### Author Topic: Quiz5 Victor's Section  (Read 1104 times)

#### Victorwoshinidie

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« on: October 30, 2019, 02:51:36 PM »
Question: (1-t)y'' + ty' - y = 2(t-1)^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: (1-t)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 1-t :
y'' + t/1-t  -  1/1-t = -2(t-1)e^-t
Then:
p(t) = t/1-t  , q(t) = -1/1-t  , g(t) = -2(t-1)e^-t

W[y1,y2](t) = (1-t)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(t-1)e^-t)/(1-t)e^t dt = -2∫te^-2tdt = (t+1/2)e^-2t
u2(t) = - ∫e^t.(-2(t-1)e^-t)/(1-t)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