For a known solution y1 of a 2nd order differential equation, this method assumes a second solution in the form Uy1 which satisfies the said differential equation. It then assumes a reduced order for U'' (w' = U'').

The differential equation becomes easy to solve, and all that is left are integration and substitutions.

Check attachments for the solution to this problem.

6 0

2 years ago

A bakery sells 12 cookies for $24.00 If each cookie costs the same amount, how much does 1 cookie cost?.

zepelin [54]

Answer:

$2.00

Step-by-step explanation:

If you divide 12 by 24 you get 2. So $2 cookies.

Hope it helps :)))

3 0

2 years ago

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The function f(x) = 5One-fifth is reflected over the y-axis. Which equations represent the reflected function? Select two option

postnew [5]

Answer:

The equations that represent the reflected function are

$f(x)=5(\frac{1}{5})^{-x}$

$f(x)=5(5)^{x}$

Step-by-step explanation:

The correct question in the attached figure

we have the function

$f(x)=5(\frac{1}{5})^{x}$

we know that

A reflection across the y-axis interchanges positive x-values with negative x-values, swapping x and −x.

therefore

$f(−x) = f(x).$

The reflection of the given function across the y-axis will be equal to

(Remember interchanges positive x-values with negative x-values)

$f(x)=5(\frac{1}{5})^{-x}$

An equivalent form will be

$f(x)=5(\frac{1}{5})^{(-1)(x)}=5[(\frac{1}{5})^{-1})]^{x}=5(5)^{x}$

therefore

The equations that represent the reflected function are

$f(x)=5(\frac{1}{5})^{-x}$

$f(x)=5(5)^{x}$

9 0

2 years ago

Read 2 more answers