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3 years ago

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Describe it, and also is it a positive or negative association? show work I will give braknliest

1 answer:

STatiana [176]3 years ago

6 0

Answer:

The relationship between the data in the scatter plot is a <u>positive or increasing trend</u>.

Step-by-step explanation:

The points are going in an upwards direction. This upward direction is an increase, which means that this data is a positive trend.

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Use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answ

olga_2 [115]

Answer:

Therefore the complete primitive is

$y=c_1 e^{2y}+c_2e^{3t}+e^{t}$

Therefore the general solution is

$y=c_1e^{2t}+c_2e^{3t}+e^t$

Step-by-step explanation:

Given Differential equation is

$y''-5y'+6y=2e^t$

<h3>Method of variation of parameters:</h3>

Let $y=e^{mt}$ be a trial solution.

$y'= me^{mt}$

and $y''= m^2e^{mt}$

Then the auxiliary equation is

$m^2e^{mt}-5me^{mt}+6e^{mt}=0$

$\Rightarrow m^2-5m+6=0$

$\Rightarrow m^2 -3m -2m +6=0$

$\Rightarrow m(m -3) -2(m -3)=0$

$\Rightarrow (m-3)(m-2)=0$

$\Rightarrow m=2,3$

∴The complementary function is $C_1e^{2t}+C_2e^{3t}$

To find P.I

First we show that $e^{2t}$ and $e^{3t}$ are linearly independent solution.

Let $y_1=e^{2t}$ and $y_2= e^{3t}$

The Wronskian of $y_1$ and $y_2$ is $\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|$

$=\left|\begin{array}{cc}e^{2t}&e^{3t}\\2e^{2t}&3e^{3t}\end{array}\right|$

$=e^{2t}.3e^{3t}-e^{2t}.2e^{3t}$

$=e^{5t}$ ≠ 0

∴ $y_1$ and $y_2$ are linearly independent.

Let the particular solution is

$y_p=v_1(t)e^{2t}+v_2(t)e^{3t}$

Then,

$Dy_p= 2v_1(t)e^{2t}+v'_1(t)e^{2t}+3v_2(t)e^{3t}+v'_2(t)e^{3t}$

Choose $v_1(t)$ and $v_2(t)$ such that

$v'_1(t)e^{2t}+v'_2(t)e^{3t}=0$ .......(1)

So that

$Dy_p= 2v_1(t)e^{2t}+3v_2(t)e^{3t}$

$D^2y_p= 4v_1(t)e^{2t}+9v_2(t)e^{3t}+ 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}$

Now

$4v_1(t)e^{2t}+9v_2(t)e^{3t}+ 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}-5[2v_1(t)e^{2t}+3v_2(t)e^{3t}] +6[v_1e^{2t}+v_2e^{3t}]=2e^t$

$\Rightarrow 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}=2e^t$ .......(2)

Solving (1) and (2) we get

$v'_2=2 e^{-2t}$ and $v'_1(t)=-2e^{-t}$

Hence

$v_1(t)=\int (-2e^{-t}) dt=2e^{-t}$

and $v_2=\int 2e^{-2t}dt =-e^{-2t}$

Therefore $y_p=(2e^{-t}) e^{2t}-e^{-2t}.e^{3t}$

$=2e^t-e^t$

$=e^t$

Therefore the complete primitive is

$y=c_1 e^{2y}+c_2e^{3t}+ e^{t}$

<h3>Undermined coefficients:</h3>

∴The complementary function is $C_1e^{2t}+C_2e^{3t}$

The particular solution is $y_p=Ae^t$

Then,

$Dy_p= Ae^t$ and $D^2y_p=Ae^t$

$\therefore Ae^t-5Ae^t+6Ae^t=2e^t$

$\Rightarrow 2Ae^t=2e^t$

$\Rightarrow A=1$

$\therefore y_p=e^t$

Therefore the general solution is

$y=c_1e^{2t}+c_2e^{3t}+e^t$

4 0

3 years ago

The length and width of a rectangle are shown below. 50 - X 2x+10 Which expression represents the area, in square units of the r

Rina8888 [55]

Answer:

<h2>hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm</h2>

Step-by-step explanation:

3 0

3 years ago

Derivative of x^3 + 5y^2=2

svetlana [45]

Answer:

Derivative- an expression representing the rate of change of a function with respect to an independent variable.

Step-by-step explanation:

7 0

3 years ago

PLEASE HELP ME WITH THIS I FINSIH MY QUARTER TONIGHT FOR SCHOOL AND I NEED TO GET THIS DONE!! Austin and his children went to th

astra-53 [7]

Answer:

$1a+2p=14\\a+p=9\\\\a=4\\p=5$

Step-by-step explanation:

$1a+2p=14\\a+p=9\\\\a+p=9\\a=9-p\\\\1(9-p)+2p=14\\9-p+2p=14\\9+p=14\\p=5\\\\a=9-(5)\\a=4$

Graphing:

Slope intercept form y=mx+b

a is y, p is x

y=-2x+14

y=-x+9

Intercept at (5,4)

8 0

3 years ago

From a survey of 100 people we found that 62 drink pepsi, 36 drink coke and 30 drink both. How many drink only Coke? How many dr

Wittaler [7]

The numbers aren't accurate, they dont add up

6 0

4 years ago