1) Solve: ¶ z/ ¶ x = ax + y
Solution: Since the differentiation is with respect to x only, we can treat y as a constant and separate the variables thus:
dz = (ax + y) dx
On integration we obtain the general solution:
z = 1/2 a x 2 + xy + y (y)
2) Solve: ¶ 2 u/ ¶ x2 - 4 [ ¶ 2 u / ¶ x ¶ y] + 4[ ¶ 2 u/ ¶ y 2 ] = 0
Solution: The auxiliary (quadratic) equation corresponding to the differential equation is:
m2 - 4 m + 4 = 0
Factoring yields: (m - 2) (m - 2) = 0
This discloses a double (e.g. repeat) root of m = 2
So the general solution is written in terms of two arbitrary functions,
each with (y + 2x):
Thus: u = f( y + 2x) + xg(y + 2x)
3) Solve the following higher order partial DE:
¶ 4 z/ ¶ x4 - 6 [ ¶ 4 z/ ¶ x 3 ¶ y] + 14 [¶ 4 z/ ¶ x2 ¶ y2 ] -
16 [ ¶ 4 z/ ¶ x ¶ y 3 ] + 8 [ ¶ 4 z/ ¶ y 4 ] = 0
Solution: Again we write out the auxiliary (quadratic) equation corresponding to the DE:
m4 - 6 m3 + 14 m2 - 16 m + 8 = 0
Factoring yields:
(m - 2) (m - 2) (m - (1 +i))(m - (1 - i))
so m = 2, 2, 1 + i, 1 - i
Hence, the general solution (noting the repeat root of m = 2) is:
z = j o (y + 2 x) + x y o (y + 2 x) + j1 (y + x + ix) + j1 (y + x - ix)
+ i[ y1 (y + x + ix) - y1 (y + x - ix)]
3) Consider a wire of length ℓ :
]-------------- ℓ ---------------------->[
for which the relevant wave equation is:
¶ 2 u/ ¶ t 2 = c 2 [¶ 2 u/ ¶ x2 ]
If the displacement satisfies:
u(0, t) = u(ℓ , t ) = 0 ( t > 0)
Suppose at time t= 0 the displacement is u(x,0) = f(x)
And: ¶ u/ ¶ t ] t= 0 = g(x)
Using the technique of separation of variables write two different general solutions for X( ℓ ) and state why one is unacceptable
Hints: i) u(0, t) = X(0) T(t) and
(ii) u(ℓ , t ) = X( ℓ ) T(t)
---------
Solution: By separation of variables we have:
u(x, t) = X(x) T(t)
Now let: dT/ dt = T ', d 2 T/ dt 2 = T"
dX/ dt = X ' d 2X/ dt 2 = X "
So that: ¶ 2 u/ ¶ t 2 = X(t) T "(t)
And:
¶ 2 u/ ¶ t 2 = X" (x) T (t)
The wave equation can then be written:
XT " = c 2 X" T
Or: T "/ c 2 T = X" / X
We set both sides equal to a constant K (separation constant):
T "/ c 2 T = K = X" / X
Two equations result:
i) X" - K X = 0
ii) T " - c 2 KT = 0
We know (hints):
a) u(0, t) = X(0) T(t) = 0
b) u(ℓ , t ) = X( ℓ ) T(t) = 0
Look at two cases:
i) K = 0
Then X" - KX = 0 becomes X" = 0
For which: X(x) = c1 X + c2
And: X(0) = 0 è c2 = 0 so X(x) = c1 X
However, X(x) = 0 ⇒ c1 = 0 and X( ℓ ) = 0
Which is unacceptable, so K cannot be zero. So next look at K > 0
ii) K > 0
Then: Let K = r 2
So that: X"(x) - r 2 X = 0
Has solution: c1 exp (r x) + c2 exp (-r x)
After eliminating c2 (as fn. of c1)
X(x) = c1 [exp ( r x) - exp (-r x) ] ⇒
X(ℓ) = c1 [exp ( r ℓ) - exp (-r ℓ) ] = 2 c1 sin n r ℓ
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