$\mathfrak{Differentiate} \space\mathfrak{my} \space \mathfrak{Day!!}$

Calculus Level 2

Isolate: $\large \displaystyle{\frac{dy}{dx} = \sqrt{xy}}$ .

y = \frac{4x^3+12x^{\frac{3}{2}}K +K}{36}

y = \frac{4x^3+12x^{\frac{2}{2}}K +K}{36}

y = \frac{3x^3+12x^{\frac{3}{2}}K +K}{36}

y = \frac{4x^3-12x^{\frac{3}{2}}K +K}{36}

1 solution

Brian Charlesworth
Mar 26, 2016

Separating variables and then integrating each side of the resulting equation yields that

$\dfrac{dy}{\sqrt{y}} = \sqrt{x} dx \Longrightarrow 2\sqrt{y} = \dfrac{2}{3}x\sqrt{x} + K \Longrightarrow y = \left(\dfrac{x\sqrt{x}}{3} + \dfrac{K}{2}\right)^{2}$

$\Longrightarrow y = \dfrac{x^{3}}{9} + \dfrac{Kx\sqrt{x}}{3} + \dfrac{K^{2}}{4} \Longrightarrow y = \boxed{\dfrac{4x^{3} + 12Kx\sqrt{x} + 9K^{2}}{36}}$ .

@Cedie Camomot I think that to be consistent, having the $K$ in the $12Kx\sqrt{x}$ term will require the answer to have the $9K^{2}$ constant term in place of the $K$ constant term in the numerator.

Brian Charlesworth - 5 years, 2 months ago

According to my research for example if you isolate the $y = \dfrac{4x^3+12Kx\sqrt{x} + 9K^2}{36}$ . So the $9K^2$ term make it into simple term $9K^2 = K$ . Make $9K^2$ as a constant because $9K^2$ is a coefficient. That is $y = \dfrac{4x^3+12Kx\sqrt{x} + K}{36}$ . Hence I consider your answer well and thanks for response.

A Former Brilliant Member - 5 years, 2 months ago

D i f f e r e n t i a t e m y D a y ! ! \mathfrak{Differentiate} \space\mathfrak{my} \space \mathfrak{Day!!} D i f f e r e n t i a t e m y D a y ! !

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$\mathfrak{Differentiate} \space\mathfrak{my} \space \mathfrak{Day!!}$