Its late anyone wanna talk?

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

I will give brainiest to whoever answers correctly !!

Lena [83]

Answer: the mark up price of the fish dinner is around 36.36%

Explanation: 18.75 - 13.75/13.75=5
5 divided by 13.75= 0.36363636.

Hope this helps!

3 0

2 years ago

Diana is planting a garden in the shape of the sector below. What is the length of fencing needed for the curred part of the gar

Lerok [7]

99/110

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3 0

3 years ago

An instructor in a college class recently gave an exam that was worth a total of 100 points. The instructor inadvertently made t

solniwko [45]

Answer:

B) Method 2 will increase the standard deviation of the students' scores.

Step-by-step explanation:

Given that an instructor in a college class recently gave an exam that was worth a total of 100 points.

The average score for his students was 43 and the standard deviation of the scores was 5 points.

And now he is considering two different strategies for rescaling the exam results of which:

Method 1 = Add 17 points to everyone's score.

Method 2 = Multiply everyone's score by 1.7 .

And we have to check what will be the impact of these methods on the standard deviation of the students' scores.

For this let us consider a simple example to understand this:

Firstly, Formula for calculating Standard Deviation = $\sqrt{\frac{\sum (X-Xbar)^{2}}{n-1}}$

Suppose,

X X - Xbar $(X - Xbar)^{2}$

3 3 - 6 = -3 -3 * -3 = 9

5 5 - 6 = -1 -1 * -1 = 1

10 10 - 6 = 4 4 * 4 = 16

Mean of above data, Xbar = $\frac{3+ 5+10}{3}$ = 6

Standard Deviation of data = $\sqrt{\frac{26}{3-1} }$ = 3.6055

Now let us suppose that we multiply each value of above data with 2 so the new data will be:

X X - Xbar $(X - Xbar)^{2}$

3*2 = 6 6 - 12 = -6 -6 * -6 = 36

5*2 = 10 10 - 12 = -2 -2 * -2 = 4

10*2 =20 20 - 12 = 8 8 * 8 = 64

Mean of new data, Xbar = $\frac{6+ 10+20}{3}$ = 12

Standard Deviation of new data = $\sqrt{\frac{104}{3-1} }$ = 7.2111

Hence, we see that when we multiply any value to the data the standard deviation will increase and in other words it will multiplied by that value which value we multiplied with each data value i.e. when we multiply each data value with 2 the standard deviation also get multiplied by as

3.6055 * 2 = 7.2111

Therefore option B is correct that Method 2 will increase the standard deviation of the students' scores.

And on the other hand Similarly by adding any constant to the data the Standard Deviation will remain same. Therefore Method 1 will have no impact on standard deviation of the students' scores.

8 0

3 years ago

Is the graph (6,-11), (-3,4), (-3,11), (6,4) a function or a relation?

matrenka [14]

The correct answer is a function

5 0

3 years ago