Quod Erat Demonstrandum

2014/06/11

M2 某題

Filed under: Additional / Applied Mathematics,NSS — johnmayhk @ 9:06 上午
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M2 學生問以下一題

The slope at any point (x,y) of a curve is given by

$\frac{dy}{dx}=y(2x+9)$.

If the curve lies above the x-axis, and it passes (0,8), find the equation of the curve.

$\frac{dy}{y}=(2x+9)dx \Rightarrow \ln y=\int (2x+9)dx$　（$y > 0$

$\frac{d(\ln y)}{dx}=?$

$\frac{d(\ln y)}{dx}=\frac{1}{y}\frac{dy}{dx}$

$\frac{dy}{dx}=y(2x+9)$

$y\frac{d(\ln y)}{dx}=y(2x+9)$

$\frac{d(\ln y)}{dx}=2x+9$

$\ln y=\int (2x+9)dx$

$\dots$