Answer Problems 1–10 without referring back to the text. Fill in the blank or answer true or false. 1. The only solution of the initial-value problem y ″ + x 2 y = 0, y (0) = 0, y ′(0) = 0 is __________.

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Answer 1

Textbook Question

Chapter 4, Problem 1RE

Answer Problems 1–10 without referring back to the text. Fill in the blank or answer true or false.

1. The only solution of the initial-value problem

y″ + x²y = 0, y(0) = 0, y′(0) = 0 is __________.

Expert Solution & Answer

To determine

To fill: The blank in the statement, “The only solution of the initial-value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is ______.”

Answer to Problem 1RE

The only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

Explanation of Solution

Theorem used:

The existence of a Unique Solution:

For a n^th order initial value problem $any(n)+...+a1y′+a0y=g(x)$ , where $an,an−1,...,a0 and g(x)$ are continuous on an interval I with $an(x)≠0$ on I.

If $x=x0$ is a point in interval I, then a solution $y(x)$ of initial value problem exists on I and the solution is unique.

Calculation:

Consider the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ . The differential equation $y″+x2y=0$ is a second order differential equation.

Compare the equation to the standard form of differential equation $any(n)+...+a1y′+a0y=g(x)$ . Here, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ .

Note that, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ are continuous on $I=(−∞,∞) and x0=(0)∈(−∞,∞)$ .

The initial value $x0=0$ is in the interval I and it satisfies the initial value problem. Thus the unique solution of the initial value problem is $y(x)=0$ .

By the existence of unique solution, the general solution of the initial value problem $y″+x2y=0$ with initial values $y(0)=0,y′(0)=0$ has a unique solution $y(x)=0$ .

Therefore, the only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

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Starting with the finished version of Example 6.2, attached, change the decision criterion to "maximize expected utility," using an exponential utility function with risk tolerance $5,000,000. Display certainty equivalents on the tree. a. Keep doubling the risk tolerance until the company's best strategy is the same as with the EMV criterion—continue with development and then market if successful. The risk tolerance must reach $ ____________ before the risk averse company acts the same as the EMV-maximizing company. b. With a risk tolerance of $320,000,000, the company views the optimal strategy as equivalent to receiving a sure $____________ , even though the EMV from the original strategy (with no risk tolerance) is $ ___________ .

2.8.1

Do not use the Residue Theorem. Thank you.

Answer 2

Textbook Question

Chapter 4, Problem 1RE

Answer Problems 1–10 without referring back to the text. Fill in the blank or answer true or false.

1. The only solution of the initial-value problem

y″ + x²y = 0, y(0) = 0, y′(0) = 0 is __________.

Expert Solution & Answer

To determine

To fill: The blank in the statement, “The only solution of the initial-value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is ______.”

Answer to Problem 1RE

The only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

Explanation of Solution

Theorem used:

The existence of a Unique Solution:

For a n^th order initial value problem $any(n)+...+a1y′+a0y=g(x)$ , where $an,an−1,...,a0 and g(x)$ are continuous on an interval I with $an(x)≠0$ on I.

If $x=x0$ is a point in interval I, then a solution $y(x)$ of initial value problem exists on I and the solution is unique.

Calculation:

Consider the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ . The differential equation $y″+x2y=0$ is a second order differential equation.

Compare the equation to the standard form of differential equation $any(n)+...+a1y′+a0y=g(x)$ . Here, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ .

Note that, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ are continuous on $I=(−∞,∞) and x0=(0)∈(−∞,∞)$ .

The initial value $x0=0$ is in the interval I and it satisfies the initial value problem. Thus the unique solution of the initial value problem is $y(x)=0$ .

By the existence of unique solution, the general solution of the initial value problem $y″+x2y=0$ with initial values $y(0)=0,y′(0)=0$ has a unique solution $y(x)=0$ .

Therefore, the only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

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Answer 3

Textbook Question

Chapter 4, Problem 1RE

Answer Problems 1–10 without referring back to the text. Fill in the blank or answer true or false.

1. The only solution of the initial-value problem

y″ + x²y = 0, y(0) = 0, y′(0) = 0 is __________.

Expert Solution & Answer

To determine

To fill: The blank in the statement, “The only solution of the initial-value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is ______.”

Answer to Problem 1RE

The only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

Explanation of Solution

Theorem used:

The existence of a Unique Solution:

For a n^th order initial value problem $any(n)+...+a1y′+a0y=g(x)$ , where $an,an−1,...,a0 and g(x)$ are continuous on an interval I with $an(x)≠0$ on I.

If $x=x0$ is a point in interval I, then a solution $y(x)$ of initial value problem exists on I and the solution is unique.

Calculation:

Consider the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ . The differential equation $y″+x2y=0$ is a second order differential equation.

Compare the equation to the standard form of differential equation $any(n)+...+a1y′+a0y=g(x)$ . Here, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ .

Note that, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ are continuous on $I=(−∞,∞) and x0=(0)∈(−∞,∞)$ .

The initial value $x0=0$ is in the interval I and it satisfies the initial value problem. Thus the unique solution of the initial value problem is $y(x)=0$ .

By the existence of unique solution, the general solution of the initial value problem $y″+x2y=0$ with initial values $y(0)=0,y′(0)=0$ has a unique solution $y(x)=0$ .

Therefore, the only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

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Answer 4

Textbook Question

Chapter 4, Problem 1RE

Answer Problems 1–10 without referring back to the text. Fill in the blank or answer true or false.

1. The only solution of the initial-value problem

y″ + x²y = 0, y(0) = 0, y′(0) = 0 is __________.

Expert Solution & Answer

To determine

To fill: The blank in the statement, “The only solution of the initial-value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is ______.”

Answer to Problem 1RE

The only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

Explanation of Solution

Theorem used:

The existence of a Unique Solution:

For a n^th order initial value problem $any(n)+...+a1y′+a0y=g(x)$ , where $an,an−1,...,a0 and g(x)$ are continuous on an interval I with $an(x)≠0$ on I.

If $x=x0$ is a point in interval I, then a solution $y(x)$ of initial value problem exists on I and the solution is unique.

Calculation:

Consider the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ . The differential equation $y″+x2y=0$ is a second order differential equation.

Compare the equation to the standard form of differential equation $any(n)+...+a1y′+a0y=g(x)$ . Here, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ .

Note that, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ are continuous on $I=(−∞,∞) and x0=(0)∈(−∞,∞)$ .

The initial value $x0=0$ is in the interval I and it satisfies the initial value problem. Thus the unique solution of the initial value problem is $y(x)=0$ .

By the existence of unique solution, the general solution of the initial value problem $y″+x2y=0$ with initial values $y(0)=0,y′(0)=0$ has a unique solution $y(x)=0$ .

Therefore, the only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

To fill: The blank in the statement, “The only solution of the initial-value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is ______.”

Answer to Problem 1RE

The only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

Explanation of Solution

Theorem used:

The existence of a Unique Solution:

For a n^th order initial value problem $any(n)+...+a1y′+a0y=g(x)$ , where $an,an−1,...,a0 and g(x)$ are continuous on an interval I with $an(x)≠0$ on I.

If $x=x0$ is a point in interval I, then a solution $y(x)$ of initial value problem exists on I and the solution is unique.

Calculation:

Consider the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ . The differential equation $y″+x2y=0$ is a second order differential equation.

Compare the equation to the standard form of differential equation $any(n)+...+a1y′+a0y=g(x)$ . Here, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ .

Note that, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ are continuous on $I=(−∞,∞) and x0=(0)∈(−∞,∞)$ .

The initial value $x0=0$ is in the interval I and it satisfies the initial value problem. Thus the unique solution of the initial value problem is $y(x)=0$ .

By the existence of unique solution, the general solution of the initial value problem $y″+x2y=0$ with initial values $y(0)=0,y′(0)=0$ has a unique solution $y(x)=0$ .

Therefore, the only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

Answer 8

Expert Solution & Answer

To determine

To fill: The blank in the statement, “The only solution of the initial-value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is ______.”

Answer to Problem 1RE

The only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

Explanation of Solution

Theorem used:

The existence of a Unique Solution:

For a n^th order initial value problem $any(n)+...+a1y′+a0y=g(x)$ , where $an,an−1,...,a0 and g(x)$ are continuous on an interval I with $an(x)≠0$ on I.

If $x=x0$ is a point in interval I, then a solution $y(x)$ of initial value problem exists on I and the solution is unique.

Calculation:

Consider the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ . The differential equation $y″+x2y=0$ is a second order differential equation.

Compare the equation to the standard form of differential equation $any(n)+...+a1y′+a0y=g(x)$ . Here, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ .

Note that, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ are continuous on $I=(−∞,∞) and x0=(0)∈(−∞,∞)$ .

The initial value $x0=0$ is in the interval I and it satisfies the initial value problem. Thus the unique solution of the initial value problem is $y(x)=0$ .

By the existence of unique solution, the general solution of the initial value problem $y″+x2y=0$ with initial values $y(0)=0,y′(0)=0$ has a unique solution $y(x)=0$ .

Therefore, the only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To fill: The blank in the statement, “The only solution of the initial-value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is ______.”

Answer 12

Answer to Problem 1RE

The only solution of the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ is $y(x)=0$ .

Answer 13

Explanation of Solution

Theorem used:

The existence of a Unique Solution:

For a n^th order initial value problem $any(n)+...+a1y′+a0y=g(x)$ , where $an,an−1,...,a0 and g(x)$ are continuous on an interval I with $an(x)≠0$ on I.

If $x=x0$ is a point in interval I, then a solution $y(x)$ of initial value problem exists on I and the solution is unique.

Calculation:

Consider the initial value problem $y″+x2y=0,y(0)=0,y′(0)=0$ . The differential equation $y″+x2y=0$ is a second order differential equation.

Compare the equation to the standard form of differential equation $any(n)+...+a1y′+a0y=g(x)$ . Here, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ .

Note that, $a2(x)=1,a1(x)=0,a0(x)=x2 and g(x)=0$ are continuous on $I=(−∞,∞) and x0=(0)∈(−∞,∞)$ .

The initial value $x0=0$ is in the interval I and it satisfies the initial value problem. Thus the unique solution of the initial value problem is $y(x)=0$ .

By the existence of unique solution, the general solution of the initial value problem $y″+x2y=0$ with initial values $y(0)=0,y′(0)=0$ has a unique solution $y(x)=0$ .