1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

8 Chapter 0 Ordinary Differential Equations

Here are the details. The second derivative ofuhas to be replaced by its ex-
pression in terms ofv, using the chain rule. Start by finding

du

dt=

d

dz

(

v

z

)

·dzdt.

Sincet= 1 /z,alsoz= 1 /t,anddz/dt=− 1 /t^2 =−z^2 .Thus

du

dt=−z

2

(zv′−v

z^2

)

=−zv′+v.

Similarly we find the second derivative

d^2 u

dt^2 =

d

dz

(du

dt

)dz

dt=

d

dz(−zv

′+v)(−z 2 )

=−z^2 (−zv′′−v′+v′)=z^3 v′′.

Finally, replace both terms of the differential equation:
( 1
z

) 4

z^3 v′′+λ^2

v

z=^0 ,

or

v′′+λ^2 v= 0.

This equation is easily solved, and the solution of the original is then found by
reversing the change of variables:

u(t)=t

(

c 1 cos(λ/t)+c 2 sin(λ/t)

)

. (23)

C. Second Independent Solution

Although it is not generally possible to solve a second-order linear homoge-
neous equation with variable coefficients, we can always find a second inde-
pendent solution if one solution is known. This method is calledreduction of
order.
Supposeu 1 (t)is a solution of the general equation
d^2 u
dt^2

+k(t)du

dt

+p(t)u= 0. (24)

Assume thatu 2 (t)=v(t)u 1 (t)is a solution. We wish to findv(t)so thatu 2
is indeed a solution. However,v(t)must not be constant, as that would not
supply an independent solution. A straightforward substitution ofu 2 =vu 1
into the differential equation leads to

v′′u 1 + 2 v′u′ 1 +vu′′ 1 +k(t)(v′u 1 +vu′ 1 )+p(t)vu 1 = 0.