Higher Engineering Mathematics, Sixth Edition

516 Higher Engineering Mathematics

and integrating

∂u

∂x

partially with respect toxgives:

u=

∫

[3x^2 sin2y+f(x)]dx

=x^3 sin2y+(x)f(x)+g(y)

f(x)andg(y)are functions that may be determined

if extra information, calledboundary conditionsor

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 solutioncan be determined by direct par-

tial integration. This is demonstrated in the following

worked problems.

Problem 1. Solve the differential equation

∂^2 u

∂x^2

= 6 x^2 ( 2 y− 1 )given the boundary conditions

that atx= 0 ,

∂u

∂x

=sin2yandu=cosy.

Since

∂^2 u

∂x^2

= 6 x^2 ( 2 y− 1 )then integratingpartially with

respect toxgives:

∂u

∂x

=

∫

6 x^2 ( 2 y− 1 )dx=( 2 y− 1 )

∫

6 x^2 dx

=( 2 y− 1 )

6 x^3

3

+f(y)

= 2 x^3 ( 2 y− 1 )+f(y)

wheref(y)is an arbitrary function.

From the boundary conditions, whenx= 0 ,

∂u

∂x

=sin2y.

Hence, sin2y= 2 ( 0 )^3 ( 2 y− 1 )+f(y)

from which, f(y)=sin2y

Now

∂u

∂x

= 2 x^3 ( 2 y− 1 )+sin 2y

Integrating partially with respect toxgives:

u=

∫

[2x^3 ( 2 y− 1 )+sin2y]dx

=

2 x^4

4

( 2 y− 1 )+x(sin 2y)+F(y)

From the boundary conditions, whenx=0,

u=cosy,hence

cosy=

( 0 )^4

2

( 2 y− 1 )+( 0 )sin 2y+F(y)

from which,F(y)=cosy

Hence, the solution of

∂^2 u

∂x^2

= 6 x^2 ( 2 y− 1 )for the given

boundary conditions is:

u=

x^4

2

(2y−1)+xsiny+cosy

Problem 2. Solve the differential equation:

∂^2 u

∂x∂y

=cos(x+y)given that

∂u

∂x

=2wheny= 0 ,

andu=y^2 whenx=0.

Since

∂^2 u

∂x∂y

=cos(x+y)then integratingpartiallywith

respect toygives:

∂u

∂x

=

∫

cos(x+y)dy=sin(x+y)+f(x)

From the boundary conditions,

∂u

∂x

=2wheny= 0 ,

hence

2 =sinx+f(x)

from which, f(x)= 2 −sinx

i.e.

∂u

∂x

=sin(x+y)+ 2 −sinx

Integrating partially with respect toxgives:

u=

∫

[sin(x+y)+ 2 −sinx]dx

=−cos(x+y)+ 2 x+cosx+f(y)

Fromtheboundaryconditions,u=y^2 whenx=0,hence

y^2 =−cosy+ 0 +cos 0+f(y)

= 1 −cosy+f(y)

from which, f(y)=y^2 − 1 +cosy

Hence, the solution of

∂^2 u

∂x∂y

=cos(x+y)is given by:

u=−cos(x+y)+ 2 x+cosx+y^2 − 1 +cosy