Math 209 - Assignment 2 Answer Key - Course Hero Page 2
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53. Let f and g be two differentiable real valued functions. Show that any function of the
2
2
∂
z
2 ∂
z
form z = f (x + at) + g(x
at) is a solution of the wave equation
= a
.
2
2
∂t
∂x
Solution:
Let u = x + at and v = x
at. Then z = f (u) + g(v) and the Chain Rule gives
∂z
df
∂u
dg
∂u
df
dg
=
+
=
+
.
∂x
du
∂x
dv
∂x
du
dv
Thus
2
2
2
∂
z
∂
∂z
∂
df
dg
d
f
d
g
=
=
+
=
+
.
(1)
2
2
2
∂x
∂x
∂x
∂x
du
dv
du
dv
Similarly
∂z
df
∂u
dg
∂v
df
dg
=
+
= a
+ a
.
∂t
du
∂t
dv
∂t
du
dv
Thus
2
2
2
2
2
∂
z
∂
∂z
∂
df
dg
d
f
d
g
d
f
d
g
2
2
2
=
=
a
+ a
= a
+ a
= a
+
.
(2)
2
2
2
2
2
∂t
∂x
∂t
∂x
du
dv
du
dv
du
dv
From Equations (1) and (2) we get
2
2
∂
z
∂
z
2
= a
.
2
2
∂t
∂x
4. A function f is called homogeneous of degree n if it is satisfies the equation f (tx, ty) =
n
t
f (x, y) for all t, where n is a positive integer. Show that if f is homogeneous of degree
n, then
∂f
∂f
x
+ y
= nf (x, y)
∂x
∂y
[Hint: Use the Chain Rule to differentiate f (tx, ty) with respect t.]
Solution:
Let u = tx and v = ty. Then
d
n 1
(f (u, v)) = nt
f (x, y) .
dt
The Chain Rule gives
∂f
du
∂f
dv
n 1
+
= nt
f (x, y) .
∂u
dt
∂v
dt
Therefore
∂f
∂f
n 1
x
+ y
= nt
f (x, y) .
(3)
∂u
∂v
Setting t = 1 in the Equation (3):
∂f
∂f
x
+ y
= nf (x, y) .
∂x
∂y
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