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March 23rd, 2017, 01:23 PM   #1
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Differential equation. Baffled!

Hi folks.

Could someone help me with this question as I'm just stumped? It's in the attachment.

Many thanks
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March 23rd, 2017, 02:02 PM   #2
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Consider the point $(x,y)$ that lies on the curve $y(x)$

The slope of the vector $\vec{OP}$ is given by $m_{OP}=\dfrac{y(x)}{x}$

Likewise the slope of the vector $\vec{AP}$ is $m_{AP}=\dfrac{y(x)}{x-2}$

the slope of the line tangent to $y(x)$ is of course $y^\prime(x)$ and the normal vector is given by $-\dfrac {1}{y^\prime(x)}$ (confirm this)

putting this all together and simplifying notation a bit we get

$\dfrac y x \cdot \dfrac {y}{x-2} = - \dfrac{1}{y^\prime}$

$y^\prime =\dfrac {dy}{dx}= -\dfrac{x(x-2)}{y^2} =\dfrac{2x-x^2}{y^2}$

To solve this we separate it

$y^2 ~dy = (2x - x^2)~dx$

and integrate both sides

$\dfrac 1 3 y^3 = x^2 - \dfrac 1 3 x^3 + C$

plugging in the point $(1,1)$ we get

$\dfrac 1 3 (1)^3 = (1)^2 - \dfrac 1 3 (1)^3 + C$

$\dfrac 1 3 = 1 - \dfrac 1 3 + C$

$C = -\dfrac 1 3$

$\dfrac 1 3 y^3 = x^2 - \dfrac 1 3 x^3 - \dfrac 1 3$

$y^3 = 3x^2 - x^3 -1$

$y(x) = \left( 3x^2 - x^3 -1 \right)^{1/3}$
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