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3 years ago

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A homogeneous second-order linear differential equation, two functions y1 and y2 , and a pair of initial conditions are given be

low. First verify that y1 and y2 are solutions of the differential equation. Then find a particular solution of the form
y = c1y1 + c2y2 that satisfies the given initial conditions.
y'' + 49y = 0; y1 = cos(7x) y2 = sin(7x); y(0) = 10 y(0)=-4
y(x)=?

1 answer:

Basile [38]3 years ago

4 0

Answer:

Step-by-step explanation:

Check part

$y= C_1y_1 + C_2y_2 = C_1cos(7x)+C_2sin(7x)$

$y'= -7 C_1sin(7x)+7C_2cos(7x)$

$y"= -49 C_1cos(7x) - 49 C_2sin(7x)$

Now, replace to the original one.

$y"+49 y = -49C_1cos(7x)-49 C_2 sin(7x) + 49 C_1cos(7x) +49 C_2sin(7x) = 0\\$

Done!!

Particular solution

$y(0) = C_1cos(0) + C_2 sin(0) = C_1= 10$

I believe that y'(0) = 4, not y(0) anymore. Since y(0) CANNOT have two different solution.

$y(0)'= -7 C_1sin(0) + 7 C_2 cos (0) = 7 C_2= -4$

$C_2 = -4/7$

The last step is to put C1, C2 into your solution. You finish it.

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How to know if a function is periodic without graphing it ?

zhenek [66]

A function $f(t)$ is periodic if there is some constant $k$ such that $f(t+k)=f(k)$ for all $t$ in the domain of $f(t)$ . Then $k$ is the "period" of $f(t)$ .

Example:

If $f(x)=\sin x$ , then we have $\sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x$ , and so $\sin x$ is periodic with period $2\pi$ .

It gets a bit more complicated for a function like yours. We're looking for $k$ such that

$\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5$

Expanding on the left, you have

$\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2$

and

$1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5$

It follows that the following must be satisfied:

$\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}$

The first two equations are satisfied whenever $k\in\{0,\pm4,\pm8,\ldots\}$ , or more generally, when $k=4n$ and $n\in\mathbb Z$ (i.e. any multiple of 4).

The second two are satisfied whenever $k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}$ , and more generally when $k=\dfrac{10n}7$ with $n\in\mathbb Z$ (any multiple of 10/7).

It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when $k$ is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

Let's verify:

$\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2$

$\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5$

More generally, it can be shown that

$f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))$

is periodic with period $\mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n)$ .

4 0

3 years ago

1 2 3 4 .... 1002 what is the sum of the sequence

Reika [66]

<span>As far as i know it is related to Gauss.
Write the sequences forward and backward first.

1 +2 +3 +.....+1002
1002+1001+1000+.....+1
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=1002x1003
But this contains the series twice.
So, the sum is = 1002x1003/2=501x1003=502503. answer</span>

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2 years ago

13 cm<br> L<br> 5 cm<br> Calculate the perimeter

aksik [14]

Answer:

36

Step-by-step explanation:

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3 years ago

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What is the discriminant of 3 x squared minus 10 x = negative 2?<br> 76<br> 94<br> 106<br> 124

abruzzese [7]

Answer:

76

Step-by-step explanation:

The discriminant is $b^{2} - 4ac$ part of the quadratic formula.

a term is 3

b term is -10

c term is 2 (add it to the side of the 3 x squared minus 10 x)

Plug the values in!

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3 years ago

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xxMikexx [17]

I think the answer is B. because the sum of the 2 smaller numbers is greater than the 3rd number.

a. 8+6 = 14 ; 14 is less than 16 ; wrong
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Hope this helped☺☺

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3 years ago