Lessons 10-20
July 26, 2022 | Author: Anonymous | Category: N/A
Short Description
Download Lessons 10-20...
Description
EQUATIONS OF ORDER ONE
Here we will study several elementary el ementary methods for solving first-order differential equations. We begin our study of the methods for solving first-order differential equations by studying an equation of the form Mdx + Ndy = 0 ; where M and N maybe functions of both x and y. Some equations of this type are are so simple that they can be put in the form A(x)dx + B(y)dy = 0.
Variable Separable Equations
Consider a differential differential equa equation tion that that can be written in the form M( x )dx + N( y )dy = 0 where M is a continuous function function of x alone and N is a continuous function of y alone. For this type of equation, all x terms t erms can be collected with dx and all y terms with dy, and a solution can be obtained by integration. Such equations are said to be variable v ariable separable equations, and the solution procedure is called separation of variables. Example:
2
1) Find the general solution of the differential equation ( x + 4 ) y’ = xy. Solution: ( x2 + 4 )
= xy
( x2 + 4 ) dy = xydx
;
xydx - ( x2 + 4 ) dy = 0
[ divide both terms by y and and ( x2 + 4 ) ]
= 0
;
∫
-
∫
=
∫ 0
Since the equations are separable, integrating yields to: ln ( x2 + 4 ) - ln( y )
= C
Simplifying further: ln ( x2 + 4 ) - ln( y )2 = 2C ; ln
= C1
;
lnA - lnB = ln and 2C = C1 = C2 ;
C3 ( x2 + 4 ) = y2 y =
( + )
x2 + 4 = C2y2
2) Find the particular solution of the equation xydx + condition of y(0) = 1.
( y2 - 1 )dy = 0 with initial
Solution:
Separating the variables: ( divide both terms by y and and
( +
( y -
)
) = 0
)dy = - x dx ;
∫( −
) dy = - ∫ 2 dx
- ln ( y ) = - + C and from the initial initial condition condition y( 0 ) = 1 - ln ( 1 ) = - e0 + C - 0
= - ( 1 ) + C thus
C = 1
Therefore, the particular solution yields to:
y2 - ln ( y2 )
= - + 2 ( 1 )
Simplifying further; y2 - ln ( y2 ) +
= 2
3) Find the general solution of the differential equation
( ) + ylnx = 0
Solution: ( y2 + 1 ) + ylnx = 0 ;
ylnx dx + ( y2 + 1 ) = 0 ( multilplied by )
Separating Separatin g the variable, we have: x lnx dx +
( ) = 0
Integrating both sides will yield to:
∫( +
) dy = + lny
;
x lnx dx + ( y +
) dy = 0
∫ dx
( by parts )
∫ = − ∫
;
du =
let u = lnx
thus:
∫ dx
lnx -
=
;
∫
v =
dv = x dx
( ) = lnx -
∫ dx
= lnx -
Therefore, final answer will yields to: + lny + lnx - = C
or
2y2 + lny4 + 2x2lnx - x2 = C1
4) Find the general solution of the differential equation sec 2x ( 3y2 + 2y + 4 ) y’ = ( y3 + 4y ) ( y3 + 4y ) dx - sec 2x ( 3y2 + 2y + 4 ) dy = 0 cos2x dx -
( ) dy = 0 ;
∫ 2 ( 2 ) sin2x
∫
∫(
) dy = ( )
∫ 0
( ) dy = sin2x ( )
)
− ∫ ( (
)
Integration by rational fraction:
∫
( )
+
=
( )
[ multiplied multiplied by y ( y2 + 4 ) ]
3y2 + 2y 2y + 4 = ( y2 + 4 ) A + ( By + C ) y 3y2 + 2y + 4 = Ay2 + 4A + By2 + Cy y2 ; y ; C ;
3 = A + B 2 = C 4 = 4A
Solving simultaneously simult aneously ;
A = 1
(1) (2) (3) B = 2
C = 2
Therefore:
∫
( ) dy = ( )
=
∫
∫
+
∫
+
∫
( )
∫
+
2 ∫
( u = y ; du = dy ) = ln y
∫
( u = y2 + 4 ;
du = 2y ) = ln ( y2 + 4 )
2 ∫
( u= y a = 2
2 ∫
= 2 tan-1 = tan-1
du = dy ) ;
∫
=
tan-1
Therefore, final solution will be: sin2x - ln y - ln ( y2 + 4 ) - tan-1 = C sin2x - ln y2 - ln ( y2 + 4 )2 - 2 tan-1 = C1
Exercise No. 5
Obtain the general/particular solution of the following variable separable differential equations: 1)
( 1 - x ) y’ = y2
2)
2xyy’ = 1 + y2
3)
xy3dx +
4)
x2dx + y ( x - 1 ) dy = 0
5)
( xy + x ) dx = ( x2y2 + x2 + y2 + 1 ) dy
when x = 2 ; y = -3
= 0 when x = -1 ; y = 1
Homogeneouss Equations Homogeneou
Some differential equations that are not separable in x and y can be make separable by a change of of variables. This This is true for differential equations equations of the the form y’ = f ( x, y ), where where f is a homogeneous function. function. The function given by f ( x, y ) is homogeneous of d degree egree n if: f ( ty, tx ) = tn f ( x, y ) where n is an integer.
A homogeneous differential equation is an equation of the form: M ( x, y ) dx + N ( x, y ) dy = 0 where M and N are homogeneous homogeneous functions of the same degree. degree. If M ( x, y ) dx + N ( x, y ) dy = 0 is homogeneous, homogeneous, then it can be transformed into a differential equation whose variables are separable by the substitution, y = vx x = vy
or
and and
dy = vdx + xdv dx = vdy + ydv
where v is a differentiable function of x or y .
Example: 1) Find the general solution of ( x2 - y2 ) dx + 3xydy = 0. Solution: Said equation are both homogeneous of degree 2, let y = vx ; dy = vdx + xdv ; by substitution yields to: ( x2 - v2x2 ) dx + 3x ( vx ) ( vdx + xdv ) = 0 x2dx - v2x2dx + 3x2v2dx + 3x3vdv = 0 Re-arranging and separating the variables yields to: x2 ( 1 + 2v2 ) dx + 3x3vdv = 0 + = 0 ( )
ln ( x ) + Simplifying the terms:
dividing the terms by ( 1 + 2v2 ) and x3 ;
and integrating both sides yield to:
ln ( 1 + 2v2 ) = C
4 ln ( x ) + 3 ln ( 1 + 2v 2 ) = C1
ln x4 + ln ( 1 + 2v 2 )3 = C1 ; x4 ( 1 + 2v2 )3 = C2
ln ( x4 ) ( 1 + 2v2 )3 = C1
;
but v =
thereby: x4 ( 1 + )3 = C2 ;
x4 ( )3 = C2
Simplifying further yields to:
( x2 + 2y2 )3 = x2C2
Alternate solution :
let :
x = vy ;
dx = ydv + vdy
( x2 - y2 ) dx + 3xydy = 0. [ v2y2 - y2 ] [ ydv + vdy ] + 3 ( vy ) ydy = 0 v2y3dv + v3y2dy - y3dv - vy2dy + 3vy2dy = 0 y2dy ( v3 + 2v ) + y3dv ( v2 - 1 ) = 0 [ divide both terms by ( v3 + 2v ) and y3 ] + dv = 0 = = + ( )
v2 - 1 = Av2 + 2A + Bv2 + Cv v2
;
1 = A + B
(1)
v
;
0 = C
(2)
C
;
-1 = 2A
(3)
Solving simultaneously will results to:
A =-
dv = - + ( )
B =
and C = 0
then integrate:
ln v + ln ( v2 + 2 )
Final solution will be: ln y -
ln v + ln ( v2 + 2 ) ( )
= C
= C2
ln y4 - ln v2 + ln ( v2 + 2 )3 = C1
; ;
but v =
y4 (
3 + 2 ) = ( ) C2
( ) = ( ) C2
;
y4 (
;
3 ) = ( ) C2
( x2 + 2y2 )3 = x2C2
2) Find the particular solution of example 1 if y ( 2 ) = -1 Solution: Substitute value of of x and y to the general solution solution { 22 + 2 ( -1 )2 }3 = ( 2 )2C2 ( ) Giving the value of C2 as = = 54 that yields to a particular solution : ( x2 + 2y2 )3 = 54x2
3) Find the general solution of ( 2xy + y2 ) dx + ( x2 - xy ) dy = 0 Solution: Let x = vy ;
dx = vdy + ydv
[ 2 ( vy ) y + y2 ] [ vdy + ydv ] + [ ( vy ) 2 - ( vy ) y ] dy = 0 2v2y2dy + 2vy3dv + vy2dy + y3dv + v2y2dy - vy2dy = 0 3v2y2dy + y3 ( 2v + 1 ) dv = 0
( ) dy + dv = 0
by separation, multiply both terms by
or 3
+ ( 2 +
dv ) = 0
:
Integrating both sides will results to:
3lny + 2lnv - = C
ln ( y3 ) (
but v =
)
-
;
dy = vdx + xdv
= C
;
lny3 + ln ( )2 - = C ln x2y -
;
= C
Alternate solution:
Let y = vx
( 2xy + y2 ) dx + ( x 2 - xy ) dy = 0 [ 2x ( vx ) + ( vx )2 ] dx
+ [ x2 - x ( vx ) ] [ vdx + xdv ] = 0
2vx2dx + v2x2dx + vx2dx + x3dv - v2x2dx - vx3dv = 0 3vx2dx + x3 ( 1 - v ) dv = 0 ( divide both terms by v and x3 )
3
( ) + = 0
;
3
+ - dv = 0
By integration it will results to:
3
3 lnx + lnv - v = C
But v =
;
or
lnx
+ lnv - v = C
lnx3 + ln ( ) - = C
ln x3 ( ) - = C
;
ln x2y -
= C
Exact Equations
A differential equation of the form M ( x, y ) dx + N ( x,y ) dy = 0 is said to be as an exact equations if M , N , and are continuous functions of x and y and if:
=
Let us now show that if this condition satisfies the equation, let ∅ ( , ) be a function for ∅ which = M. The function ∅ is the result of integrating i ntegrating Mdx with respect to while holding y constant. Now
∅ ∅ = ; hence if = , then also = .
Let us integrate both sides of this last l ast equation with respect to x, holding y fixed. In the integration with respect to x, the “arbitrary constant “ maybe any funct function ion of y. Let us call it B’(y), for ease in indicating its integral. Then integration of it with respect to x yields to: ∅ = N + B’(y)
Now a function F can be exhibited, namely, F =
∅ ( x,y )
- B( y )
for which:
∅ ∅ dx + dy - B’( y ) dy = M dx + [ N + B’( y ) } dy - B’( y ) dy = M dx + N dy
dF =
Hence, it is exact equation. Example: 1) Solve the general solution of the given differential equation: 3x ( xy - 2 ) dx + ( x3 + 2y ) dy = 0
Solution: = 3x2
and
= 3x2
thus it is exact.
Therefore, its solution is F = c, where where
= M = 3x 2y - 6x
and
= N = x3 + 2y .
= 3x2y - 6x. Integration of both sides with respect to x, holding y constant yields to F = x3y - 3x2 + T( y ), where the usual arbitrary constant in indefinite integration is now necessarily a function f unction T( y ), as yet y et unknown. To determine T( y ), we use the fact that the function F with T( y ) must also satisfy the equation = x3 + 2y . Hence; Hence; x3 + T’( y ) = x3 + 2y and T’( y ) = 2y.
Let us attempt to determine F from M from the equation
No arbitrary constant is needed in obtaining T( y ), since one is being introduced on the right in the solution F = c. Then, T( y ) = y 2. Thus, F = x3y - 3x2 + y2. Finally, a set of solutions of the said equation is defined by:
x3y - 3x2 + y2 = C
2) Solve the general solution of the given differential equation: ( 3x2y - 2ycosx 2yc osx + 1 ) dx - ( 2si 2sinx nx - x 3 - y ) dy = 0
Solution: = 3x2 - 2cosx
and
= - 2cosx + 3x2
( exact )
Simpler solution will be, integrate i ntegrate both sides with respect to each derivatives,
∫( 3x2y [ 3
- 2ycosx + 1 ) dx -
∫( 2sinx
- x3 - y ) dy =
∫ 0
y - 2y ( sin x ) + x ] - ( 2ysinx - x3y - ) = C
( x3y - 2ysinx + x ) + ( x 3y - 2ysinx +
) = C
Eliminate one of the term/s common to both bo th x and y and then simplify: 2x3y - 4ysinx + 2x + y 2 = C
Exercise No. 6: Find the general/particular solution of the given differential equation:
1)
y’ = ( x - 2y )
2)
( 6x + y2 ) dx + y ( 2x - 3y ) dy = 0
3)
( y2 - 2xy + 6x ) dx - ( x2 - 2xy 2xy + 2 ) dy = 0
4)
x ( 3xy - 4y3 + 6 ) dx + ( x3 - 6x2y2 - 1 ) dy = 0
5)
( cosxcosy - cotx ) dx = sinxsiny dy
6)
[ 2x + ycos( y cos( xy ) ] dx + xcos( xy ) dy = 0
7)
( y -
8)
( x3siny - x2 + 4y ) dy = ( 3x2cosy + 2xy - 3 ) dx
+
) = xy’
;
y ( 4 ) = -2
;
y ( -1 ) = 2
View more...
Comments
