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Naddika [18.5K]
3 years ago
13

Anybody good at this? x - ( 9x -10) + 11 = 12x + 3 ( -2 + 1/3)

Mathematics
1 answer:
kenny6666 [7]3 years ago
8 0
X - (9x -10) + 11 = 12x + 3( -2 + 1/3)
x - 9x + 10 + 11 = 12x - 5  ...........................(1/3-2) = -5/3 => x 3 = -5
-8x + 21 = 12x - 5
26 = 20x
x = 13/10
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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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So we distribute an integrating factor i(x,y) across both sides of the ODE to get

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

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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
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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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option 2

option 3

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