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Snezhnost [94]
3 years ago
10

Does 4x+5y=0 represent a direct variation ?

Mathematics
2 answers:
omeli [17]3 years ago
7 0

4x+5y=0

subtract 4x from each side

5y = -4x

divide by 5

y = -4/5 x

y = kx where k = -4/5

this is a direction variation

trapecia [35]3 years ago
6 0

\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 4x+5y=0\implies 5y=-4x\implies y=-\cfrac{4}{5}x\qquad \boxed{k=-\cfrac{4}{5}}\qquad \text{\Large\checkmark}

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2 years ago
Solve the Inequality.<br>-9-3x &gt; 2(25 + 2x) + 4<br>The solution is​
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3 years ago
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Step-by-step explanation:

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Determine the values of the constants r and s such that i(x, y) = x rys is an integrating factor for the given differential equa
garri49 [273]
\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.

M_y=21xy^2+6
N_x=2xy^2-1

So we distribute an integrating factor i(x,y) across both sides of the ODE to get

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

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

i_yM+iM_y=i_xN+iN_x\implies i(M_y-N_x)=i_xN-i_yM

Suppose i(x,y)=x^ry^s. Then substituting everything into the PDE above, we have

x^ry^s(19xy^2+7)=rx^{r-1}y^s(x^2y^2-x)-sx^ry^{s-1}(7xy^3+6y)
19x^{r+1}y^{s+2}+7x^ry^s=rx^{r+1}y^{s+2}-rx^ry^s-7sx^{r+1}y^{s+2}-6sx^ry^s
19x^{r+1}y^{s+2}+7x^ry^s=(r-7s)x^{r+1}y^{s+2}-(r+6s)x^ry^s
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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
4 0
3 years ago
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Step-by-step explanation:

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