Please help with problem 14 and 16Solution16 y 5y 6y 18t2

Please help with problem 14 and 16

Solution

16. y\'\' + 5y\' + 6y = 18t²

The auxiliary equation is given by:

D² + 5D + 6 = 0

(D + 3)(D+2) = 0

D = -3 , D = -2

The complementary solution is given by :

y_c = C₁e^-3t + C₂e^-2t

Let the Particular solution be of the form : y_p= at²+bt+c

y\'_p = 2at + b

y\'\'_p = 2a

Substituting the values in the given equation and comparing co-efficients:

2a + 5(2at + b) + 6(at²+bt+c) = 18t²

6at² + (10a+6b) t + (2a + 5b+6c) = 0

t² : 6a = 18   ==> a = 3

t: 10a + 6b = 0 => 10 x 3 + 6b = 0 => b = -5

constant: 2a + 5b + 6c = 0

==> 2x3 + 5 x (-5) + 6c = 0

c = 19/6

The particular solution y_p= 3t² -5t + (19/6)

The general solution is given by: y = y_c + y_p = C₁e^-3t + C₂e^-2t + 3t² -5t + (19/6)