Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

_Answers_463
2780. Paraboloid of revolution. Solution. By virtue of symmetry tUe soughtfor
mirror is a surface of revolution. The coordinate origin is located in the
source of light; the x-axis is the direction of the pencil of rays. If a tangent
at any point M (x, y) of the curve, generated by the desired surface being cut
by the xi/-plane, forms with the x-axis an angle q>, and the segment connecting
the origin with the point M (x, y) forms an angle a, then tan a = tan = 2q>
= - . . But tana = ;
^ y
tancp = j/'. The desired differential equation is
1 tan 2
q>
x
2
y = */*/' 2jq/' and its solution is y* = 2Cx-\-C*. The plane section is a parabola.
The desired surface is a paraboloid of revolution. 2781. (x y) 2
C*/ = 0.
2782. x 2 = 2
C(2# + C). 2783. (2(/ x 2
)-=Cx 2 . Hint. Use the fact that the area
X
_
is equal to \ y dx. 2784. y = Cx x In |x |. 2785. f/ = Cx + x 2 . 2786. t/ =
~--* 4 . + 2787. x 2 Y 1 + !/
respect t>x and ~. 2788. x = Q/ 2
2 + arc sin x)
2
2793. = r/ xln-. 2794. x 2 ^=
?807. -
x 2 + C 2
// =2 2808. lnU| .
2 + cos y = C. Hint, The equation
. 2789. y=^+ ab~ e *
is linear with
. 2790. # =
= 0. 2799. x = 01n . 2800. + - = 1. 2801.
= C. 2803.
2 arc tan- = C. 2806. x*' =
2809. ~ + ~==C. 2810.
c. 2811. (xsiii[/-j-f/cost/~ sin^)c v =tC. 2812. (A 2 C 2 + 1
2C|/) = 0; singular integral .v 2
2 = 1/ 0. 2813. General integral
(y + C) 2 =x 8 ; there is HO singular integral. 2814. General integral "o"""^~r"^ x
x( x ~+C J=0; there is no singular integral. 2815. General integral
ry2.|-C 2 ^2Cx; singular integral x2
1 1/^3
2 = / 0. 2816. -~- sin x. 2817.
f/^-^cosxi
A
- + C,
Singular solution: t/ = 0. 2820. 4y = x 1 + p 1 , In |p x| = C+ .
2821. In V r p 2 2 + i/ + arctan~ = C, x^\n y
^ p
. Singular solution: y = e