1) Solve: ¶ z/ ¶ x
= ax + y
Soln.:
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) For the 2nd order
PDE:
am2 f" (y
+ mx) + b m f" (y + mx) + cf" (y + mx)
= 0
Let the distinct roots
be m1 and m2. What would the solutions be?
Soln. Given distinct roots of m1 and m2, the
solutions will be:
f(y + m1 x) and g(y
+ m2 x)
Where f and g are
arbitrary functions of their respective arguments. Then we can write the solution as the linear
combination:
f(y + m1 x) + g(y
+ m2 x)
3) Same PDE as (2)
but now let there be a double root, e.g. m o.
What would
the solutions be?
Soln.
We can write the solution
for the double root as:
f(y + m o x) where f is again an arbitrary function of
its argument. Further, in this case the solution: xg (y + m o x) also applies where g is an arbitrary
function. Then the solution can be
expressed as the linear combination:
f(y + m o x) + xg (y
+ m o x)
4) Apply
condition (iii) to the same equation what would the solution be?
Soln.:
Condition (iii) specifies: a = 0,
b ≠ 0 Then
the relevant quadratic equation reduces to bm + c = 0. And hence will have only
one root. Denote this single root by m1
then the solution will be f(y + m1 x)
where f is an arbitrary function of its argument. Further, it is also valid to write the full
solution as: f(y + m1 x) +
g(x)
Where g is an arbitrary function of x only.
5) 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)
No comments:
Post a Comment