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Elina [12.6K]
3 years ago
15

The product of 2x^2+6x-8 and x+3 in standard form

Mathematics
1 answer:
anzhelika [568]3 years ago
6 0

Answer:

2x³ + 12x² + 10x - 24

Step-by-step explanation:

(2x² + 6x - 8)(x + 3)                         Distribute

2x³ + 6x² + 6x² + 18x - 8x - 24      Combine like terms

2x³ + 12x² + 10x - 24                      This expression is in standard form

If this answer is correct, please make me Brainliest!

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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
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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.

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}
N_x=7x^6-6x^5y^{-2}

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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