Methods of Substitution and Bernoulli’s Equations - (2.5)
1. Bernoulli’s Equation:
The differential equation:
dy
! P!x" y " f!x" y n ! !""
dx
where n is any real number, is called a Bernoulli’s equation. Note that
a. when n " 0, the equation is a linear differential equation in y;
dy
! P!x"y " f!x",
dx
b. when n " 1, the equation can be rewritten as
dy
dy
! P!x"y " f!x"y #
! !P!x" ! f!x""y " 0
dx
dx
a linear homogeneous differential equation and is separable.
In both cases, we know how to solve the differential. Now let us consider the cases where n # 0, 1.
Let u " y 1!n . Then by implicit differentiation, we have
du " !1 ! n" y 1!n!1 dy " !1 ! n" y !n dy .
dx
dx
dx
!n
Multiply !1 ! n" y to the equation !"" :
dy
! !1 ! n" y !n P!x" y " !1 ! n" y !n f!x" y n
!1 ! n"y !n
dx
dy
! !1 ! n"y !n!1 P!x" " !1 ! n"f!x"
!1 ! n"y !n
dx
du ! !1 ! n"P!x" u " !1 ! n" f!x"
dx
It is a linear differential equation in u with P!x" " !1 ! n"P!x", and f !x" " !1 ! n"f!x". Solve for u first,
and then let
y " u 1/!1!n"
dy
Steps for solving a Bernoulli equation:
! P!x" y " f!x" y n
dx
a. Let u " y 1!n . Express the equation as a linear differential equation in u:
du ! !1 ! n"P!x" u " !1 ! n" f!x"
dx
b. Solve u in terms of x :
i. h!x" " $!1 ! n"P!x"dx, integrating factor: e h!x"
ii. k!x" " $ e h!x" !1 ! n"f!x"dx
iii. u!x" " e !h!x" #C ! k!x"$
c. The general solution y : y 1!n " e !h!x" #C ! k!x"$ or y "
Example Solve the differential equation x
e !h!x" #C ! k!x"$
1/!1!n"
dy
! y " x2y2.
dx
The equation is
dy
! 1x y " xy 2 . n " 2. Let u " y 1!2 " y !1 . Solve du ! !!1" 1x u " !x.
dx
dx
a. h!x" " $ ! 1x dx " ! ln|x|, integrating factor is e !ln|x| " 1x
b. k!x" " $ e !ln|x| !!x"dx " $!!1"dx " !x
c. The general solution for u : u " e ln|x| !C ! x" " x!C ! x" " Cx ! x 2
1
1
d. The general solution for y : y " 1u "
Cx ! x 2
Example Solve the initial value problem: x 2
dy
! 2xy " 3y 4 , y!1" "
dx
1
2
.
The equation is
dy
! 2x y " 32 y 4 , n " 4, 1 ! n " !3. Let u " y !3 . Solve du ! 6x u " ! 92
dx
dx
x
x
6
6 ln|x|
6
a. h!x" " $ x dx " 6 ln|x|, the integrating factor is: e
"x
b. k!x" " $ x 6 ! x92 dx " !9 $ x 4 dx " ! 95 x 5
c. The general solution for u : u " e !6 ln|x| C ! 95 x 5 " 16 C ! 95 x 5 " C6 ! 9
5x
x
x
C
9
!3
d. The general solution for y : y " 6 !
5x
x
!3
C
9
1
1
e. When x " 1, y " 2 . 2
! , C " 8 ! 9 " 49
"
5
5
5
1
The solution for the initial value problem: y !3 " 496 ! 9
5x
5x
dy
" 2ty!y 3 ! 1".
dt
dy
! P!t" y " f!t" y n
Note that it is also separable. Rewrite it as the form:
dt
dy
3!1 ! t 2 "
" 2ty!y 3 ! 1" " 2ty 4 ! 2ty,
dt
dy
2t
2t
y"
y4
!
2
dt
3!1 ! t 2 "
3!1 ! t "
Let u " y 1!4 " y !3 . Solve u from
Example Solve the differential equation: 3!1 ! t 2 "
u$ ! 3
2t
3!1 ! t 2 "
u " !3
2t dt " ln!1 ! t 2 "
1 ! t2
2
b. k!t" " $ ! 2t 2 e ln 1!t dt " ! $ 2tdt " !t 2
!1 ! t "
2
c. The general solution for u : u " e !ln 1!t !C ! t 2 "
2t
3!1 ! t 2 "
a. h!t" " $
d. The general solution for y :
y !3 " e !ln
1!t 2
!C ! t 2 " "
1 !C ! t 2 "
1 ! t2
2. Reduction to Separable:
For example, the differential equation of the form
dy
" f!Ax ! By ! C"
dx
is not separable in y and x. Let u " Ax ! By ! C. Then
du " A ! B dy , dy " 1 du ! A
B dx
dx dx
dx
dy
" f!Ax ! By ! C" # 1 du ! A " f!u"
B dx
dx
du " Bf!u" ! A,
1
du " dx separable in u and x.
dx
Bf!u" ! A
2
Example Solve the initial value problem:
dy
" 2 ! y ! 2x ! 3 , y!0" " 1.
dx
Let u " y ! 2x ! 3.
du " dy ! 2, dy " 2 ! y ! 2x ! 3 # du ! 2 " 2 ! u , du "
dx
dx
dx
dx
dx
u,
1 du " dx.
u
The last equation is separable in u and x.
$ 1u du " 2 u ! C 1 , $ dx " x ! C 2
The general solution for u : u " 12 !x ! C", or u " 14 !x ! C" 2
The general solution for y : y ! 2x ! 3 " 14 !x ! C" 2 or y " 2x ! 3 ! 14 !x ! C" 2
Solve the initial value problem: 1 " !3 ! 14 C 2 , C 2 " 16, C " 4.
The solution of the initial value problem: y " 2x ! 3 ! 14 !x ! 4" 2
Example Solve the differential equation:
dy
" sin!x ! y".
dx
Let u " x ! y. Then
du " 1 ! dy , dy " sin!x ! y" # du ! 1 " sin!u" # du " sin!u" ! C,
1
du " dx
dx
dx
dx
dx
dx
sin!u" ! 1
The last equation is separable.
$ sin u1 ! 1 du " tan u ! sec u ! C 1 , $ dx " x ! C 2
The general solution for u : tan u ! sec u " x ! C
The general solution for y : tan!x ! y" ! sec!x ! y" " x ! C
Example Solve the initial value problem:
3x ! 2y
dy
,
"
3x ! 2y ! 2
dx
y!!1" " !1
Let u " 3x ! 2y. Then
du " 3 ! 2 dy , dy " 3x ! 2y # 1 du ! 3 " u ,
3x ! 2y ! 2
2 dx
u!2
dx
dx
dx
du ! 3 " 2u , du " 2u ! 3 " 5u ! 6 , u ! 2 du " dx
u!2
u!2
u!2
5u ! 6
dx
dx
The last equation is separable.
$ 5uu !!26 du " $ 5 ! u !4 2 du " 5u ! 4 ln|u ! 2| ! C 1 , $ dx " x ! C 2
The general solution for u : 5u ! 4 ln|u ! 2| " x ! C
The general solution for y : 5!3x ! 2y" ! 4 ln|3x ! 2y ! 2| " x ! C
Solve the initial value problem: when x " !1, y " !1 :
5!3!!1" ! 2!!1"" ! 4 ln|3!!1" ! 2!!1" ! 2| " !1 ! C, C " !24 ! 4 ln 3
The solution of the initial value problem : 5!3x ! 2y" ! 4 ln|3x ! 2y ! 2| " x ! 24 ! 4 ln 3
3