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3 years ago

PLEASE HELP ASAP!!!! 100% CORRECT ANSWERS ONLY PLEASE!!!!

Mathematics

2 answers:

Setler [38]3 years ago

7 0

That would be the last choice.
The formula for the circumference = 2 pi r.

jolli1 [7]3 years ago

6 0

Your answer is
( Find the length of the diameter and the length of the circumference of the circle. Divide the length of the circumference by the length of the diameter. Set up an equation showing the ratio of the circumference to the diameter equal to π . Then rearrange the equation by solving it for the circumference. Substitute 2 times the radius for the diameter. )
give me brainlyesttt!!!!!

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Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

3 years ago