## Thursday, November 10, 2022

### Solutions To More Advanced Differential Operator Problems

1) Solve the differential equation below using an appropriate differential operator:

y"  -  4y   + 4 = x2x

Solution:

Write:

( -   2) 2  y   =  xe 2x

Divide both sides by e 2x

( -   2) 2  y  e -2x   =  x

Or: D [ e -2x] y =   x

Integrate twice to get, first.

e -2x  y =  x /  2  +  c1

But Rem (See original post from Oct. 12th):

(D-2  x n )   =  ( xn+1 )   (n + 1) (n +2)

So on next integration (for n=1):

( xn+1 )   (n + 1) (n +2)  =  x /(2)(3) =  x / 6

Then for general soln.:

y = (x /  6  +  c1x + c2 ) e 2x

2) Solve the operator form of the differential equation below:

(D- 1) 3  y =  xx

Divide both sides by e x

( -   1) 3  y  -x   =  x

Or: D [ e -x] y =   x

Integrate twice to get, first.

e -x  y =  x /  2  +  c1

But Rem (See original post from Oct. 12th):

(D-2  x n )   =  ( xn+1 )   (n + 1) (n +2)

So on next integration:

x/  6  +  c1 x + c2

And for higher order (k) in general:

(D-k  x n )   = (xn+k ) / (n+ 1) (n + 2)...(n  +   k)

Then for n= 1, k = 3:

(xn+k )  /  (n+ 1) (n + 2)...(n  +   k)  =

(x1+3 )/(1+ 1) (1 + 2)(1  +   3)  =  x/24

y = (x/24   x3 /  6  +  c1 2  + c2 e x

3)  Consider the differential equation:

dy2/dx2  + k2 y = 0

a) Show the equation in operator form, with operator specified.

Soln.  This DE can also be written in operator form:

(D 2  +  k2) y    = 0

Where: D =  d/ dx  (y )  or:  d(y) / dx

b) Obtain the general solution from factorization of the full operator differential equation.

Soln.  On factoring:

(D - ik) (D + ik)y = 0

Which has the general soln.: