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monitta
3 years ago
12

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

in parentheses. For example, (a−b)/(1+n).
Mathematics
1 answer:
Nata [24]3 years ago
4 0

Answer:

The solution is y=-\frac{12}{12x-30x^2+1}.

Step-by-step explanation:

A first order differential equation y'=f(x,y) is called a separable equation if the function f(x,y) can be factored into the product of two functions of x and y:

f(x,y)=p(x)h(y)

where p(x) and h(y) are continuous functions.

We have the following differential equation

y'=(1-5x)y^2, \quad y(0)=-12

In the given case p(x)=1-5x and h(y)=y^2.

We divide the equation by h(y) and move dx to the right side:

\frac{1}{y^2}dy\:=(1-5x)dx

Next, integrate both sides:

\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C

Now, we solve for y

-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}

We use the initial condition y(0)=-12 to find the value of C.

-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}

Therefore,

y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}

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Step-by-step explanation:

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ahh this is the correct answer.

so if it helps don't forget to like and Mark me.

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1.) What are the zeros of the polynomial? f(x)=x^4-x^3-16x^2+4x+48.
Lerok [7]

Answer:

3.) \displaystyle [x - 2][x^2 + 2][x + 4]

2.) \displaystyle 2\:complex\:solutions → x^2 + 3x + 6 >> -\frac{3 - i\sqrt{15}}{2}, -\frac{3 + i\sqrt{15}}{2}

1.) \displaystyle 4, -3, 2, and\:-2

Step-by-step explanation:

3.) By the Rational Root Theorem, we would take the Least Common Divisor [LCD] between the leading coefficient of 1, and the initial value of −16, which is 1, but we will take 2 since it is the <em>fourth root</em> of 16; so this automatically makes our first factor of \displaystyle x - 2.Next, since the factor\divisor is in the form of \displaystyle x - c, use what is called Synthetic Division. Remember, in this formula, −c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:

2| 1 2 −6 4 −16

↓ 2 8 4 16

__________________

1 4 2 8 0 → \displaystyle x^3 + 4x^2 + 2x + 8

You start by placing the <em>c</em> in the top left corner, then list all the coefficients of your dividend [x⁴ + 2x³ - 6x² + 4x - 16]. You bring down the original term closest to <em>c</em> then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that 1 in your quotient can be an x³, the 4x² follows right behind it, bringing 2x right up against it, and bringing up the rear, 8, giving you the quotient of \displaystyle x^3 + 4x^2 + 2x + 8.

However, we are not finished yet. This is our first quotient. The next step, while still using the Rational Root Theorem with our first quotient, is to take the Least Common Divisor [LCD] of the leading coefficient of 1, and the initial value of 8, which is −4, so this makes our next factor of \displaystyle x + 4.Then again, we use Synthetic Division because \displaystyle x + 4is in the form of \displaystyle x - c:

−4| 1 4 2 8

↓ −4 0 −8

_____________

1 0 2 0 → \displaystyle x^2 + 2

So altogether, we have our four factors of \displaystyle [x^2 + 2][x + 4][x - 2].

__________________________________________________________

2.) By the Rational Root Theorem again, this time, we will take −1, since the leading coefficient & variable\degree and the initial value do not share any common divisors other than the <em>special</em><em> </em><em>number</em> of 1, and it does not matter which integer of 1 you take first. This gives a factor of \displaystyle x + 1.Then start up Synthetic Division again:

−1| 1 3 5 −3 −6

↓ −1 −2 −3 6

__________________

1 2 3 −6 0 → \displaystyle x^3 + 2x^2 + 3x - 6

Now we take the other integer of 1 to get the other factor of \displaystyle x - 1,then repeat the process of Synthetic Division:

1| 1 2 3 −6

↓ 1 3 6

_____________

1 3 6 0 → \displaystyle x^2 + 3x + 6

So altogether, we have our three factors of \displaystyle [x - 1][x^2 + 3x + 6][x + 1].

Hold it now! Notice that \displaystyle x^2 + 3x + 6is unfactorable. Therefore, we have to apply the Quadratic Formula to get our two complex solutions, \displaystyle a + bi[or zeros in this matter]:

\displaystyle -b ± \frac{\sqrt{b^2 - 4ac}}{2a} = x \\ \\ -3 ± \frac{\sqrt{3^2 - 4[1][6]}}{2[1]} = x \\ \\ -3 ± \frac{\sqrt{9 - 24}}{2} = x \\ \\ -3 ± \frac{\sqrt{-15}}{2} = x \\ \\ -3 ± i\frac{\sqrt{15}}{2} = x \\ \\ -\frac{3 - i\sqrt{15}}{2}, -\frac{3 + i\sqrt{15}}{2} = x

__________________________________________________________

1.) By the Rational Root Theorem one more time, this time, we will take 4 since the initial value is 48 and that 4 is the root of the polynomial. This gives our automatic factor of \displaystyle x - 4.Then start up Synthetic Division again:

4| 1 −1 −16 4 48

↓ 4 12 −16 −48

___________________

1 3 −4 −12 0 → \displaystyle x^3 + 3x^2 - 4x - 12

We can then take −3 since it is a root of this polynomial, giving us the factor of \displaystyle x + 3:

−3| 1 3 −4 −12

↓ −3 0 12

_______________

1 0 −4 0 → \displaystyle x^2 - 4 >> [x - 2][x + 2]

So altogether, we have our four factors of \displaystyle [x - 2][x + 3][x + 2][x - 4],and when set to equal zero, you will get \displaystyle 4, -3, 2, and\:-2.

I am delighted to assist you anytime.

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3 years ago
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Tatiana [17]

Answer:

Step-by-step explanation:

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4 0
2 years ago
How to solve these 3 problems
Bingel [31]

Answer:

  1. a. decay; b. growth; c. decay; d. neither
  2. r = 4; a = 1; y = 1·4^x
  3. a. an = 3(5^(n-1)); b. f(x) = (3/5)(5^x); c. exponential growth; d. y-intercept: 3/5; first term: 3.

Step-by-step explanation:

There are two kinds of exponential problems here.

  1. exponential functions of the form f(x) = a·b^x
  2. exponential sequences of the explicit form an = a1·r^(n-1)

The second problem gives you a table that suggests the sequence form, but it asks for the exponential function form. The third problem does something similar.

__

<h3>1.</h3>

In an exponential function of the form f(x) = a·b^x, the function grows if b>1 and decays if b<1. Using this check, we can easily answer ...

  a. 0.4 < 1 . . . decay

  b. 1.3 > 1 . . . growth

  c. 1/2 < 1 . . . decay

  d. 1 = 1 . . . neither growth nor decay; the function is constant: j(x) = 1.

__

<h3>2.</h3>

The value of x is given starting at 1, so we can consider this a geometric sequence. The common ratio is r = 16/4 = 4. The first term is a1 = 4, so the explicit formula for the sequence is ...

  an = 4·4^(n-1)

When this is expanded to get rid of the constant in the exponent, we have ...

  an = 4·(4^n)·(4^-1) = 1·4^n

We recognize this form as matching the functional form f(x) = a·r^x. The multiplier of the exponential factor is a=1. In summary, ...

  r = 4; a = 1; f(x) = 1·4^x

__

<h3>3.</h3>

The first term of this geometric sequence is a1 = 3. The common ratio is r = 15/3 = 5. Using the explicit formula, we have ...

 a. explicit form: an = 3·5^(n-1)

Using the method of question 2 to write the functional form, we find ...

  an = 3(5^n)(5^-1) = (3/5)(5^n)

  b. functional form: f(x) = (3/5)(5^x)

  c. function family: exponential growth functions

 d. y-intercept: (3/5) . . . . read this from the f(x) form

     1st term: the first term listed in the given sequence is 3

_____

<em>Additional comment</em>

The "y-intercept" of a sequence is irrelevant (undefined), as the sequence term numbering starts with 1, not 0. The domain of the explicit formula is <em>natural numbers</em>, which does not include 0.

Similarly, the "first term" of a function f(x) needs further definition. Here, we've answered the question by saying the first term is f(1). There is no conventional definition of a "first term" for a continuous function.

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