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4 years ago

Please help with this

Mathematics

1 answer:

Sunny_sXe [5.5K]4 years ago

6 0

Answer:

Step-by-step explanation:

The triangle is reflected across the y-axis and moved down 3 units.

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3 years ago

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

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

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3 years ago

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The answer is $90x because 65+25 = 90

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2 years ago

If i make 1.5 million dollars per day, plus an extra 250k interest every 31 hours, how much money do i make per day?

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Answer: Alot

Step-by-step explanation:

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3 years ago

It is believed that 11% of all Americans are left-handed. A college needs to know the number of left-handed desks to place in th

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Answer:

a) State the random variable

Random variable : x

which refers to a randomly selected student from the college that is left-handed.

b) state population parameter

population parameter : P

which is the percentage of all students from the college that are left handed

c) state the hypotheses

The hypothesis are;

Null hypothesis H₀ : p = 0.11

Alternative hypothesis H₁ : p > 0.11

d) State the Type I error in the context of this problem.

Type - I Error: Rejecting that the % of all the students from the college that are left-handed is 11% when actually the % is really 11%

(Reject H₀ when H₀ is true)

e) State the Type 11 error in the context of this problem

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(Fail to reject H₀ when H₀ is false)

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3 years ago