PERTEMUAN VI
First Order Linear Equations
A first order linear differential equation has the following form:
The general solution is given by
where
called the integrating factor. If an initial condition is given, use it to find
the constant C.
Here are some practical steps to follow:
1. If the differential equation is given as
,
rewrite it in the form
,
where
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2. Find the integrating factor
.
3. Evaluate the integral
4. Write down the general solution
.
5. If you are given an IVP, use the initial condition to find the constant
C.
Example: Find the particular solution of:
Solution: Let us use the steps:
Step 1: There is no need for rewriting the differential equation. We have
Step 2: Integrating factor
.
Step 3: We have
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.
Step 4: The general solution is given by
.
Step 5: In order to find the particular solution to the given IVP, we use the
initial condition to find C. Indeed, we have
.
Therefore the solution is
.
Example : Solve the following initial value problem for t > 0
Answer: This is a linear equation. Let us follow these steps for solving such
equations:
1. We have to divide by 2t
2. We get the integration factor u(t) by
3. The general solution is given by
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.
Since
Therefore, we have
4. The solution to the given initial value problem may be obtained by
using the initial condition y(2) = 4. We have
,
which gives
. Therefore, the solution is
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Example: Find the solution to
Answer: This is a linear equation. First we have to rewrite the equation with
no function in front of y'. We get
,
which may also be rewritten as
.
Hence, the integrating factor is given by
Therefore, the general solution can be obtained as
Since we have
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we get
The initial condition
implies
,
which gives C=-1. Therefore, the particular solution to the initial value
problem is
Solving a Linear Ordinary Differential Equation
http://mss.math.vanderbilt.edu/cgi-bin/MSSAgent/~pscrooke/MSS/linode.def
ODE: dy/dx +
1/x
y=
x
.
Solve
The differential equation is:
The general solution is:
1
dy/dx + -y = x.
x
y(x) =
3
x
c1 + -3
-------.
x
Persamaan Differensial Bernaulli
A differential equation of Bernoulli type is written as
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This type of equation is solved via a substitution. Indeed, let
.
Then easy calculations give
which implies
This is a linear equation satisfied by the new variable v. Once it is solved,
you will obtain the function
.
Note that if n > 1, then we have to add the solution y=0 to the solutions
found via the technique described above.
Let us summarize the steps to follow:
(1) Recognize that the differential equation is a Bernoulli equation. Then
find the parameter n from the equation;
(2) Write out the substitution
;
(3) Through easy differentiation, find the new equation satisfied by the new
variable v. You may want to remember the form of the new equation:
(4) Solve the new linear equation to find v;
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(5) Go back to the old function y through the substitution
;
(6) If n > 1, add the solution y=0 to the ones you obtained in (4).
(7) If you have an IVP, use the initial condition to find the particular
solution.
Example: Find all the solutions for
Solution: Perform the following steps:
(1) We have a Bernoulli equation with n=3;
(2) Consider the new function
;
(3) The new equation satisfied by v is
;
(4) This is a linear equation
:
4.1 the integrating factor is
4.2 we have
4.3 the general solution is given by
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5 Back to the function y: we have
which gives
6 All solutions are of the form
The form of a Bernoulli ODE is: dy/dx + P(x)y = Q(x)y r.
http://mss.math.vanderbilt.edu/cgi-bin/MSSAgent/~pscrooke/MSS/bernoulli.def
P(x) =
Q(x) =
r =
1/x
x
2
Solve
The general solution is:
y(x) =
1
--------(c - x) x
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Contoh :
1. Selesaikan P.D
-2
Jawab : y
dy
dx
- y = xy2
dy 1
 x
dx y
misalkan z = y-1 atau
dz
1 dy
- 2
dx
y dx
kemudian substitusi kepersamaan didapat :
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-
dz
- z  x atau
dx
dz
z-x
dx
z e  -  x e  c
dx
dx
zex = -xex + ex + c atau z = y-1 = - x + 1 + ce-x
dy
dx
2. Selesaikan P.D
-3
Jawab : y
- y = xy3
dy 1
 x
dx y 2
misalkan z = y-2 atau
dz
2 dy
- 3
dx
y dx
kemudian substitusi kepersamaan didapat :
-
1 dz
- z  x atau
2 dx
ze 
2dx
 -  2xe 
2dx
dz
 2z  - 2x
dx
 c
ze2x = - xe2x + ½ e2x + c
atau
z = y-1 = ½ - x + ce-2x
3. Selesaikan P.D
-3
Jawab : y
dy
dx
- y = xy3
dy 1
 x
dx y 2
misalkan z = y-2 atau
dz
2 dy
- 3
dx
y dx
kemudian substitusi kepersamaan didapat :
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-
1 dz
- z  x atau
2 dx
ze 
2dx
dz
 2z  - 2x
dx
 -  2xe 
2dx
 c
ze2x = - xe2x + ½ e2x + c
atau
4. Selesaikan P.D
z = y-1 = ½ - x + ce-2x
x
Jawab :
dy
+ y lny = xyex
dx
x
dy
+ y lny = xyex
dx
kedua ruas kiri dan kanan
dibagi dengan xy
y-1
dy
+ x-1 lny = ex
dx
misalkan z = ln y, jadi
dz 1 dy

dx y dx
dz z

 ex
dx x
1
ze
ze
lnx
=
e
 x dx
1
 x dx
 c
dx  c 
 xe
 e e
x
1
x
e
 x dx
xz = xex – ex + c
x
dx  c
atau
xlny = xex – ex + c
5. Selesaikan P.D
(1 + y2 ) dx = ( arc tg y – x ) dy
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(1  y2 )
dx
 x  arc tg y
dy
dx
x
arc tg y


dy 1  y 2
1 y2
dy
xe
 1 y2
dy
arc tgy  1 y2
 
e
dy  c
1 y2
x e arc tg y   arc tg y e arc tgy darc tgy  c
 arc tgy e arc tg y - e arc tg y  c atau
x  arc tg y - 1  ce - arc tg y
dy + y tg x dx = y2 sec x dx
6. Selesaikan P.D
y -2
dy
 y -1 tg x  sec x
dx
misalkan y-1 = z jadi
dz
dy
  y -2
dx
dx
-
dz
 tgx z  sec x
dx
dz
- tgx z  - sec x
dx
ze 
- tg x dx
 -  sec x e 
z e ln cos x  - 
- tg x dx
dx  c
1
cos x dx  c  - x  c
cos x
cos x
-xc
y
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dy
1
1
 y  ( 1 - 2x ) y 4
dx
3
3
7. Selesaikan P.D
y -4
dy
1
1
 y -3  ( 1 - 2x )
dx
3
3
: y-4
misalkan z = y-3 ,
dy
dz
 - 3y -4
dx
dx
-
1 dz
1
1
 z  ( 1 - 2x )
3 dx
3
3
- dx
- dx
ze   -  ( 1 - 2x ) e  dx  c
ze -x 
 (2x - 1) e
-x
dx  c
z ex = (1 – 2x – 2 ) e-x + c
atau
z = y-3 = ( -1 – 2x ) + cex
SOAL-SOAL
III-15
TERIMA KASIH
III-16