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**Unformatted text preview: **Math 415 Quiz 5B The diﬀerential equation t2 y + 5ty + 3y = 0 has y1 (t) = t−1 as one solution. Use the method of reduction
of order to ﬁnd a second solution.
Since y1 (t) = t−1 is a solution, we assume that y2 is of the form v (t)y1 (t) for some unknown function v (t).
Plugging this into the diﬀerential equation gives us
0 = t2 (v y1 + 2v y1 + vy1 ) + 5t(v y1 + vy1 ) + 3vy1
2
1
3
1
2
1
= t2 ( v − 2 v + 3 v ) + 5 t( v − 2 v ) + v
t
t
t
t
t
t
= tv + 3v .
So we have a ﬁrst-order linear diﬀerential equation in v . For clarity, set w(t) = v (t). Then we have the
diﬀerential equation tw + 3w = 0, or in the normalized form w + 3 w = 0. This has integrating factor
t
µ(t) = t3 , i.e. t3 w + 3t2 w = (t3 w) = 0, and so w(t) = D
t3 . Which means that v (t) = Going back to our original assumption, we see that y2 (t) = v (t)y1 (t) = C1
t3 + C2
t. w(t) dt = C1
t4 + C2 . Since we only need one solution, which together with y1 forms a linearly independent pair, we can say that y2 (t) = t−3 . Find the solution of the initial value problem:
y + y − 6y = −9te−t ,
y (0) = 0,
y (0) = 1.
The homogeneous equation y + y − 6y = 0 has the general solution C1 e2t + C2 e−3t . Note that e−t is not
a solution to this version. That tells us that we can assume our speciﬁc solution is Y (t) = (At + B )e−t .
Plugging this into the diﬀerential equation gives us
−9te−t = (At + B − 2A)e−t + (−At + A − B )e−t − 6(At + B )e−t
or
−9t = −6At + (−A − 6B ).
So −9 = −6A and 0 = −A − 6B , or A = 3/2 and B = −1/4 and the general solution will be
y (t) = C1 e2t + C2 e−3t + 6t − 1 −t
e.
4 Looking at our initial conditions tells us that
0 = C1 + C2 − 1
4 1 = 2C1 − 3C2 + 1 31
+
24 which we can put into standard form and then solve to ﬁnd that C1 = 0 and C2 = 1/4. Making our ﬁnal
solution
y (t) = 1 −3t
(e
+ 6te−t − e−t ).
4 2 ...

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