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 Differential Equations Ordinary and Partial Differential Equations Math Forum

March 23rd, 2017, 01:23 PM   #1
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Joined: Oct 2014
From: england

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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
Attached Images IMG_20170323_212141_888.jpg (94.6 KB, 18 views) March 23rd, 2017, 02:02 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,427 Thanks: 1314 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 Tags baffled, differential, equation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jiasyuen Calculus 7 March 26th, 2015 01:21 PM Sonprelis Calculus 6 August 6th, 2014 10:07 AM PhizKid Differential Equations 0 February 24th, 2013 10:30 AM tsl182forever8 Differential Equations 2 March 14th, 2012 10:15 PM tsl182forever8 Differential Equations 2 March 14th, 2012 03:12 PM

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