Solutions To More Advanced Differential Operator Problems

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

y"  -  4y   + 4 = xe ^2x

Solution:

Write:

(D  -   2) ²  y   =  xe ^2x

Divide both sides by e ^2x

(D  -   2) ²  y  e ^-2x   =  x

Or: D [ e ^-2x] y =   x

Integrate twice to get, first.

 e ^-2x  y =  x ² /  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 ³ /(2)(3) =  x ³ / 6  

Then for general soln.:

 y = (x ³ /  6  +  c1x + c2 ) e ^2x

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

 (D- 1) ³  y =  xe ^x 

^{Divide both sides by e x}

^{(D  -   1) 3  y  e -x   =  x}

Or: D [ e ^-x] y =   x

Integrate twice to get, first.

 e ^-x  y =  x ² /  2  +  c1

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

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

So on next integration:

  x³ /  6  +  c1 x²  + c2

And for higher order (k) in general:

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

Then for n= 1, k = 3: 

(x^n+k )  /  (n+ 1) (n + 2)...(n  +   k)  = 

(x¹⁺³ )/(1+ 1) (1 + 2)(1  +   3)  =  x⁴ /24  

Leading to the general soln:

y = (x⁴ /24   - x³ /  6  +  c1 x 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.: