Lesson test: Second order equations and superposition

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Second order equations and superposition

1.[2p]
Why does a second order equation need two initial conditions?

Its solution space has dimension two, so two numbers are needed to select one member

Correct

Because it is solved by integrating twice with a constant each time, and the constants are independent of the equation

The answer is: Its solution space has dimension two, so two numbers are needed to select one member

Because the Wronskian involves two functions

The answer is: Its solution space has dimension two, so two numbers are needed to select one member

Because the forcing and the transient must be fixed separately

The answer is: Its solution space has dimension two, so two numbers are needed to select one member

2.[1p]
Superposition applies to the equation $y ′′ + \sin y = 0$ .

True

The answer is: False

False

Correct

3.[2p]
What is the Wronskian of $y_{1} = e^{3 t}$ and $y_{2} = e^{- 2 t}$ at $t = 0$ ?

CorrectNot quite: -5

4.[3p]
The general solution of $y ′′ - y' - 6 y = 0$ is $y = c_{1} e^{3 t} + c_{2} e^{- 2 t}$ . For $y (0) = 0$ and $y' (0) = 5$ , what is $y (1)$ ?

CorrectNot quite: 19.95

5.[3p]
In reduction of order, why does the term in $v$ drop out when $y = v y_{1}$ is substituted?

Its coefficient is $L [y_{1}]$ , which is zero because $y_{1}$ already solves the equation

Correct

Because $v$ is assumed small

The answer is: Its coefficient is $L [y_{1}]$ , which is zero because $y_{1}$ already solves the equation

Because the Wronskian is constant

The answer is: Its coefficient is $L [y_{1}]$ , which is zero because $y_{1}$ already solves the equation

Because $y_{1}$ is never zero

The answer is: Its coefficient is $L [y_{1}]$ , which is zero because $y_{1}$ already solves the equation

6.[2p]
Abel's identity gives $W = C e^{- \int p d t}$ . What does that tell you?

The Wronskian is either identically zero or never zero, so independence can be tested at one point

Correct

The Wronskian always decays to zero

The answer is: The Wronskian is either identically zero or never zero, so independence can be tested at one point

The Wronskian equals the forcing term

The answer is: The Wronskian is either identically zero or never zero, so independence can be tested at one point

Two solutions are independent only where $p$ is continuous

The answer is: The Wronskian is either identically zero or never zero, so independence can be tested at one point

7.[3p]
Which statements about $L [y] = g$ are correct?

Select all that apply

Every solution is a particular solution plus the general solution of $L [y] = 0$

Correct

Initial conditions must be applied after the particular solution has been included

Correct

If $L [y_{1}] = g_{1}$ and $L [y_{2}] = g_{2}$ then $L [y_{1} + y_{2}] = g_{1} + g_{2}$

Correct

The difference of two particular solutions is another particular solution

The answer is: Every solution is a particular solution plus the general solution of $L [y] = 0$ , Initial conditions must be applied after the particular solution has been included, If $L [y_{1}] = g_{1}$ and $L [y_{2}] = g_{2}$ then $L [y_{1} + y_{2}] = g_{1} + g_{2}$

8.[3p]
Match each term to what it names.

Fundamental set
Complementary function
Particular solution
Wronskian

two independent solutions of the homogeneous equation
any one solution of the driven equation
the determinant that tests independence
the general homogeneous solution

Show the answer

Fundamental set: two independent solutions of the homogeneous equation Complementary function: the general homogeneous solution Particular solution: any one solution of the driven equation Wronskian: the determinant that tests independence

9.[2p]
$y_{1} = e^{2 t}$ solves $y ′′ - 4 y' + 4 y = 0$ . Reduction of order gives a second solution $t^{n} e^{2 t}$ . What is $n$ ?

CorrectNot quite: 1

1.[2p] Why does a second order equation need two initial conditions?

2.[1p] Superposition applies to the equation y′′+siny=0.

3.[2p] What is the Wronskian of y1=e3t and y2=e−2t at t=0?

4.[3p] The general solution of y′′−y′−6y=0 is y=c1e3t+c2e−2t. For y(0)=0 and y′(0)=5, what is y(1)?

5.[3p] In reduction of order, why does the term in v drop out when y=vy1 is substituted?

6.[2p] Abel's identity gives W=Ce−∫pdt. What does that tell you?

7.[3p] Which statements about L[y]=g are correct?

8.[3p] Match each term to what it names.

9.[2p] y1=e2t solves y′′−4y′+4y=0. Reduction of order gives a second solution tne2t. What is n?

1.[2p]
Why does a second order equation need two initial conditions?

2.[1p]
Superposition applies to the equation $y ′′ + \sin y = 0$ .

3.[2p]
What is the Wronskian of $y_{1} = e^{3 t}$ and $y_{2} = e^{- 2 t}$ at $t = 0$ ?

4.[3p]
The general solution of $y ′′ - y' - 6 y = 0$ is $y = c_{1} e^{3 t} + c_{2} e^{- 2 t}$ . For $y (0) = 0$ and $y' (0) = 5$ , what is $y (1)$ ?

5.[3p]
In reduction of order, why does the term in $v$ drop out when $y = v y_{1}$ is substituted?

6.[2p]
Abel's identity gives $W = C e^{- \int p d t}$ . What does that tell you?

7.[3p]
Which statements about $L [y] = g$ are correct?

8.[3p]
Match each term to what it names.

9.[2p]
$y_{1} = e^{2 t}$ solves $y ′′ - 4 y' + 4 y = 0$ . Reduction of order gives a second solution $t^{n} e^{2 t}$ . What is $n$ ?