Question

Solution of the D.E y’’ + 3y’ + 2y = 12x^2 when solved using the method of undetermined coefficients is ________

(a) y = c1 e^x +  c2 e^2x +  2 –   11x +  x^2

(b) y = c1 e^ – x +  c2 e^ – 2x +  18 +  21x +  3x^2

(c) y = c1 e^x +  c2 e^ – 2x +  11 +  18x +  2x^2

(d) y = c1 e^ – x +  c2 e^ – 2x +  21 –   18x +  6x^2

I got this question in an interview.

My doubt stems from Method of Undetermined Coefficients topic in portion Linear Differential Equations – Second and Higher Order of Engineering Mathematics

Answer 1

The correct answer is (d) y = c1 e^ – x +  c2 e^ – 2x +  21 –   18x +  6x^2

Explanation: We have (D^2 + 3D + 2)y = 12x^2

A.E is m^2 + 3m + 2 = 0 –> (m + 1)(m + 2) = 0 –> m =  – 1, – 2

yc = c1 e^ – x +  c2 e^ – 2x and ∅(x) =  12x^2 and 0 is not a root of the A.E,

we assume P.I in the form yp = a + bx + cx^2….(1)to find a,b & c such that

yp’’ + 3yp’ + 2yp = 12x^2….(2), yp‘ = b + 2cx, yp” = 2c now (2) becomes

2c + 3(b + 2cx) + 2(a + bx + cx^2) = 12x^2

(2a + 3b + 2c) + (2b + 6c)x + (2c)x^2 = 12x^2

2a + 3b + 2c = 0, 2b + 6c = 0, 2c = 12 –> c = 6, b =  – 18, a = 21 hence (1) becomes

yp = 21 – 18x + 6x^2 thus complete solution is

y = yc + yp –> c1 e^ – x +  c2 e^ – 2x +  21 –   18x +  6x^2.