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2 years ago

What is the domain of the function for this situation?

Mathematics

1 answer:

raketka [301]2 years ago

8 0

Answer:

There is no problem. If you put the problem then i'll help

Step-by-step explanation:

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3 years ago

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2 years ago

Read 2 more answers

Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$

- Therefore, the complete solution to the given ODE can be expressed as:

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}$

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6 0

2 years ago

A stadium has 47,000 seats. Seats sell for $28 in Section A, $24 in Section B, and $20 in Section C. The number of seats in

mote1985 [20]

A:23,500 seats
B and C: both have 11,750 seats

7 0

2 years ago

I need help pls due tomorrow pls I just need part B done plsss

lara [203]

Part A

1 day = 1/4 hours of practice

7 days = 7/4 hours of practice (multiply both sides by 7)

1 week = 7/4 hours of practice

1 week = (4+3)/4 hours of practice

1 week = (4/4)+(3/4) hours of practice

1 week = 1+(3/4) hours of practice

1 week = 1 & 3/4 hours of practice

side note: 1 & 3/4 = 1.75

=======================================

Part B

Take the result from part A, and multiply it with 60

So we'll have 60 times 1&3/4 on the left side on the first line, then 60*(1+3/4) on the right side of this same line.

The rest of the lines look like this

(60*1) + (60*3/4)

60 + 60*3/4

60 + 180/4

60 + 45

105 minutes

8 0

2 years ago