Modul X
 log x  c  sin 1 u
log x  c  sin 1 y
x
 Table : 16
 sin log x  c 
x
ans y  x sin log x  c 
9 x tan y x  y  xy'  0.....................1
y  ux
dy
du
u
x..................................2 
dx
dx
ux
x tan
 ux  xy'  0
x
tan u  u  y '  0
y
atau y '  u  tan u..........................3
2  3, jadi : u  du x  u  tan u
dx
du
x   tan u
dx
1
1
var
sep
.

dx 
du  0
x
tan u
1
dx  cot u.du  0
x
 log x  c  log sin u 
 log xsin u   c  log c
Ans
xsin u   c
x sin y
c
x
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
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10  x cos y x  y sin y x  y   y sin y x  x cos y x  xy'...........1




y  ux
dy
du
 u  x .....................................................................2 
dx
dx
Pers 1 dirubah menjadi :
 x cos y  y sin y  y   y sin y  x cos y  y '
x
x x 
x
x

x cos u  ux sin u u  ux sin u  x cos u  y '..........................3
substitusikan 2  ke 3 didapat :
cos u  u sin u u  u sin u  cos u u  xu'
u cos u  u 2 sin u  u 2 sin u  u cos u  xu' u sin u  cos u 
2u cos u  xu' u sin u  cos u 
du
2u cos u
x

dx u sin u  cos u
1
u sin u  cos u
dx 
du
x
2u cos u
2
var
sep
.
 : dx 
x
1
 sin u 

du  du.......................................3
u
 cos u 
jadi 2 log x  c   log cos u  log u
atau 2 log x  log cos u  log u  c
Exact equations, and why we cannot solve very many differential equations
When we began our study of differential equations, the only equation we could solve
was dy/dt= g (t) . We then enlarged our inventory to include all linear and separable
equations. More generally, we can solve all differential equations which are, or can be
put, in the form
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
2
for some function  (t,y) . To wit, we can integrate both sides of Equation (1) to obtain
that
and then solve for y as a function of t from Equation (2).
Example 1: The equation 1 + cos(t+y) + cos(t+y)dy/dt = 0 can be written in the form
d/dt[t+sin(t+y)] = 0. Hence,
Example 2 ; The equation cos (t+y) + [1+cos (t+y) ] dy/dt = 0 can be written in the form
d/dt [y+sin{t+y)] = 0. Hence,
We must leave the solution in this form though, since we cannot solve for y explicitly as
a function of time.
Equation (1) is clearly the most general first order differential equation that we can solve.
Thus, it is important for us to be able to recognize when a differential equation can be
put in this form. This is not as simple as one might expect. For example, it is certainly
not obvious that the differential equation
can be written in the form d/dt(y3 +t2 +ty+sin y+cos t) = 0.
To find all those differential equations which can be written in the form (1), observe, from
d
  y
 (t , y (t )) 

.Hence the
dt
t y t
d
 (t , y (t ))  0 .
differential equation M.(t,y) + N(t,y)dy/dt = 0 can be written in the form
dt
the chain rule of partial di t fferentiation, that
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
3
if and only if there exists a function
=

.
y
d

 (t , y (t )) such that M(t,y) =
.and N{t,y)
dt
t
This now leads us to the following question. Given two functions M(t,y) and N(t,y),
does there exist a function Φ(t,y) such that M(t,y) =


. and N(t,y) =
.?
t
y
Unfortunately, the answer to this question is almost always no as the following theorem
shows.
Theorem 1; Let M(t,y) and N(t,y) be continuous and have continuous partial
derivatives with respect to t and y in the rectangle R consisting of those points (t,y)
such tha a < t < b and c < y < d. There exists a function Φ (t,y) such that M(t,y) =
and N(t,y) =

y
if, and only if,
Proof; Observe that M(t,y) =
M N

y
t

.
t

. for some function Φ(t,y) if, and only if,
t
where h (y) is an arbitrary function of y. Taking partial derivatives of both sides of
Equation (3)-with respect to y, we obtain that
Now h' (y) is a function of y alone, while the right hand side of Equation (4) appears to
be a function of both t and y. But a function of y alone cannot be equal to a function of
both t and y. Thus Equation (4) makes sense only if the right hand side is a function of
y alone, and this is the case if, and only if,
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
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N(t,y)
Definition;
-
The differential equation
M(t,y) + N(t,y)
is said to be exact if
dy
 0 .= 0
dt
M N

y
t
The reason for this definition, of course, is that the left hand side of Equation (6) is the
exact derivative of a known function of t and y if
M N

y
t
Remarks: 1. It is not essential, in the statement of Theorem 1, that
in a rectangle. It is sufficient if
M N

y
t
M N
in any region R which contains no "holes".

y
t
That is to say, if C is any closed curve lying entirely in R, then its interior also lies
entirely in R.
1. The differential equation dy/dt = f(t,y) can always be written in the form M(t,y) +
N(t,y)dy/dt =0 by setting M(t,y) = -f(t,y) and N(t,y) = 1.
2. It is customary to say that the solution of an exact differential equation is given by
Φ(t,y) = constant. What we really mean is that the equation Φ (t,y) = c is to be
solved for y as a function of t and c. Unfortunately, most exact differential
equations cannot be solved explicitly for y as a function of t. While this may
appear to be very disappointing, we wish to point out that it is quite simple, with
the aid of a computer, to compute y(t) to any desired accuracy
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
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In practice, we do not recommend memorizing Equation (5). Rather, we will follow one of
three different methods to obtain Φ (t,y).
Example 3:
Find the general solution of the differential equation
3 y  e t  (3t  cos y )
dy
0
dt
Solution: Here M(t,y) = 3y + et and N(t,y) = 3t + cosy,
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
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This equation is exact since
M
N
 3..and ..
 3 . Hence,
y
t
there exists a function Φ(t,y) such that


.and (ii) 3t + cosy=
.
t
y
(i) 3y + et=
We will find Φ(t,y) by each of the three methods outlined above.
First Method: From (i) Φ(t,y) = et+ 3ty + h(). Differentiating this equation with respect to
y and using (ii) we obtain that
h’(y) + 3t = 3t + cos y.
Thus, h(y) = sin y and Φ(t,y) = et + 3ty + sin y. (Strictly speaking, h(y) = sin y + constant.
However, we already incorporate this constant of integration into the solution when we
write Φ(t,y) = c. ) The general solution of the differential equation must be left in the form
et + 3ty + sin y = c since we cannot find y explicitly as a function of t from this equation.
Second Method: From (ii) , Φ(t,y) = 3ty + sin y + k (t) .
Differentiating this expression with respect to t, and using (i) we obtain that
3y + k' (t) = 3y + et.
Thus, k(t) = et and Φ(t,y) = 3ty + sin y + et .
Third Method: From (i) and (ii)
Φ(t,y) = et + 3ty + h (y) and Φ(t,y) = 3ty + sin y + k(t).
Comparing these two expressions for the same function Φ (t,y) it is obvious that
= sin y and k{t) = et . Hence
Φ(t,y) = et
Example 4:
+ 3ty + sin y.
Find the solution of the initial value problem
3t2y + 8ty2 + (t3+8t2y+12y2)
Solution:
h(y)
Here
y
. = 0,
t
M(t,y) = 3t2y + 8ty
y(2) = 1.
and
N(t,y) = t3+8t2y+12y2 . This equation is exact since
N
M
.= 3t2 + 16ty and
= 3t2 + 16ty.
t
y
Hence, there exists a function Φ(t,y) such that
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
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(i) 3t2y + 8ty2 =


. and (ii) t3 + 8t2y + 12y2 =
.
t
y
Again, we will find
Φ(t,y)
by each of three methods.
First Method: From (i) , Φ(t,y) = t3y + 4t2 y2+ h(y). Differentiating this equation with
respect to y and using (ii) we obtain that
t3 + 82 y+ h’y)= t3 + 82 y+ 12 y2
Hence, h(y) = 4y3and the general solution of the differential equation is Φ(t,y) = t3y
+4t2 y2+ 4 y3=c Setting t = 2 and y = 1 in this equation, we see that c
=28.
Thus, the solution of our
equation t3y + 4t2 y2+ 4y3 =28
initial
value
problem is
defined implicitly by
the
Second Method:
From (ii) , Φ(t,y) = t3y + 4t2 y2+ 4y3 +k(t).
Differentiating this expression with respect to t and using (i) we obtain that
3t2y + 8t y2 +k’(t)= 3t2y + 8t y2
Thus k(t) = 0 and Φ(t,y) = t3y + 4t y2 + 4y3.
Third Mejthod: From (i) and (i i)
Φ(t,y) = t3y + 4t2y2+h(y) and Φ(t,y) = t3y+4t2y2+4y3+k (t).
Comparing these two expressions for the same function
and k(t) = 0. Hence,
Φ(t,y) we see that h(y) = 4y3
Φ(t,y) = t3y + 4t2y2 + 4y3.
In most instances, as Examples 3 and 4 illustrate, the third method is the simplest to
use. However, if it is much easier to integrate N with respect to y than it is to integrate
M with respect to t, we should use the second method, and vice-versa.
Example___5; Find the solution of the initial value problem
4t3et+y + t4et+Y + 2t + (t4et+y+2y)dy/dt= 0, y(0) = 1.
Solution: This equation is exact since
 4 t+y

(4t3et+y + t4et+Y + 2t)= (t4 + t4 )et+Y =
t e +2y)
t
y
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
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Hence, there exists a function Φ(t,y) such that
(i) 4t3et+y + t4et+y+2t = =


and (ii) t4et+y+2y = =
t
y
Since it is much simpler to integrate t4et+y+2y
with respect to y than it is to integrate
(4t3et+y + t4et+Y + 2t)
with respect to t, we use the second method.
From (ii),
Φ(t,y)= t4et+Y + y2+k(t)
Differentiating this expression with respect to t and using (i) we obtain
(t4+4t3)et+y + k’(t) = 4t3et+Y + t4et+Y + 2t.
Thus,
2
k(t) = t2 and the general solution of the differential equation is
Φ(t,y) = t4et+Y + y2 + t2 = c.
Setting t = 0 and y = 1 in this equation yields c = 1.
Thus, the solution of our initial value problem is defined implicitly by the equation
Φ(t,y) = t4et+Y + y2 + t2 = 1.
PUSAT PENGEMBANGAN BAHAN AJAR-UMB
Dr. Ir. Abdul Hamid M.Eng. MATEMATIKA TEKNIK
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