Unformatted text preview: f sin2 (x) > 1. Never occurs.
Equation is parabolic if sin2 (x) = 1 ) x = 2 + k ⇡ , k
Equation is elliptic if sin2 (x) < 1 uy = tan(x) Z
Z 6. uxx 10uxy + 16uyy xux + sin(y )u = 0.
3
(a) 2 = , uyy 1 ,
3. uxx + a uxy1+ b = =00 c = 16 ) b2 4ac = 36 > 0. PDE is hyperbolic.
4
(b) Characteristic equations: the equation is hyperbolic. This means that
2
Notice that b
4ac = 1 > 0, so
p
we can solve it by the dy
method bof characteristics. To ﬁnd b pb2 4ac
the characteristics,
+ b2 4ac
dy
=
=
we need to solve the equations
dx
2a
dx
2a
p
p
=22
=8
dy
b± b
4ac
2± 4 3
31
Integrating:
=
=
=,.
dx
2
2
y = 2a 2 x + k 1
y = 2 8 x + k2
Solving the ﬁrst of y + 2xwe ﬁnd
these, = k1
y + 8 x = k2
dy ⇠ 3 + x,
(c) Let = = y ) 2y =⌘3= + C 8x, then=⇠x x +2, , yi.e. 3x ⌘x2y = ,K⌘. = 1 and
=8 y
=
x y + =) 2y 3 = C1 ⇠ = 1,
1
dx
2
2
all second partial derivatives are zero. Noting that d = x, e = 0, f =
1
1
sin(y ) g = 0, and
ˆ
The second ,one gives x = 6 (⌘ ⇠ ), y = 3 (4⇠ ⌘ ), we obtain g = 0 and B = 2(1)(2)(8) 10 x
((2)(1) + (8)(1)) + 2(16)(1)(1)...
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 Fall '14
 Exponential Function, Sin, Complex number, B2 Spirit

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