(x)+C
B 1/f(x)+C1/f(x)+C
C intf(y)dyintf(y)dy
D intf(x)dx+Cintf(x)dx+C
When derivative depends only on xx, integrate both sides with respect to xx. This directly gives y=intf(x)dx+Cy=intf(x)dx+C, the simplest separable-type situation.
If dy/dx=g(y)dy/dx=g(y), method is
A Separate variables
B Integrating factor
C Exactness test
D Variation method
If dy/dx=g(y)dy/dx=g(y), rewrite as dy/g(y)=dxdy/g(y)=dx. This separates variables and allows integration on both sides to obtain the implicit or explicit solution.
A homogeneous linear ODE has Q(x)Q(x) as
A Polynomial only
B Exponential only
C 0
D Constant
A first-order linear equation y′+P(x)y=Q(x)y′+P(x)y=Q(x) is homogeneous when Q(x)=0Q(x)=0. Then the solution depends only on the integrating factor and one constant.
In y′+P(x)y=0y′+P(x)y=0, solution form is
A y=C+intPdxy=C+intPdx
B y=Ce−intPdxy=Ce−intPdx
C y=CeintQdxy=CeintQdx
D y=Px+Cy=Px+C
Solve y′/y=−P(x)y′/y=−P(x) by separation:ln∣y∣=−intP(x)dx+Cln∣y∣=−intP(x)dx+C. Exponentiating gives y=Ce−intPdxy=Ce−intPdx.
Bernoulli equation basic form is
A Mdx+Ndy=0Mdx+Ndy=0
B dy/dx=F(y/x)dy/dx=F(y/x)
C y′+Py=Qy′+Py=Q
D y′+Py=Qyny′+Py=Qyn
Bernoulli’s equation is y′+P(x)y=Q(x)yny′+P(x)y=Q(x)yn, where nneq0,1nneq0,1. It becomes linear after substitution v=y1−nv=y1−n, making it solvable using integrating factor.
In Bernoulli, substitution usually is
A v=ynv=yn
B v=xyv=xy
C v=y1−nv=y1−n
D v=y/xv=y/x
Dividing by ynyn gives a form involving y1−ny1−n. Setting v=y1−nv=y1−n converts the Bernoulli equation into a linear first-order equation in vv.
General solution represents a
A Single slope
B Family of curves
C Single point
D Only constant
Because it contains arbitrary constants, the general solution represents an entire family of curves. Each choice of constants gives a specific curve, and conditions select one from the family.
A particular solution is obtained by
A Differentiating twice
B Setting x=0x=0
C Removing derivatives
D Applying conditions
To get a particular solution, use initial or boundary conditions to determine constants in the general solution. This converts the family into one unique solution matching the given data.
To verify a proposed solution, you should
A Substitute into DE
B Change independent variable
C Differentiate only once
D Compare degrees only
Verification means compute derivatives of the proposed function and substitute into the original differential equation. If the equation is satisfied identically, the function is a valid solution.
An implicit solution is usually written as
A x=g(y)x=g(y) only
B F(x,y)=CF(x,y)=C
C y=mx+cy=mx+c
D y=f(x)y=f(x) always
An implicit solution expresses the relation between xx and yy without explicitly solving for yy. Many exact and separable equations naturally lead to forms like F(x,y)=CF(x,y)=C.
An explicit solution is written as
A F(x,y)=CF(x,y)=C
B Mdx+Ndy=0Mdx+Ndy=0
C y′′=0y′′=0
D y=f(x)y=f(x)
An explicit solution directly expresses yy in terms of xx. Sometimes implicit solutions can be rearranged into explicit form, but not always.
Which is a separable differential equation
A Mdx+Ndy=0Mdx+Ndy=0
B y′′+y=0y′′+y=0
C dy/dx=xydy/dx=xy
D y′+xy=1y′+xy=1
dy/dx=xydy/dx=xy can be rewritten as dy/y=xdxdy/y=xdx, which separates variables. Then integrating givesln∣y∣=x2/2+Cln∣y∣=x2/2+C, so it is separable.
Equation dy/dx=x/ydy/dx=x/y is
A Not solvable
B Separable
C Exact always
D Second order
Rewrite dy/dx=x/ydy/dx=x/y as y dy=x dxydy=xdx. Variables are separated, so integrate both sides to get y2/2=x2/2+Cy2/2=x2/2+C.
Equation dy/dx=(x+y)/xdy/dx=(x+y)/x is
A Homogeneous type
B Exact type
C Second order
D Not an ODE
dy/dx=(x+y)/x=1+y/xdy/dx=(x+y)/x=1+y/x, which depends on y/xy/x. Therefore it is homogeneous in x,yx,y and can be solved using substitution y=vxy=vx.
Exact equation corresponds to differential of
A Two constants
B Linear factor
C Only derivative
D Single function
Exactness means Mdx+NdyMdx+Ndy equals the total differential dphi(x,y)dphi(x,y) of one function. Then dphi=0dphi=0 gives the solutionphi(x,y)=Cphi(x,y)=C.
If integrating factor depends only on xx, it ismu(x)mu(x) where
A mu(x)=sinymu(x)=siny
B mu(x)=x+ymu(x)=x+y
C mu(x)Mdx+mu(x)Ndymu(x)Mdx+mu(x)Ndy exact
D mu(x)=y/xmu(x)=y/x
An integrating factormu(x)mu(x) is chosen so that after multiplication, the new equation becomes exact. Thenpartial(muM)/partialy=partial(muN)/partialxpartial(muM)/partialy=partial(muN)/partialx holds.
Growth model differential equation is commonly
A dy/dt=k+ydy/dt=k+y
B dy/dt=kydy/dt=ky
C dy/dt=ktdy/dt=kt
D dy/dt=y/kdy/dt=y/k
Exponential growth or decay is modeled by dy/dt=kydy/dt=ky, where rate is proportional to current amount. k>0k>0 gives growth and k<0k<0 gives decay, widely used in populations and radioactivity.
A slope field represents
A Exactness condition
B Degree of equation
C Direction of solutions
D Number of constants
A slope field is a diagram showing small line segments with slope dy/dxdy/dx at many points. It visually indicates how solution curves behave without solving the equation explicitly.