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March 23rd, 2017, 01:23 PM  #1 
Newbie Joined: Oct 2014 From: england Posts: 22 Thanks: 0  Differential equation. Baffled!
Hi folks. Could someone help me with this question as I'm just stumped? It's in the attachment. Many thanks 
March 23rd, 2017, 02:02 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,199 Thanks: 611 
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 $m_{AP}=\dfrac{y(x)}{x2}$ 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 $\dfrac y x \cdot \dfrac {y}{x2} =  \dfrac{1}{y^\prime}$ $y^\prime =\dfrac {dy}{dx}= \dfrac{x(x2)}{y^2} =\dfrac{2xx^2}{y^2}$ 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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