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i tried it by finding dy/dx [dy/dx = 2A(x-A)] then comparision with above equation by finding (dy/dx)^2/y got the value of A and substitued the value of A but the answer does not tally. If any of you are solving by using determinants please explain. also let me know why my method is not giving me the solution

2006-12-08 03:22:59 · 2 answers · asked by Mathematishan 5 in Science & Mathematics Mathematics

I am looking at a solution without d^2/dy2
x,y,dy/dx, (dy/dx)^2 are admissible. mathematically speaking one constant A menas we can have a solution without going to the second derivative level.

2006-12-09 02:44:42 · update #1

2 answers

y= A (x-A)^2
differentiate
dy/dx = A 2(x-A)
square both sides

(dy/dx)^2 = 4A^2(x-A)^2
= 4Ay...1


differentiate onece again

2(dy/dx)d^2/dx^2 = 4A dy/dx

or 2(dy/dx)(d^2/dx^2) = 4A = 4(dy/dx)^2/y

or 2(d^2/dx^2) = 4(dy/dx)/y

2006-12-08 03:40:37 · answer #1 · answered by Mein Hoon Na 7 · 0 0

First of all, there are of course different diff. eq. possible that will have the same function as solution.
The easiest way I see is to differentiate twice, which gives
y' = 2A (x-A) and y'' = 2 A. Substitute these in y=A(x-A)^2 to get
4 y = y' (2x-y''), but of course other solutions are possible.

2006-12-08 11:40:52 · answer #2 · answered by cordefr 7 · 0 0

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