Solution To Variation of Parameters Differential Equation

 Problem:  

Solve using variation of parameters:

y"  - y  =  x ²

Solution:

 The complementary function is:

y _c  = c ₁ e ^x +  c ₂  e ^-x

Let  c ₁ ,  c ₂    be functions of x, i.e.

y _p    = c _{1 (x)} e ^x +  c _{2 (x)}  e ^-x

Differentiate to obtain:

y'_p  =  c ₁ e ^x -  c ₂  e ^-x  +  c '₁ e ^x +  c' ₂  e ^-x

Impose the condition that: 

    c '₁ e ^x +  c' ₂  e ^-x  = 0

With this condition imposed on  y'_p  differentiate again to get:

y''_p  =  c ₁ e ^x +  c ₂  e ^-x    +  c '₁ e ^x -  c' ₂  e ^-x

Substitute eqns, for  y''_p  and y _p    back into original complementary function eqn. to get:

c ₁ e ^x +  c ₂  e ^-x  +  c '₁ e ^x -  c' ₂  e ^-x   -  c ₁ e ^x -  c ₂  e ^-x   =  x ²

Simplifying:  

 c '₁ e ^x -  c' ₂  e ^-x   = x ²

We now have two equations, in c '₁   and c '₂  that form a linear system to be solved, i.e.

1) c '₁ e ^x +  c' ₂  e ^-x  = 0

2)  c '₁ e ^x -  c' ₂  e ^-x   = x ²

We add these two equations to obtain:

2 c '₁ e ^x = x ²

Whence:

d c ₁ / dx   =  ½ x ² e ^-x 

Or:  d c ₁  =  ½ x ² e ^-x dx

Integrate to get:

c ₁  =  - ( 1  +   x  +  ½ x ²) e ^-x 

from eqn.  (1) in the linear pair we get:

 c' ₂    =  -  c' ₁  e ^2x  =       -  ½ x ² e ^x

^and:

c ₂  =  - ( 1  +   x  +  ½ x ²) e ^x 

^{The general solution is then:}

^{y    =}  

(c ₁  +   x  -  ½ x ²) e ^x    + (c ₂  -  x  -  ½ x ²) e ^-x