Mark earned $1340 for a 40-hour week. Use the formula P = RH where (P is the pay,

Mathematics

1 answer:

algol [13]2 years ago

6 0

Mark earns 33.5 dollars per hour.

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Consider the differential equation <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D6%20y%5E%7

Nutka1998 [239]

As a Bernoulli equation:

$x^2\dfrac{\mathrm dy}{\mathrm dx}=6y^2+6xy\iff x^2y^{-2}\dfrac{\mathrm dy}{\mathrm dx}-6xy^{-1}=6$

Let $z=y^{-1}\implies\dfrac{\mathrm dz}{\mathrm dx}=-y^{-2}\dfrac{\mathrm dy}{\mathrm dx}$ . The ODE becomes

$-x^2\dfrac{\mathrm dz}{\mathrm dx}-6xz=6$
$x^6\dfrac{\mathrm dz}{\mathrm dx}+6x^5z=-6x^4$
$\dfrac{\mathrm d}{\mathrm dx}[x^6z]=-6x^4$
$x^6z=-6\displaystyle\int x^4\,\mathrm dx$
$x^6z=-\dfrac65x^5+C$
$z=-\dfrac6{5x}+\dfrac C{x^6}$
$y^{-1}=-\dfrac6{5x}+\dfrac C{x^6}$
$y=\dfrac1{\frac C{x^6}-\frac6{5x}}$
$y=\dfrac{5x^6}{C-6x^5}$

With $y(3)=6$ , we get

$6=\dfrac{5(3)^6}{C-6(3)^5}\implies C=\dfrac{4131}2$

so the solution is

$y=\dfrac{5x^6}{\frac{4131}2-6x^5}=\dfrac{10x^6}{4131-12x^5}$

which is valid as long as the denominator is not zero, which is the case for all $x\neq\sqrt[5]{\dfrac{4131}{12}}$ .