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Jobisdone [24]
2 years ago
14

What is the correct algebraic expression to these ?!!! Quickkkk

Mathematics
2 answers:
sammy [17]2 years ago
6 0
There you go. Hope it help.
agasfer [191]2 years ago
5 0

Answer:

see below

Step-by-step explanation:

1)

\frac{4/5}{3/2}=\frac{4}{5}*\frac{2}{3}

so 1 goes with 3

2)

\frac{1/4}{1/7} =\frac{1}{4}*7

so 2 goes with 2

3)

\frac{7}{1/4} =7*4

so 3 goes with 1

4)

and that leaves 4 with 4

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alexandr1967 [171]

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Pleaseeeee ı dont find the answer
mote1985 [20]

Answer:  y(x) = \sqrt{\frac{7x^{14}}{-2x^7+9}}\\\\

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

Explanation:

The given differential equation (DE) is

y'-\frac{7}{x}y = \frac{y^3}{x^8}\\\\

Which is the same as

y'-\frac{7}{x}y = \frac{1}{x^8}y^3\\\\

This 2nd DE is in the form y' + P(x)y = Q(x)y^n

where

P(x) = -\frac{7}{x}\\\\Q(x) = \frac{1}{x^8}\\\\n = 3

As the instructions state, we'll use the substitution u = y^{1-n}

We specifically use u = y^{1-n} = y^{1-3} = y^{-2}

-----------------

After making the substitution, we'll end up with this form

\frac{du}{dx} + (1-n)P(x)u = (1-n)Q(x)\\\\

Plugging in the items mentioned, we get:

\frac{du}{dx} + (1-n)P(x)u = (1-n)Q(x)\\\\\frac{du}{dx} + (1-3)*\frac{-7}{x}u = (1-3)\frac{1}{x^8}\\\\\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\

We can see that we have a new P(x) and Q(x)

P(x) = \frac{14}{x}\\\\Q(x) = -\frac{2}{x^8}

-------------------

To solve the linear DE \frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\, we'll need the integrating factor which I'll call m

m(x) = e^{\int P(x) dx} = e^{\int \frac{14}{x}dx} = e^{14\ln(x)}

m(x) = e^{\ln(x^{14})} = x^{14}

We will multiply both sides of the linear DE by this m(x) integrating factor to help with further integration down the road.

\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\m(x)*\left(\frac{du}{dx} + \frac{14}{x}u\right) = m(x)*\left(-\frac{2}{x^8}\right)\\\\x^{14}*\frac{du}{dx} + x^{14}*\frac{14}{x}u = x^{14}*\left(-\frac{2}{x^8}\right)\\\\x^{14}*\frac{du}{dx} + 14x^{13}*u = -2x^6\\\\\left(x^{14}*u\right)' = -2x^6\\\\

It might help to think of the product rule being done in reverse.

Now we can integrate both sides to solve for u

\left(x^{14}*u\right)' = -2x^6\\\\\displaystyle \int\left(x^{14}*u\right)'dx = \int -2x^6 dx\\\\\displaystyle x^{14}*u = \frac{-2x^7}{7}+C\\\\\displaystyle u = x^{-14}*\left(\frac{-2x^7}{7}+C\right)\\\\\displaystyle u = x^{-14}*\frac{-2x^7}{7}+Cx^{-14}\\\\\displaystyle u = \frac{-2x^{-7}}{7}+Cx^{-14}\\\\

u = \frac{-2}{7x^7} + \frac{C}{x^{14}}\\\\u = \frac{-2}{7x^7}*\frac{x^7}{x^7} + \frac{C}{x^{14}}*\frac{7}{7}\\\\u = \frac{-2x^7}{7x^{14}} + \frac{7C}{7x^{14}}\\\\u = \frac{-2x^7+7C}{7x^{14}}\\\\

Unfortunately, this isn't the last step. We still need to find y.

Recall that we found u = y^{-2}\\\\

So,

u = \frac{-2x^7+7C}{7x^{14}}\\\\y^{-2} = \frac{-2x^7+7C}{7x^{14}}\\\\y^{2} = \frac{7x^{14}}{-2x^7+7C}

We're told that y(1) = 1. This means plugging x = 1 leads to the output y = 1. So the RHS of the last equation should lead to 1. We'll plug x = 1 into that RHS, set the result equal to 1 and solve for C

\frac{7*1^{14}}{-2*1^7+7C} = 1\\\\\frac{7}{-2+7C} = 1\\\\7 = -2+7C\\\\7+2 = 7C\\\\7C = 9\\\\C = \frac{9}{7}

So,

y^{2} = \frac{7x^{14}}{-2x^7+7C}\\\\y^{2} = \frac{7x^{14}}{-2x^7+7*\frac{9}{7}}\\\\y^{2} = \frac{7x^{14}}{-2x^7+9}\\\\y = \sqrt{\frac{7x^{14}}{-2x^7+9}}\\\\

We go with the positive version of the root because y(1) is positive, which must mean y(x) is positive for all x in the domain.

3 0
2 years ago
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