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3 years ago

13

What is the answer to this problem? Thank you!

1 answer:

Misha Larkins [42]3 years ago

8 0

what are the statements?

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Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

Now, substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0$

Using the characteristic equation:

$\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0$

Factor out $e^{\lambda x}$

$e^{\lambda x}(\lambda ^2 +4) =0$

Where:

$e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda$

Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

$y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2$

Evaluating $y'(0)=2$ :

$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

$y(x)=sin(2x)+2cos(2x)$

5 0

3 years ago

Each side of a square is increasing at a rate of 5 cm/s. at what rate is the area of the square increasing when the area of the

topjm [15]

<span>This problem is solved using the chain rule. the area of the square is f(t) and the length of the side is g(t) f(t)=g(t)^2 g'(t)=5 Using the chain rule f'(t)=2*g(t)*g'(t) The value of g(t) is sqrt(49) which is 7. g'(t) is given as 5 cm/s f'(t)=2*7*5=14*5=70cm^2/s</span>

3 0

2 years ago

Lua is creating a rectangular prism. The base of her prism is shown below. She plans to have a height of 7 cubes.

jenyasd209 [6]

Answer:

Step-by-step explanation:

7 0

2 years ago

Please HELP THIS IS A TEST AND IT NEEDS TO GET DONE

bonufazy [111]

Wouldn’t it be the same as Z? Lol

7 0

3 years ago

Find all solutions of each equation on the interval 0 ≤ x<2π.

Aleksandr [31]

Answer:

$x = 0 \: or \:x = \pi$

Step-by-step explanation:

The given equation is

$\tan^{2} (x) \sec^{2} (x) + \sec^{2} (x) - \tan^{2} (x) = 2$

Subtract 2 from both sides and factor by grouping to get:

$\sec^{2} (x)(\tan^{2} (x) + 2) -1( \tan^{2} (x) + 2)$

$(\tan^{2} (x) + 2) (\sec^{2} (x) - 1) = 0$

By the zero product principle:

$(\tan^{2} (x) + 2) = 0 \: or \: (\sec^{2} (x) - 1) = 0$

$(\tan^{2} (x) = - 2 \: or \: \sec^{2} (x) = 1$

When

$\sec^{2} (x) = 1 \implies\cos^{2} (x) = 1$

$\implies \: \cos(x) = \pm1$

This implies

$x = 0 \: or \:x = \pi$

When

${ \tan }^{2} (x) = - 2$

We have

$\tan(x) = \pm \sqrt{ - 2}$

hence x is not defined for all real numbers

6 0

2 years ago

Read 2 more answers