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3 years ago

HELP I NEED HELP ASAP

Mathematics

1 answer:

Orlov [11]3 years ago

5 0

Answer:

Step-by-step explanation:

Process of elimination shows that the most reasonable answer is A.

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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

Find a decomposition of a=⟨−5,−1,1⟩ into a vector c parallel to b=⟨−6,0,6⟩ and a vector d perpendicular to b such that c+d=a.

dezoksy [38]

The projection of vector A parallel to vector B is $\langle -3, 0, 3\rangle$ and the projection of vector A perpendicular to vector B is $\langle -2, -1, -2\rangle$ .

In this question, we need to determine all projections of a vector with respect to another vector. In this case, the projection of vector A parallel to vector B is defined by this formula:

$\vec a_{\parallel , \vec b} = \frac{\vec a \,\bullet\,\vec b}{\|\vec b\|^{2}}\cdot \vec b$ (1)

Where $\|\vec b\|$ is the norm of vector B.

And the projection of vector A perpendicular to vector B is:

$\vec a_{\perp, \vec b} = \vec a - \vec a_{\parallel, \vec b}$ (2)

If we know that $a = \langle -5, -1, 1 \rangle$ and $\vec b = \langle -6, 0, 6 \rangle$ , then the projections are now calculated:

$\vec a_{\parallel, \vec b} = \frac{(-5)\cdot (-6)+(-1)\cdot (0)+(1)\cdot (6)}{(-6)^{2}+0^{2}+6^{2}} \cdot \langle -6, 0, 6 \rangle$

$\vec a_{\parallel, \vec b} = \frac{1}{2}\cdot \langle -6, 0, 6 \rangle$

$\vec a_{\parallel, \vec b} = \langle -3, 0, 3\rangle$

$\vec a_{\perp, \vec b} = \langle -5, -1, 1 \rangle - \langle -3, 0, 3 \rangle$

$\vec a_{\perp, \vec b} = \langle -2, -1, -2\rangle$

We kindly invite to check this question on projection of vectors: brainly.com/question/24160729

7 0

3 years ago

For his business, gil has determined that the time it takes to finish a job varies iversely with the number of workers. this can

Vinvika [58]

I believed its 4 days

5 0

4 years ago

Let $PQR$ be a triangle such that $\angle P = \angle Q + \angle R.$ Among all three exterior angles of triangle $PQR,$ what is t

saw5 [17]

Answer:

The exterior angle to the vertex P, measuring 90°

Step-by-step explanation:

The sum of the internal angles of any triangle needs to be equal 180 degrees. So, we can write two equations:

mP = mQ + mR

mP + mQ + mR = 180°

Substituting (mQ + mR) by mP in the second equation, we have:

mP + mP = 180°

mP = 90°

this right angle is the bigger angle in the triangle, because the sum of all three angles needs to be 180°.

Each exterior angle is the supplementary angle of the respective internal angle, so the bigger the internal angle, the smaller the external angle.

So if the bigger internal angle is mP = 90°, the smaller exterior angle is 180 - 90 = 90°.

5 0

3 years ago

Ashley recently opened a store that uses only natural ingredients. She wants to advertise her products by distributing bags of s

Elenna [48]

I will rewrite the question for better understanding:

Ashley recently opened a store that uses only natural ingredients. She wants to advertise her products by distributing bags of samples in her neighborhood. It takes Ashley 2/3 of a minute to prepare one bag. It takes each of her friends 75% longer to prepare a bag. How many hours will it take Ashley and 4 of her friends to prepare 1575 bags of samples?

Answer:

5.3 hours

Explanation:

1) Time it takes Ashley to preprate one bag:

2/3 min

2) Time it takes each friend of Ashley: 75% more than 2/3 min

75% × 2/3 min = 0.75 × 2/3 min = 3/4 × 2/3 min= 2/4 min = 1/2 min = 0.5 min

2/3 min + 1/2 min = 7/6 min

3) Time it takes Ashley and the 4 friends working along to prepare one bag:

Convert each time into a rate, since you can set that the total rate of Ashley along with her four friends is equal to the sum of each rate:

Rate of Ashley: 1 bag / (2/3) min = 3/2 bag/min

Rate of each friend: 1 bag / (7/6) min = 6/7 bag/min

Rate of Ashley and 4 friends = 3/2 bag/min + 4 × 6/7 bag/min = (3/2 +24/7) bag/min = 69/14 bag/min

4) Time of prepare 1575 bags of samples:

time = number of bags / number of bags per min = 1,575 bags / (69/14) bags/min = 319.56 min

5) Convert minutes to hours:

356.56 min × 1 hour / 60 min = 5.3 hours

3 0

3 years ago