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**Unformatted text preview: **Math 415 Name: Quiz 4A Find a diﬀerential equation whose general solution is y (t) = C1 e2t + C2 e−3t .
The solutions could have come from the characteristic equation 0 = (r − 2)(r + 3) = r2 + r − 6. So the
diﬀerential equation y + y − 6y = 0 would work. If the Wronskian of f (t) and g (t) is t2 et and f (t) = t, ﬁnd g (t).
The Wronskian of two functions can be written as f g − gf so with the given information we have tg − g =
t2 et , a ﬁrst-order, linear diﬀerential equation in the unknown g . After dividing by t we ﬁnd that the
integrating factor is t−1 , so we get the equation
1
g
t = et . So g (t) = tet + Ct for some constant C . 1 Math 415 Name: Quiz 4B Verify that y1 (t) = 1 and y2 (t) = t1/2 are solutions of the diﬀerential equation yy + (y )2 = 0 for t > 0.
Show that C1 + C2 t1/2 is not a solution for this equation for all C1 & C2 .
y1 y1 + (y1 )2 = 1 · 0 − (0)2 = 0 and y2 y2 + (y2 )2 = t1/2 ( −1 t−3/2 ) + ( 1 t−1/2 )2 = 0, so both y1 and y2 are
4
2
solutions to the given equation. But if we let y3 (t) = C1 + C2 t1/2 then
y3 y3 + (y3 )2 = (C1 + C2 t1/2 )( −1
1
−C1 C2 −3/2 1
2
C2 t−3/2 ) + ( C2 t−1/2 )2 =
t
+ (−C2 + C2 )t−1 ,
4
2
4
4 which is not equal to zero for all t unless C2 = 0, or C2 = 1 and C1 = 0. Determine the value of α, if any, for which all solutions of y + (3 − α)y − 2(α − 1)y = 0 tend to zero as
t → ∞. For which values of α, if any, do solutions become unbounded (positively or negatively) as t → ∞.
The characteristic equation for the given diﬀerential equation is 0 = r2 +(3−α)r −2(α−1) = (r +2)(r −(α−1)).
So the general solution is y (t) = C1 e−2t + C2 e(α−1)t . The e−2t term goes to zero as t approaches inﬁnity,
so we only need to concern ourselves with the second term. If α − 1 < 0 (i.e. α < 1), then the solution will
approach zero as t → ∞. If α > 1 then the solution will become unbounded as t → ∞ (whether to positive
or negative inﬁnity depends on C2 ). 1 ...

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