Submitted by maths123 on Sat, 04/25/2015 - 08:17
Forums:
After differentiating * in (3ii),
I obtain y''=0 and
$\ a^2 y' + x(y-xy')\ $ = 0
the above is a differential equation
$\ y' = \frac {xy}{x^2 - a^2}\ $
I get the solution as
$\ y^2 = A (x^2 - a^2) \ $
I have tried substituting the above into * given in question but this has not helped to figure out the constant
How do I show my solution for the differential equation is x^2 + y^2 = a^2 - as given in question ?
Have you considered looking
Have you considered looking back at the property given in the first part of 3(ii): i.e $C$ has the property that the distance $a$ is the same for all points $P$ on $C$?
Isn't this property
Isn't this property equivalent to the * equation - since I used this property to actually derive * ?
Hi
Hi,
I like your method so far. I presume you got the equation of the straight line by substituting back into (*). It is possible (I'm not claiming it is the most efficient way) to find $A$ by substituting back into (*) also.
Every curve satisfying (*) must satisfy the resulting differential equations, but not every curve satisfying the resulting differential equations must satisfy (*), essentially because when we differentiate (*) we lose information about $ y $. So when you substitute the solution back into (*) you are able to find which values of the constant are valid.
To make substituting back into (*) easier perhaps you might like to try differentiating $y^2 = A (x^2-a^2)$ implicitly first (i.e. $ y' = Ax/y $) and then substitute since it avoids square roots (i.e. $ y' = Ax/y $).
Thank you; Substituting back
Thank you; Substituting back in * using your method, I get A=-1 as required