Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz5 => Topic started by: Victorwoshinidie on October 30, 2019, 02:51:36 PM

Question: (1t)y'' + ty'  y = 2(t1)^2 e^t , 0 < t < 1 , y1(t) = e^t, y2(t) = t.
Answer: y1(t) = e^t, y1'(t) = e^t, y1''(t) = e^t, y2(t) = t , y2'(t) = 1 , y2''(t) = 0
Substitute back in to the homogeneous equation: (1t)y'' +ty'  y = 0
Verified that y1(t) and y2(t) both satisfy the corresponding homogeneous equation .
And the complementary solution yc(t) = c1e^t + c2
Now divide both sides of the original eqaution by 1t :
y'' + t/1t  1/1t = 2(t1)e^t
Then:
p(t) = t/1t , q(t) = 1/1t , g(t) = 2(t1)e^t
W[y1,y2](t) = (1t)e^t
Since the particular solution has the form:
Y(t) = u1(t)y1(t) + u2(t)y2(t)
and
u1(t) =  ∫y2(t)g(t)/W[y1,y2](t) dt = ∫t.(2(t1)e^t)/(1t)e^t dt = 2∫te^2tdt = (t+1/2)e^2t
u2(t) =  ∫e^t.(2(t1)e^t)/(1t)e^t .dt = 2 ∫e^t = 2e^ t
Therefore,
Y(t) = (t+1/2)e^2t.e^t + ( 2e^t) t = (1/2 t)e^t
Hence, the general solution:
y(t) = yc(t) + Y(t) = c1e^t + c2t + (1/2  t) e^t
Therefore , the particular solution of the given nonhomogeneous equation is
Y(t) = (1/2  t)e^t