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March 23rd, 2017, 02:23 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, 03:02 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,757 Thanks: 900 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 $\dfrac 1 3 (1)^3 = (1)^2 - \dfrac 1 3 (1)^3 + C$ $\dfrac 1 3 = 1 - \dfrac 1 3 + C$ $C = -\dfrac 1 3$ $\dfrac 1 3 y^3 = x^2 - \dfrac 1 3 x^3 - \dfrac 1 3$ $y^3 = 3x^2 - x^3 -1$ $y(x) = \left( 3x^2 - x^3 -1 \right)^{1/3}$ Thanks from topsquark, nicevans1, Country Boy and 1 others

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