Transcript First-Order Differential Equations Part 1 First-Order Differential Equations Types: • Variable Separable • Linear Equations • Exact Equations • Solvable by Substitutions.

First-Order
Differential Equations
Part 1
First-Order Differential Equations
Types:
• Variable Separable
• Linear Equations
• Exact Equations
• Solvable by Substitutions
Variable Separable
• The simplest of all differential equations
are those of the first order with separable
variables.
• A first-order differential equation of the
form
dy
 g( x)h( y )
dx
is said to be separable or to have
separable variables.
Variable Separable
• To solve variable separable first-order
differential equations, proceed as follows:
dy
 g( x)h( y )
dx
dy

 g( x)dx
h( y )
 p( y )dy  g( x)dx
  p( y )dy   g( x)dx
 H( y )  G ( x )  c
Let 1/h(y) = p(y)
H(y) and G(x) are
antiderivatives of
p(y) and g(x),
respectively.
Example
Solve:
(1  y )dx  (1  x )dy  0
2
2
Solution:
(1  y 2 )dx  (1  x 2 )dy  0


1
 (1  y )dx  (1  x )dy  0 
(1  y 2 )(1  x 2 )
dy
dx


2
1 y
1  x2
dy
dx

 
2
2
1 y
1 x
 arctan (y )   arctan (x )  c
2
2
Alternate
Solution:
(1  y 2 )dx  (1  x 2 )dy  0


1
 (1  y )dx  (1  x )dy  0 
(1  y 2 )(1  x 2 )
dy
dx



1  y2
1  x2
2
2
dy
dx


 1  x2
1  y2
 arctan (y )   arctan (x)  arctanc

 tanarctan (y )  tan arctan (x)  arctanc
tanarctanc  tanarctan (x )
y
1  tanarctanc tanarctan (x)
cx
y
1  cx
 y  cxy  c  x
 y  x  c(1  xy)
c
xy
1  xy
Example
Solve:
1  y dx  1  x dy  0,
2
2
Solution:
 1  y dx 
2


1
1 y
2
dy 
1
1 y
2

1  x dy  0 
2
1
1 x
dy  
2
1
1  y 2 1  x2
dx  0
1
1 x
 arcsi ny  arcsi nx  C
2
3
y ( 0) 
2
dx
Example
3
Initial Boundary Condition: y (0) 
2
Solving for C:
arcsiny  arcsinx  C
 3
  arcsin(0)  C
arcsin

2



  0C
3

C
3
Example
3
Initial Boundary Condition: y (0) 
2
Solving for C:
arcsi ny  arcsi nx  C

arcsi ny  arcsi nx 
3
Example
Alternate Form of Final Answer:
1  y dx  1  x dy  0
2
2

arcsiny  arcsinx 
3

 si n (arcsiny )  si n (arcsinx  )
3


 y  si n (arcsinx ) cos( )  cos(arcsinx ) si n ( )
3
3
1
1  x2
3
 y  x( )  (
)( )
2
1
2
x
3
y 
1  x2
2 2
Linear Equations
A first-order differential equation of the form
dy
a1 ( x )
 a 0 ( x )y  g ( x )
dx
is said to be a linear equation in the
dependent variable y.
When g(x) = 0, the linear equation is said to
be homogeneous; otherwise, it is
nonhomogeneous.
Linear Equations
We can divide both sides of the equation by
the lead coefficient a1(x):
dy
a1 ( x )
 a 0 ( x )y  g( x )
dx
dy a 0 ( x )
g( x )


y
dx a1 ( x )
a1 ( x )
dy

 P ( x )y  f ( x )
dx
Linear Equations
Standard form of a linear 1st-order DE:
dy
 P( x ) y  f ( x )
dx
This differential equation has the property that
its solution is the sum of two solutions:
y = yc + yp
Linear Equations
Now, yc is a solution of the associated
homogeneous equation
dy
 P( x )y  0
dx
and yp is a particular solution of the
nonhomogeneous equation:
dy
 P( x ) y  f ( x )
dx
Linear Equations
Proof:
d
y  P ( x )y  f ( x )
dx
?
d

y c  y p  P( x ) y c  y p  f ( x )
dx
?
 dy c
  dy p

 P ( x )y c   
 P ( x )y p   f ( x )
 dx
  dx





?
 dy c
  dy p

 P ( x )y c   
 P ( x )y p   f ( x )
 dx
  dx

0
f (x)  f (x)


Now, the previous homogeneous equation is
also separable:
dy
 P ( x )y  0
dx
dy

 P( x )dx  0
y
dy

   P( x )dx
y
 l n y  l nk    P( x )dx
 l n (yk )    P( x )dx
 yk  e   P ( x )dx
1   P ( x )dx
y e
k
 y  ce  P ( x )dx
Le t y c  y  ce
  P ( x ) dx
 y c  cy1
wh e re y 1  e
  P ( x ) dx
Now, let yp = u(x)y1:
 
d
y p  P ( x )y p  f ( x )
dx
d
u(x)y 1   P(x)u(x)y 1   f (x)

dx
dy 1
d
u(x)  P(x)u(x)y 1   f (x)
 u( x )
 y1
dx
dx
d
 dy 1

u(x)  f (x)
 u( x ) 
 P( x )y 1   y 1
dx
 dx

d
u(x)  f (x)

u( x)0  y 1
dx
d
u(x)  f (x)

y1
dx
Separating variables and integrating gives
d
u( x)  f ( x)
y1
dx
f (x)
 du 
dx
y 1 (x)
f (x)
u
dx
y 1 ( x)
f ( x)
u
dx
  P ( x ) dx
e
P ( x ) dx

 u  e
f ( x )dx
Now, going back to yp = uy1:
y p  uy1
P ( x ) dx
  P ( x ) dx 




 yp   e
f ( x)dx e



Now, going back to y = yc + yp:
y  yc  yp

 y  ce
 P ( x )dx

e
 P ( x )dx

P ( x )dx

e
f (x)dx
Remember this special term called the
“integrating factor”:
e
 P ( x ) dx
We can use the integrating factor as follows:
 y  ce  P ( x )dx  e   P ( x )dx e  P ( x )dx f ( x)dx e  P ( x )dx



P ( x ) dx
P ( x ) dx
General


 ye
 c e
f ( x)dx

Solution
RECALL AGAIN
Standard form of a linear 1st-order DE:
dy
 P( x ) y  f ( x )
dx
Left-hand side of the
standard form, to be used
for deriving the solution
(see next slide)
A Simpler Derivation
This derivation hinges on the fact that the left
hand side of the 1st-order differential equation (in
standard form) can be recast into the form of the
exact derivative of a product by multiplying both
sides of the equation by a special function (x).
Left side of standard form
of 1st order, linear DE
multiplied by .
dy
d
y 
 P( x ) y 
dx
dx
Derivative of a product
Left-hand side of the
standard form of a 1st
order linear D.E.
of two variables
dy
d
d
?

 P( x)y    y   y   
dx
dx
dx
A Simpler Derivation
The derivation then involves solving the
encircled elements as follows:
d
 P
dx
d

 Pdx


 ln |  | P( x )dx  c1
 ( x )  e
 P ( x )dx  c1
A Simpler Derivation
Continuing, we have:
( x )  e
 P ( x )dx  c1
 ( x ) 
c1  P ( x )dx
e e
 ( x )  c 2
P ( x )dx

e
A Simpler Derivation
Even though there are infinite choices of (x),
all produce the same result. Hence, to simply,
we let c2 = 1 and obtain the integrating factor.
( x )  c 2
P ( x )dx

e
 ( x ) 
P ( x )dx

(1)e
 ( x ) 
P ( x )dx

e
A Simpler Derivation
• This is what we have derived so far. We
multiply both sides of the standard form of
the 1st-order equation by the integrating
factor (x).
• We can then integrate both sides of the
resulting equation and solve for y, resulting
in a one-parameter family of solutions.
A Simpler Derivation
dy
 P( x )y  f ( x )
dx
P ( x )dx dy
P ( x )dx
P ( x )dx



e
e
P( x )y  e
f ( x)
dx
d   P ( x )dx 
P ( x )dx


ye
e
f ( x)


dx 

P ( x )dx
  P ( x )dx 


d  ye
 e
f ( x )dx





P ( x )dx

ye

y e
  P ( x )dx


P ( x )dx

e
f ( x )dx  c
P ( x )dx
  P ( x )dx

e
f ( x )dx  ce
Solving a Linear, 1st-Order DE
1. Put the differential equation in standard
form.
2. From the standard from, identify P(x) and
then find the integrating factor P ( x ) dx

e
3. Multiply the standard form equation by
the integrating factor.
Solving a Linear, 1st-Order DE
4. The left hand side of the resulting
equation is automatically the derivative of
the integrating factor and y:
P ( x ) dx
d   P ( x )dx 

e
y

e
f
(
x
)

dx 
5. Integrate both sides of this last equation.
Example
Find the general solution of:
(x  3y)dx  xdy  0
5
Solution:
Step 1:


1
( x  3y )dx  xdy  0  
xdx
5
dy x  3y


0
dx
x
dy 3
4

 yx
dx x
5
Example
Solution:
Step 2:
P(x): include
the negative
sign if
present
dy 3
4
 yx
dx x
Integrating Factor:
P ( x )dx


e
e
3
(  )dx
x
e
 3 ln x
x
3
Example
Solution:
Step 3:  3 dy
x  dx  x 
x 
3
Step 4:
Recall:
3
 
3
3
4
y x
x
x
dy
4
 3x y  x
dx
d
dv
du
uv  u  v
dx
dx
dx
Example
Solution:
Step 3:
Derivative of y
x 
3
Step 4:
Thus:
dy
4
 3x y  x
dx
Derivative of x –3


d 3
x y x
dx
Solution:
Step 5:
Example


d 3
x y x
dx
 d x  3 y  xdx




  d x  3 y   xdx
2
x
 x3y 
 c'
2
2
 3

x
 x y 
 c'  2x 3
2


 2y  x 5  2c' x 3
 2y  x  C x
5
3
Exercises
Solve the following variable separable
differential equations.
1. y'  x exp(y  x ); when x  0, y  0
2
2. (2a  r )dr  r sind; when   0, r  a
2
2
3
3. xy3dx  (y  1)e xdy  0
4.
y'  cos x cosy
5.
dx  t(1  t ) sec xdt
2
2
2
Answers:
1) y  ln 2  ln(1  e
 x2
)
r
2) r cos   a  r ln
a
2
3) e ( x  1) 
x
2
2
2y  1
2y
2
c
4) 4 ln | sec y  tan y | 2x  sin 2x  c
5) 2x  sin 2x  (t  1)  c
2
2
Exercises
Solve the following 1st-order, linear differential
equations.
1.
(x  3y)dx  xdy  0
2.
y'  cscx  y cot x
3.
5
( y  x  x ycot x)dx  x dy  0
4.
2ydx  (x  1)(dx  dy)
5.
(2x  3)y'  y  2x  3
2
Answers:
1) 2y  x  cx
5
3
2) y  c sin x  cos x
3) x ysin x   x cos x  sin x  c
4) ( x  1)y  ( x  1)( x  2 ln | x  1 |  c)
5) 2y  2x  3 ln | 2x  3 |