Solution to DE Problem with Undetermined Coefficients

Find the general solution of: y”’ – 3y” + 3y’ – y = 48 x exp (x)

The auxiliary equation is:

 m³  - 3m² + 3 m – 1 = 0

This can be factored on inspection to give:

(m – 1)(m – 1) (m – 1) = 0

With the triplicate repeated solution: m = 1, 1, 1.

 

 

Then the complementary function is:

 

y _c = (c o + c1x + c2 x²)  exp (x)

 

 

Normally, the assumed form of particular integral would be:

 

 

y _p  = (A + Bx) exp (x)

 

 

However, these are already contained in    y _p  so it’s necessary to multiply by  x³, i.e. the exponent must be one greater than that peculiar to y _c . So:

 

y _p  = (A x³ +B x⁴ ) exp (x)

 

 

y’ _p  =  (3A x² + 4B x³ + A x³ + B x⁴  ) exp (x) = [3A x² + (4B + A) x³ + B x⁴ ) exp (x)

y” _p  = [ 6A x + (12B  + 6A) x² + (8B + A) x³ + B x⁴ ) exp (x)

 

y”’ _p  = [ 6A  + (24B + 18A) x + (36B + 9A) x² + (12B + A) x³ + B x⁴ ] exp (x)

 

 

Now, substitute each of the above into the DE and collect like terms to get:

 

(6A  + 24 Bx) exp (x) = (0 + 48x) exp (x)

 

 

Then:  6A = 0 and 24B = 48, or A = 0 and B = 2

 

 

The general solution is then:

 

y  = (c o + c1x + c2 x² +  2 x⁴ ) exp (x)