automobile 3rd sem aem ppt.2016
TRANSCRIPT
Ordinary Differential Equation
Presented by:-1.Patel Nirmal 1508601020202.Patel Yash 150860102021 3.Prajapati vivek 1508601020224.Savant Nikhil 1508601020235.Tandel mehar 1508601020266.Vaghdodiya kalpesh 150860102027
Definitions Differential equation is an equation involving an
unknown function and its derivatives.Ordinary Differential equation is differential
equation involving one independent variable and its differentials are ordinary.
Partial Differential equation is differential equation involving two or more independent variables and its differentials are partial.
Order of Differential equation is the order of the highest derivative appearing in the equation.
Degree of Differential equation is the power of highest derivative appearing in the equation.
particular solution of a differential equation is any one solution.
The general solution of a differential equation is the set of all solutions.
Solutions of First Order Differential Equations
1- Separable Equations2- Homogeneous
Equation3- Exact Equations4- Linear Equations5- Bernoulli Equations
1- Separable Equations (separation variable)
General form of differential equation is (x ,y) dx + (x ,y) dy = 0By separation variable Then 1 (x) 2 (y) dx + 1(x) 2(y) dy = 0
by integrating we find the solution of this equation.
Ex) find general solution for
0)()(
)()(
2
2
1
1 dyyydx
xx
cxy
dyy
dxx
dyxxydx
lnln
nintegratioby 1102
2- Homogeneous EquationThe condition of homogeneous function is
f (x , y) = f (x ,y)and n is Homogeneous degree
(x ,y) dy + (x ,y) dx = 0
and , is Homogeneous function and have the same degree
so the solution is put y = xz , dy = x dz + z dx and
substituting in the last equation
the equation will be separable equation, so separate variables and then integrate to find the solution.
n
3- Exact Equations
(x ,y) dy + (x ,y) dx = 0
only. timeonefactor repeated take: note
issolution general its and
isequation exact be oequation t ofcondition required The
cdydx
yx
dyy
dxx
yx
xy
)( exp)(
)( exp)(
1
1
. following asfactor integralby it multiply exact be it toconvert to
exact,not be illequation w The
if yx
integral factor is
Examples i ) (2x + 3cosy) dx + (2y – 3x siny) dy = 0Solutionit is exact so, (2x + 3cosy) dx = x + 3x cosy (2y - 3x siny) dy = y + 3x cosyThe solution is x + 3x cosy + y = c
ii) (1 – xy) dx + (xy – x ) dy = 0
exactnot its So,
2)()1( 2
xyy
xxyxxxy
dxxyxxThen
dxx
yxy
xy
)2( exp)(
)( exp)( Since
)(1
1
cyxyx
cdyxydxy
dyxydxyx
x
xxdx
x
x
2ln
)(
exact isequation s thi0)(1
ii)equation by value thisgmultiplyin(by 1
lnexplnexpexp
1
11
note :- we took the repeated factor one time only
4- Linear Equations Linear Equation form is
the integral factor that convert Linear Equations to exact equation is :-
= exp p(x) dx by multiplying integral factor by Linear Equation
form
so the general solution is :- y = Q dx + c
)()( xQ y x Pdxdy
exact isequation his t)()( xQ y xP dxdy
cxx
dxxxx
dxxxyxx
cdxQy
issolutiongeneralxx
xxxdx
xxQxxpxQyxpdxdy
solutiondxdyyxxyyEx
sin21sin
)cossin(cos
cos).tan(sec)tan(sec
tansec
tanseclnexpsecexp
cos)( , sec)( )()(
, cossec )
2
2
2
5- Bernoulli Equation Bernoulli Equation form is
nyxQ y x Pdxdy )()(
before. told weassolution its andequation linear is this
)()1()()1(
)()()1(
1
)1( then Put b)
)()(1yover Equation Bernoulli Divide a)
Equation Bernoulli solve To
)1(
)1(
n
xQnzxpndxdz
xQxpdxdz
n
dxdyyn
dxdzyz
xQ yx Pdxdy
y
n
nn
note :- if n = 0 the Bernoulli Equation will be linear equation.
if n = 1 Bernoulli Equation will be separable equation
the general solution will be
2
22
1
2
lnexpln2expexpexp
linear isequation thissin62
2
.sin3
solution
sin3 )
x
xxdxpdx
x xz
dxdz
dxdyy
dxdzthen y and put z
y xxy –
dxdy
xxy –
dxdyy Ex
x
c x x x x y x
cxdx x y x
)cossincos(6
sin622
22
Solution of 1st order and high degree differential equation :-
1- Acceptable solution on p. 2- Acceptable solution on y. 3- Acceptable solution on x. 4- Lagrange’s Equation. 5- Clairaut’s Equation. 6- Linear homogeneous differential Equations with
Constant Coefficients. 7- Linear non-homogeneous differential Equations
with Constant Coefficients.
1- Acceptable solution on pif we can analysis the equation then the
equation will be acceptable solution on p
equation. theofsolution general theis thisand0))((
0 0
lnln2ln lnlnln
2
02 0
02 002
023 )
22
1
22
1
21
222
cyxcxy
cyxorcxy
cxyorcxyx
dxy
dyorx
dxy
dy
ydxdyxory
dxdyx
yxporyxpyxpyxp
SolyxpypxEx
2- Acceptable solution on yIf we can not analysis the equation then the
equation will be acceptable solution on y or x
firstly , to solve the equation that acceptable solution on y
there are three steps :-1- Let y be in term alone . 2- By differentiation the equation with respect to x
and solve the differential equation .3- By deleting p from two equations (the origin
equation and the equation that we got after second step) if we can not delete it the solution called the parametric solution .
2
2
2
2
3212
32
32
xrespect to ation withdifferentiby 32
32
, 223 )
xp
dxdp
xp
dxdpxp
dxdy
xppxy
Solutiondxdy p
xppxyEx
22
2
2
2
by gmultiplyin , 222
3by gmultiplyin , 34
32
32
31
xdxdp
xpx
xpp
dxdp
xpx
xpp
3
2
2
2
2
22
322
61
equationorigin on pabout ngsubstitutiequation twofrom p delete to
ln21ln
332
2 2
2 2
02 02
02)2(
)2(2)2(
)2(22
xy
xpxpcxy
xdx
pdpdxxdy
dxdpxpx
dxdp
dxdpxporpx
dxdpxppx
dxdppxxpxp
dxdppxxppx
3- Acceptable solution on x
secondly, to solve the equation that acceptable solution on x
there are three steps :-1- Let x be in term alone . 2- By differentiation the equation with respect to y
and solve the differential equation .3- By deleting p from the two equations (the origin equation and the equation that we got after
second stepif we can not delete it the solution called the
parametric solution .
solution. parametric thethis so equations last tow thefrom p deletenot can we
equation)origin (the 43
21
)3(
)31(1
)31(1
1but , 3
y respect to ation withdifferentiby
, )
3
42
3
2
2
2
3
ppx
ppy
dpppdy
ppdydp
dydpp
p
dydx
pdydpp
dydp
dydx
dxdypppxEx
4- Lagrange’s Equation
Lagrange’s Equation form y = x g (p) + f (p)
cpxp
dppxppe
pdp
px
dpdx
px
dpdx
dxdp
px
dxdppxp
dxdppxp p
dxdpp
dxdpx p
dxdy
pxp y
p
32
2
factor integral 2 exp
equation aldifferentilinear 22
22)22(1
)22()22(2
222
2 Ex)
32
222ln2
5- Clairaut’s EquationClairaut’s Equation is special case of Lagrange’s
EquationClairaut’s Equation form :-
y = x p + f (p)
0)(0
0
)
2
2
2
paxr o
dxdp
dxdp
pax
dxdp
paxpp
dxdp
pa
dxdpx p
dxdy
pa p x y Ex
(parabola)solution single 4
22
2
2
22
2222
2
2222
axy
ax a x a , yax pa xp y
papx
pa x p y
ca x cy
xap & c p
6 - Linear homogeneous Differential Equations with Constant Coefficients
L(D) y = f (x) non-homogeneous but L(D) y = 0 homogeneous then L() = 0 assistant equation
Roots of this equation are 1 , 2 , 3 ,……,n
This roots take different forms as following:-
constant are ,.....,,, ,
)()........(
3210
22
110
n
nnnn
aaaaadxdD
xfyaDaDaDa
1- if roots are real and different each other then the complement solution is
xn
xxc
neCeCeCy .........2121
2- if roots are real and equal each other then complement solution is
3- if roots are imaginary then complement solution is
).........( 121
rn
xc xCxCCey
)sincos( 21 xCxCey xc
Examples:-
xxc eCeCCy
yDD
yy
321
321
23
3
1,1,00)1)(1(
0)1(0)(
0)(
0)1
xxc eCeCy
yDD
yyy
221
21
2
2,10)2)(1(
0)23(0)23(
023)2
axCaxCyaia
yaD
yay
c sincos
0)(
0)(
0))(3
21
22
22
xCxCeCy
i
yDD
xc sincos
, 10)1)(1(
0)22)(4
321
21
2
2