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for large 15. V Solution. It appears that any solution of this differential equation eventually
grows Without bound: as t increases, the change in y(t) becomes very large and
positive. c. Find the general solution of the differential equation and use it to determine
‘how solutions behave as t —> oo‘. _ Solution. In this section it is outlined how to solve differential equations of
the form $111 + p(t)y = g(t), and the equation in this problem is almost in that form.
To get it exactly in the correct form, one divides every term in the equation by
2, yielding the equation % + %y = $752. Note that this problem looks much like
Example 2, on page 33. CaHTlN "50 R.lp.5l
/ As in Example 2, one uses the integrating factor Mt) = elm”, that is (since '
p(t) = %), ,u(t) = eit. By multiplying every term in the equation by this integrating factor, one changes the equation to: The left half of that equation is, by the product rule, the derivative with
respect to t, of 6%‘y(t). The differential equation can then be written as _ gﬂeéty) = 3ft2‘e t </‘ tel» By integrating both sides of the expression with respect to t, one derives, eventually, that: 6%‘3/ # 6t2851t  24te%t + 486% + C where C is a constant of integration. One can then divide all terms in the expression by 6%t and thus get an explicitform for y(t): y(t) = 6t2 — 24: + 48 + (Je—%t Note that the constant of integration must also be divided by the expression
cit — forgetting to do this is a common source of errors.
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Problem. Find the value of yo for Which the solution of the initial value problem y’—y=1+3sin(t) y(0) =90 remains ﬁnite as t —> 00. Solution. One has to ﬁnd the general solution for this equation. The solution
will contain some constant of integration, and by judicious choice of that constant
I the entire expression can be limited to some ﬁnite value for all time t. Once that
constant is found, one places it back into the expression for the general solution to
ﬁnd the value yo = l V Note that this problem is of the same form as Example 2: 53f+ay = g(t), Where
a is a constant, in this case —1. To solve this equation, one uses the integrating
factor “(15) 2 cf “dt = 6". Multiply every term in this equation by the integrating factor, and one derives: ti—ty' —— e”ty = e't + 36“ sin(t) The left side of this expreSsion is, by the product rule, the derivative with . respect to t of (e‘ty(t)). So one can rewrite the equation as: diy(e—ty(t)) = 6”: + 3et sin (t) By then integrating both sides of the expression with respect to time t, one
gets:
3 '3 e_ty(t) = —e‘t — Ee't cos (t) —— 56": sin (t) + C With C', as mentioned above, some constant of‘integration. By dividing every
term in the equation by e“ one gets:
3 3 .' . t
y(t) = —1 — —2—cos (t) — —2sm (t) + Ce Note that one must divide the constant of integration 0 by e‘t  forgetting to do so is a common source of error.
Now one looks at what will happen as t grows very large. None of the terms —,1, *5; cos (t), nor —3 sin (t) grow very large, and their sum will always be ﬁnite.
If the constant of integration 0 is any nonzero number, though, then the term 06‘ will grow without bound. The only way to make the solution stay ﬁnite, then, is for C to equal 0. g
The only solution of the differential equation to stay ﬁnite, then, is y(t) = ——1 — gcos (t) — gsin The initial condition choosing this solution yo = y(0) =
—1—%cos(0)— gsin(0)= —1—%= —3—.
So if yo = ~3, then the solution will stay ﬁnite as t —) oo. 2.! p. 1"! Section 2.1, Problem 32 Problem. Construct a ﬁrst order linear differential equation whose solutions
have the required behavior as t —> 00. Then solve your equation and conﬁrm that
the solutions do indeed have the speciﬁed property. The property is: all solutions
are asymptotic to the line y = 3 — t as t —> 00. Solution. As the phrasing of the problem suggests, there is no unique solution
to this problem — an inﬁnite number of solutions could be imagined. This allows a
certain ﬂexibility in creating an answer. I Suppose one has a ﬁrst order differential equation whose solutions are asymp
totic to the line y = 3 — t as t —> 00. From the work on other problems in this
section, one would expect the exact solution to look like y(t) =2 3 — t + 06““, with
C' a constant of integration, and a some positive‘constant. One knows the rightmost
expression must be of the form 6"” because it is the solution of a ﬁrst order linear
equation, forcing there to be an exponential, and it must be an exponential of a
negative function of t, else the solutions would diverge from the line y = 3 — t as
t —+ 00. It is assumed to be a constant function because that is the easiest function
to write which satisﬁes the requirements of the problem. Then y’(t) .= —1 —— aCe‘“, and y’ + ay = —1 — aC'e’“t + 3a — at + 006"“ =
3a — 1 — at. A differential equation with a solution of the speciﬁed properties is
y’ + ay = 3a — 1 — at, where a is some positive constant, for example, if a = 2, then
2/ + 2y = 5 — 2t. To verify that all solutions have this property, one solves the equation in the
same manner as one solves Examples 2 and 3 in this section. From the equation
21’ + ay = 3a — 1 — at, one gets the integrating factor Mt) = e“t and multiplies every
term in the equation by the integrating factor. ‘80 eaty’ + aeaty = 30.6” — e“t — ate“.
The left side of the expression is, by the product rule, the derivative with respect
to t of eaty, so one can rewrite the equation %(e“‘y) = 3%“ — a“? — ate“. One can then integrate both sides of the expression with respect to t, and
derive eaty = 3e“t — ﬁe“ — teat + ﬁe“ + C, With C a constant of integration. Divide every term, including the constant C', by the expression 8‘“ and one derives: Jfau‘rlN 060 l lggl 9.5] 1 1
y(t)=3—E“t+E+C€—at=3—t+CB—at This equation approaches the line y = 3 — t as t —> oo. ...
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 Spring '04
 Yoon
 Differential Equations, Equations

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