Expert: Paul Klarreich - 2/28/2006

Question
Hello,
Thank you for your last solution, and for your correction on my use of English.
I, however, have two questions this time around.
1)How can i find the angle between two planes?
2)How do i solve the differential equation
(1+x^2)d^2y/dx^2 + 1 + (dy/dx)^2 = 0
Your effort is greatly appreciated.

Answer
Hi, Pharell,
Hello,
Thank you for your last solution, and for your correction on my use of English.
I, however, have two questions this time around.

1) How can I find the angle between two planes?

2) How do i solve the differential equation
(1+x^2)d^2y/dx^2 + 1 + (dy/dx)^2 = 0

Your effort is greatly appreciated.
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1. Suppose that  A1x + B1y + C1z = D1   is the equation of a plane.  Then the vector <A1, B1, C1> is perpendicular (normal) to the plane.  So you take your two equations:

A1x + B1y + C1z = D1  and  A2x + B2y + C2z = D2

and write the normal vectors:

<A1, B1, C1> and <A2, B2, C2>

The angle between these normal vectors will also be the angle between the planes.  To find the angle t between any two vectors V1 and V2, use the fact that:
       V1 dot V2
cos t = ---------
       |V1| |V2|

Yes, the numerator is the dot- or inner-product of the vectors.  
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2. For your differential equation, this is a nonlinear D.E. and I don't know of any 'method' for solving one, except by being brilliant.  (also known as being lucky)

We can do a little massaging of the equation:

(1+x^2)d^2y/dx^2 + 1 + (dy/dx)^2 = 0

into: [DON'T FORGET THE FIXED-SIZE FONT HERE.]

     -(1 + y'^2)
y'' = -----------
      (1 + x^2)

Now wouldn't it be loverly if the right side simplified nicely.  That it would do if y'^2 = x^2, or if  y' = +- x, and then the right side is -1.

If y' = +- x,  then y'' = +- 1.  Looks as if we want the minus solution.

So we can take y' = - x.  Integrating that gives

y = -x^2/2 + C as the solution(s).

OK, that's sneaky and not very helpful, but it's the best I can do.