Use the method of variation of parameters when solving part b.
a. Verify that y_1( t)=e^t and y_2( t) = t are solutions to the homogeneous equation.
b. Find a particular solution to the differential equation.
1 Answer

a) Direct substitution yields
(1  t) e^t + t e^t  e^t = 0, and (1  t) * 0 + t * 1  t = 0.

b) Assume that y = e^t * u + t * v.
Differentiating,
y' = e^t u + e^t u' + v + tv'.
Setting e^t u' + tv' = 0 (*), we obtain
y' = e^t u + v.
Differentiating again,
y'' = e^t u' + e^t u + v'.
Substitute into the DE:
(1  t)(e^t u' + e^t u + v') + t(e^t u + v)  (e^t u + tv) = 2(1  t)^2 e^(t)
==> e^t u' + v' = 2(1  t) e^(t) (**).
From (*) and (**), we have two equations in (*) and (**):
e^t u' + tv' = 0
e^t u' + v' = 2(1  t) e^(t).
Solving for u' and v' yields
u' = 2te^(2t) , and v' = 2e^(t)
Integrating, we obtain
u = te^(2t) + (1/2)e^(2t) + A, and v = 2e^(t) + B.
Hence, the general solution is
y = e^t [te^(2t) + (1/2)e^(2t) + A] + t [2e^(t) + B]
...= (Ae^t + Bt) + e^(t) (t + 1/2).
So, a particular solution is y = e^(t) (t + 1/2).
I hope this helps!