516 Higher Engineering Mathematics
and integrating
∂u
∂x
partially with respect to x gives:
u =
[3x
2 sin 2y + f (x)] dx
= x
3 sin 2y + (x)f (x) + g(y)
f (x) and g(y) are functions that may be determined
if extra information, called boundary conditions or
initial conditions, are known.
53.3 Solution of partial differential
equations by direct partial
integration
The simplest form of partial differential equations
occurs when a solution can be determined by direct partial integration. This is demonstrated in the following
worked problems.
Problem 1. Solve the differential equation
∂ 2 u
∂x 2 = 6x 2 (2y − 1) given the boundary conditions
that at x = 0,
∂u
∂x
= sin 2y and u =cos y.
Since
∂ 2 u
∂x 2 = 6x 2 (2y − 1) then integrating partially with
respect to x gives:
∂u
∂x
=
6x
2
(2y − 1)dx = (2y − 1)
6x
2 dx
= (2y − 1)
6x 3
3
+ f ( y)
= 2x
3
(2y − 1) + f ( y)
where f (y) is an arbitrary function.
From the boundary conditions, when x = 0,
∂u
∂x
= sin 2y.
Hence,
sin 2y = 2(0) 3 (2y − 1) + f ( y)
from which,
f ( y) = sin 2y
Now
∂u
∂x
= 2x 3 (2y − 1) + sin 2y
Integrating partially with respect to x gives:
u =
[2x
3
(2y − 1) + sin 2y]dx
=
2x 4
4
(2y − 1) + x(sin 2y) + F( y)
From the boundary conditions, when x = 0,
u = cos y, hence
cos y =
(0) 4
2
(2y − 1) + (0)sin 2y + F( y)
from which, F( y) = cos y
Hence, the solution of
∂ 2 u
∂x 2 = 6x 2 (2y − 1) for the given
boundary conditions is:
u =
x
4
2
(2y − 1) + x sin y + cos y
Problem 2. Solve the differential equation:
∂ 2 u
∂x∂ y
= cos(x + y) given that
∂u
∂x
= 2 when y = 0,
and u = y
2 when x = 0.
Since
∂ 2 u
∂x∂ y
= cos(x + y) then integrating partially with
respect to y gives:
∂u
∂x
=
cos(x + y)d y = sin(x + y) + f (x)
From the boundary conditions,
∂u
∂x
= 2 when y = 0,
hence
2 = sin x + f (x)
from which, f (x) = 2 − sin x
i.e.
∂u
∂x
= sin(x + y) + 2 − sin x
Integrating partially with respect to x gives:
u =
[sin(x + y) + 2 − sin x]dx
= −cos(x + y) + 2x + cos x + f (y)
From the boundary conditions, u = y 2 when x = 0, hence
y
2
= −cos y + 0 + cos 0 + f ( y)
= 1 − cos y + f ( y)
from which, f (y) = y
2
− 1 + cos y
Hence, the solution of
∂ 2 u
∂x∂ y
= cos(x + y) is given by:
u = −cos(x + y) + 2x + cos x + y
2
− 1 + cos y