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:

am^{2} 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/ ¶ x^{2}
- 4 [ ¶ ^{2} u / ¶ x ¶ y]
+ 4[ ¶ ^{2} u/ ¶ y ^{2} ]
= 0

**Solution**:
The auxiliary (quadratic) equation corresponding to the differential equation
is:

m^{2}
- 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)

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