Modul VI
4-2 Application in the field of science and engineering
(continued)
6-1.Reaksi antara dua partikel A  B  C
Sebelum reaksi jumlah partikel A  a, dan partikel B  b
Jumlah partikel C selama reaksi yaitu x bertambah menurut pers.diff.:
dx
 kdt
( a  x)(b  x )
k : ..kons tan ta
Hitunglah jumlah molekul C, sebagai fungsi x dari waktu.
6-2Suatu partikel dengan masa m ditembakkan ke udara dan mendapat gaya hambatan
yang sebanding dengan besar kecepatannya yaitu sebesar m(v),lihat Gamb.
Gambar soal 6-2
Tunjukkanlah bahwa hubungan antara besarnya sudut  yaitu sudut antara vector
kecepatan dan garis vertical ,adalah dalam bentuk pers.diff.:
1 dv
 (v )

 cot 
v d
g sin 
 (v)  a  bv n
Juga hitunglah saat:
a, b,..dan..n : kons tan ta
6-3.Suatu gas di-ionisasi , dalam proses tsb muatan electron dan ion dibuat sama
jumlahnya sebesar n.Tetapi agar supaya muatan electron dan ion tsb dapat terikat
menjadi muatan netral dalam suatu nucleous ,maka besarnya muatan electron dan ion
tersebut jumlahnya haruslah berbanding non-linear sebesar n2,disini  adalah
konstanta pengikat.
Sebelum di-ionisasi ,gas dalam keadaan netral. Bila pada saat t0,jumlah ion adalah
konstan sebesar I,dan diberikan kondisi bahwa perubahan jumlah muatan electron dan
ion adalah berdasarkan pers.diff.:
dn
 I  n 2
dt
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Hitunglah jumlah muatan electron dan ion tersebut yg seharusnya bila diketahui “initial
value” nya adalah saat t=0, maka n=0.
6-4Sepotong karet yg ujungnya diikat,digantungkan diatas langit2 dan pada ujung lain
karet tsb ditarik dengan suatu gaya sebesar F, sehingga karet tsb mengalami
displacement (mulur) sepanjang x (satuan panjang).Bila besarnya gaya F adalah
dikondisikan/disyaratkan sbb:
0  t  T  F  Fo t
T  t  2T  F  F0 (2T  t )
saat:
2T  t  F  0
( F0 : kons tan ta)
maka hitunglah displacement karet tsb sebagai fungsi dari waktu t ,juga lukiskan kurva
displcement nya tsb dalam suatu grafik dengan x sebagai dependent variable.
6-5. Suatu kolam percobaan model (model basin ) berisi air tawar dengan volume V
pada temperatur awal =0C.Kemudian air dialirkan keluar sebesar volume v per detik
dan pada saat yang bersamaan basin diisi air dengan volume yang sama yaitu v per
detik tetapi berbeda temperatur ,yaitu pada temp.T1C.Tunjukkanlah bahwa temperatur
air di-basin untuk sebarang waktu adalah:
T1 (1  e

v
t
V
)
Dengan catatan bahwa tidak terjadi perpindahan kalor dari air dlm basin kekeadaan
sekelilingnya.
6-6.Prinsip hukum Stephan-Boltezmann menyatakan bahwa jumlah pancaran panas dari
suatu benda berbanding lurus dengan selisih pangkat empat dari temperatur mutlaknya
T dan temperatur sekitarnya T0 atau dengan kata lain sebagai fungsi dari waktu dapat
ditulis:
Mc
dT
 kA(T
dt
4
 T04 )
Turunkanlah dari persamaan diatas hubungan antara sebarang waktu t dan temperatur
T darit=0 danT=T1.
6-7.Sungai Musi di Palembang terkenal sangat lebar bahkan sampai mencapai lebar
100 meter.Seorang bocah mencoba berenang menyeberang sungai tsb tegak lurus dari
tepi kanan yg satu ketepi kiri yg lain (lihat Gambar );bila kecepatan aliran sungai Musi
tsb sama dengan kecepatan renang si bocah tersebut,tiba dititik manakah ditepi sisi kiri
si bocah tadi.
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5-1 Homogeneous and Non- Homogeneous Diffrential Equations
Definition: The equation
…..(5.1)
is called the homogeneous first order linear differential equation, and Equation (3-3) is
called the nonhomogeneous first order linear differential equation for b(t) not indentically
zero.
Fortunately, the homogeneous equation (5-1) can be solved quite easily. First, divide
both sides of the equation by y and rewrite it in the form
…….(5.2)
Secondly, observe that
…….(5-3)
where by In Іy(t)| we mean the natural logarithm of }y(t)|. Hence Equation (5-1) can be
written in the form
…(5-5)
But this is "essentially" Equation (5-2) since we can integrate both sides of Equation (55) to obtain that
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where c1, is an arbitrary constant of integration. Taking exponentials of both sides
yields
…..(5-6)
Now, y(t) e 
is a continuous function of time and Equation (5-6) states that its
absolute value is constant. But if the absolute value of a continuous function g(t) is constant then g itself must be constant. To prove this observe that if g is not constant,
then there exist two different times t1, and t2 for which g(t1) = c and g(t2) = -c. By the
intermediate value theorem of calculus g must achieve all values between -c and +c
which is impossible if |g(t)| = c. Hence, we obtain the equation
a ( t ) dt
y(t) e 
a ( t ) dt
Now, y(t) is a continuous function of time and Equation (5-6) states that its absolute
value is constant. But if the absolute value of a continuous function g(t) is constant then
g itself must be constant. To prove this observe that if g is not constant, then there
exist two different times t, and t_ for which g(t,) = c and g(t~) = -c. By the intermediate
value theorem of calculus g must achieve all values between -c and +c which is impossible if |g(t)| = c. Hence, we obtain the equation
…….(5-7)
Equation (5-7) is said to be the general solution of the homogeneous equation since
every solution of Equation (5-4) must be of this form. Observe that an arbitrary constant
c appears in Equation (5-7). This should not be too surprising. Indeed, we will always
expect an arbitrary constant to appear in the general solution of any first order
differential equation. To wit, if we are given
dy
and we want to recover y(t), then we
dt
must perform an integration, and this, of necessity, yields an arbitrary constant. Observe
also that Equation (5-4) has infinitely many solutions; for each value of c we obtain a
distinct solution y(t).
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In applications, we are usually not interested in all solutions of Equation (5-4). Rather,
we are looking for the specific solution y(t) which at some initial time t has the value
y,.. Thus, we want to determine a function y(t) such that
……(5-8)
Equation (5-8) is referred to as an initial value problem for the obvious reason that of the
totality of all solutions of the differential equation, we are looking for the one solution which initially (at time t0 ) has the value y0 . . To find this solution we integrate both
sides of Equation (5-5) between t0 and t. Thus
of both sides of this equation we obtain that
The function inside the absolute value sign is a continuous function of time. Thus, by the
argument given previously, it is either identically +1 or identically -1. To determine which one it is, evaluate it at the point t-; since
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Now, at first glance this problem would seem to present a very serious difficulty in that
we cannot integrate the function es directly. However, this solution is equally as valid
and equally as useful as the solution to Example 5-3. The reason for this is twofold.
Firstly, there are very simple numerical schemes to evaluate the above integral to any
degree of accuracy with the aid of a computer. Secondly, even though the solution to
Example 5-3 is given explicitly, we still cannot evaluate it at any time t without the aid of
a table of trigonometric functions and some sort of calculating aid, such as a slide rule,
electronic calculator or digital computer.
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We now return to the nonhomogeneous equation
It should be clear from our analysis of the homogeneous equation that the way to solve
the nonhomogeneous equation is to express it in the form
and then to integrate both sides to solve for "something". However„- the expression
does not appear to be the derivative of some simple expression. The
next logical step in our analysis therefore should be the following: Can we make the left
hand side of the equation to be
d
of "something"? More precisely, we can multiply
dt
both sides of Equation (5-3) by any continuous function y (t) to obtain the equivalent
equation
….(5-9)
(By equivalent equations we mean that every solution of Equation (5-9) is a solution of
Equation (5-3) and vice-versa.)
Thus,we can choose
y (t)
so that
Is the drivative of some simple expression? The answer to this qestion is yes, and is
obtained by observing that
Hence,  (t )
dy
 a(t )  (t ) y will be equal to the derivative of  (t ) if and only if
dt
d (t )
 a (t )  (t )
dt
But this is a first order linear homogeneous equation for μ(t), i.e.
d
 a(t )   0 which
dt
we already know how to solve and since we only need such function
constant c in quation (5-7) equal to one and take
μ(t) we set the
Now, for this μ(t), Equation (5-9) can be written as
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……(5-10)
TO obtain the general solution of the nonhomogeneous equa-ion (5-3), that is, to find all
solutions of the nonhomogene-as equation, we take the indefinite integral (anti-derivaive) of both sides of Equation (5-10) which yields
Alternately, if we are interested in the specific solution of Equation (5-3) satisfying
the initial condition y(tn) = y0 that is, if we want to solve the initial value problem
then we can take the definite integral of both sides of Equation (5-10) between t0 and t
to obtain that
Or
Remarks:
1. Notice how we used our knowledge of the solution of the homogeneous equation to
find the function μ(t) which enables us to solve the nonhomogeneous equation. This is
an excellent illustration of how we use our knowledge of the solution of a simpler
problem to solve a harder problem.
2. The function y(t) = e is called an integrating factor for the nonhomogeneous equation
since after multiplying both sides by y(t) we can immediately integrate the equation to
find all solutions.
3. The reader should not memorize formulae (11) and (12). Rather, we will solve all
nonhomogeneous equations by first multiplying both sides by M(t), by writing the new
left hand side as the derivative of p(t)y(t), and then by integrating both sides of the
equation.
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4.An alternate way of solving the initial value problem
dy
+ a(t)y = b(t), y(tQ) = yQ is to
dt
find the general solution (11) of Equation (3) and then use the initial condition y(tQ) = yn
to evaluate the constant c. If the function p(t)b(t) cannot be integrated directly, though,
then we must take the definite integral of Equation (10) to obtain Equation (12), and this
equation is then approximated numerically.
Example 5: Find the general solution of the equation
dy
- 2ty = t.
dt
Solution: Here a(t) = -2t so that
Example 6:Find the solution of initial problem:
Multiplying both sides of the equation by μ(t) we obtain that:
Hence,
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Example 7:Find the solution of initial problem:
Multiplying both sides of the equation by μ(t) we obtain that:
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