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2 years ago

PLEASE HELP

Mathematics

1 answer:

Lera25 [3.4K]2 years ago

3 0

Answer:

To construct the Square using a compass and straight edge, you have written the steps not in proper order. I am rearranging the steps for you

1. Construct horizontal H J¯¯¯¯¯

2. Construct a circle with point H as the center and a circle with point J as a center with each circle having radius HJ .

3.Label the point of intersection of the two circles above HJ¯¯¯¯¯ , point K, and the point of intersection of the two circles below HJ¯¯¯¯¯ , point L.

4.Construct KL¯¯¯¯¯ , the perpendicular bisector of HJ¯¯¯¯¯ , intersecting HJ¯¯¯¯¯ at point M.

5.Construct a circle with point M as the center with radius MJ .

6.Label the point of intersection of circle M and KL¯¯¯¯¯ closest to point K, point N, and the point of intersection of circle M and KL¯¯¯¯¯ closest to point L, point O.

7.Construct HN¯¯¯¯¯¯ , NJ¯¯¯¯¯ , JO¯¯¯¯¯ , and OH¯¯¯¯¯¯ to complete square HNJO .

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$\underbrace{y(7xy^2+6)}_{M(x,y)}\,\mathrm dx+\underbrace{x(xy^2-1)}_{N(x,y)}\,\mathrm dy=0$

For the ODE to be exact, we require that $M_y=N_x$ , which we'll verify is not the case here.

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$N_x=2xy^2-1$

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$iM\,\mathrm dx+iN\,\mathrm dy=0$

Now for the ODE to be exact, we require $(iM)_y=(iN)_x$ , which in turn means

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so that our integrating factor is $i(x,y)=x^5y^{-2}$ . Our ODE is now

$(7x^6y+6x^5y^{-1})\,\mathrm dx+(x^7-x^6y^{-2})\,\mathrm dy=0$

Renaming $M(x,y)$ and $N(x,y)$ to our current coefficients, we end up with partial derivatives

$M_y=7x^6-6x^5y^{-2}$
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as desired, so our new ODE is indeed exact.

Next, we're looking for a solution of the form $\Psi(x,y)=C$ . By the chain rule, we have

$\Psi_x=7x^6y+6x^5y^{-1}\implies\Psi=x^7y+x^6y^{-1}+f(y)$

Differentiating with respect to $y$ yields

$\Psi_y=x^7-x^6y^{-2}=x^7-x^6y^{-2}+\dfrac{\mathrm df}{\mathrm dy}$
$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

Thus the solution to the ODE is

$\Psi(x,y)=x^7y+x^6y^{-1}=C$

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