1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

2 Chapter 0 Ordinary Differential Equations

This equation can be solved by isolatinguon one side and then integrating:

1

u

du

dt =k(t),

ln|u|=

∫

k(t)dt+C,

u(t)=±eCe

∫k(t)dt

=ce

∫k(t)dt

. (4)

It is easy to check directly that the last expression is a solution of the differential
equation for any value ofc.Thatis,cis an arbitrary constant and can be used
to satisfy an initial condition if one has been specified.

Example.
Solve the homogeneous differential equation

du

dt=−tu.

The procedure outlined here gives the general solution

u(t)=ce−t^2 /^2

for anyc. If an initial condition such asu( 0 )=5isspecified,thencmust be
chosen to satisfy it(c= 5 ).

The most common case of this differential equation hask(t)=kconstant.
The differential equation and its general solution are

du

dt=ku, u(t)=ce

kt. (5)

Ifkis negative, thenu(t)approaches 0 astincreases. Ifkis positive, thenu(t)
increases rapidly in magnitude witht. This kind of exponential growth often
signals disaster in physical situations, as it cannot be sustained indefinitely.

B. Second-Order Equations

It is not possible to give a solution method for the general second-order linear
homogeneous equation,

d^2 u

dt^2 +k(t)

du

dt+p(t)u=^0. (6)

Nevertheless, we can solve some important cases that we detail in what follows.
The most important point in the general theory is the following.