Advanced Engineering mathematics

1. Ordinary Differential Equation
Presented by:-
1.Patel Nirmal 150860102020
2.Patel Yash 150860102021
3.Prajapati vivek 150860102022
4.Savant Nikhil 150860102023
5.Tandel mehar 150860102026
6.Vaghdodiya kalpesh 150860102027

2. 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.

3. Solutions of First Order Differential Equations
1- Separable Equations
2- Homogeneous Equation
3- Exact Equations
4- Linear Equations
5- Bernoulli Equations

4. 1- Separable Equations (separation variable)
General form of differential equation is
 (x ,y) dx + (x ,y) dy = 0
By 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
 dy
y
y
dx
x
x




cxy
dy
y
dx
x
dyxxydx



lnln
nintegratioby
11
02

5. 2- Homogeneous Equation
The 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

6. 3- Exact Equations
(x ,y) dy + (x ,y) dx = 0
only.timeonefactorrepeatedtake:note
issolutiongeneralitsand
isequationexactbeoequation tofconditionrequiredThe
cdydx
yx






 


7.  
 











dyy
dxx
yx
xy
)(exp)(
)(exp)(
1
1






.followingas
factorintegralbyitmultiplyexactbeit toconvertto
exact,notbeillequation wThe
if
yx 



 
integral factor is

8. Examples
i ) (2x + 3cosy) dx + (2y – 3x siny) dy = 0
Solution
it is exact so,
 (2x + 3cosy) dx = x + 3x cosy
 (2y - 3x siny) dy = y + 3x cosy
The solution is
x + 3x cosy + y = c
ii) (1 – xy) dx + (xy – x ) dy = 0

9. exactnotitsSo,
2
)()1( 2
xy
y
xxy
x
x
xy






 
 








dxxyxxThen
dxx
yxy
xy
)2(exp)(
)(exp)(Since
)(
1
1

 

     
 
c
y
xyx
cdyxydxy
dyxydxy
x
x
xxdx
x
x














2
ln
)(
exactisequationsthi0)(
1
ii)equationbyvaluethisgmultiplyin(by
1
lnexplnexpexp
1
11


note :- we took the repeated factor one time
only

10. 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
)()( xQyxP
dx
dy

exactisequationhist)()( xQyxP
dx
dy
 

11.  
cxx
dxxxx
dxxxyxx
cdxQy
issolutiongeneral
xx
xxxdx
xxQxxpxQyxp
dx
dy
solution
dx
dy
yxxyyEx












sin
2
1
sin
)cossin(cos
cos).tan(sec)tan(sec
tansec
tanseclnexpsecexp
cos)(,sec)()()(
,cossec)
2
2
2




12. 5- Bernoulli Equation
Bernoulli Equation form is
n
yxQyxP
dx
dy
)()( 
before.toldweassolutionitsandequationlinearisthis
)()1()()1(
)()(
)1(
1
)1(thenPutb)
)()(
1
yoverEquationBernoulliDividea)
EquationBernoullisolveTo
)1(
)1(
n
xQnzxpn
dx
dz
xQxp
dx
dz
n
dx
dy
yn
dx
dz
yz
xQyxP
dx
dy
y
n
n
n








note :-
if n = 0 the
Bernoulli Equation
will be linear
equation.
if n = 1 Bernoulli
Equation will be
separable equation

13. the general solution will be
2
22
1
2
lnexpln2expexpexp
linearisequationthissin62
2
.sin3
solution
sin3)
x
xxdxpdx
x
x
z
dx
dz
dx
dy
y
dx
dz
theny andput z
yx
x
y
–
dx
dy
x
x
y
–
dx
dy
yEx
x










cxxxxyx
cxdxxyx


)cossincos(6
sin6
22
22

14. 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.

15. 1- Acceptable solution on p
if we can analysis the equation then the equation
will be acceptable solution on p
  
equation.theofsolutiongeneraltheisthisand
0))((
00
lnln2lnlnlnln
2
020
020
02
023)
2
2
1
2
2
1
21
222









cyxcxy
cyxorcxy
cxyorcxy
x
dx
y
dy
or
x
dx
y
dy
y
dx
dy
xory
dx
dy
x
yxporyxp
yxpyxp
Sol
yxpypxEx

16. 2- Acceptable solution on y
If 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 .

17. 2
2
2
2
3
21
2
3
2
3
2
xrespect toation withdifferentiby
3
2
3
2
,223)
x
p
dx
dp
x
p
dx
dp
xp
dx
dy
x
p
pxy
Solution
dx
dy
p
x
p
pxyEx



2
2
2
2
2
bygmultiplyin,222
3bygmultiplyin,
3
4
3
2
3
2
3
1
x
dx
dp
x
p
x
x
p
p
dx
dp
x
p
x
x
p
p















18. 3
2
2
2
2
22
322
6
1
equationoriginonpaboutngsubstitutiequationtwofrompdeleteto
ln
2
1
ln
3
3
2
2
2
22
0202
02)2(
)2(2)2(
)2(22
xy
xpxpc
x
y
x
dx
p
dp
dxxdy
dx
dp
xpx
dx
dp
dx
dp
xporpx
dx
dp
xppx
dx
dp
pxxpxp
dx
dp
pxxppx
















19. 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 step
if we can not delete it the solution called the
parametric solution .

20. solution.parametricthethis
soequationslast towthefrompdeletenotcanwe
equation)origin(the
4
3
2
1
)3(
)31(
1
)31(
1
1
but,3
yrespect toation withdifferentiby
,)
3
42
3
2
2
2
3
ppx
ppy
dpppdy
ppdy
dp
dy
dp
p
p
dy
dx
pdy
dp
p
dy
dp
dy
dx
dx
dy
pppxEx










21. 4- Lagrange’s Equation
Lagrange’s Equation form
y = x g (p) + f (p)
c
p
xp
dppxppe
p
dp
p
x
dp
dx
p
x
dp
dx
dx
dp
p
x
dx
dp
pxp
dx
dp
pxpp
dx
dp
p
dx
dp
xp
dx
dy
pxpy
p










3
2
2
factorintegral2exp
equationaldifferentilinear2
2
2
2
)2
2
(1
)22()22(2
222
2Ex)
3
2
222ln2


22. 5- Clairaut’s Equation
Clairaut’s Equation is special case of Lagrange’s Equation
Clairaut’s Equation form :-
y = x p + f (p)
0)(0
0
)
2
2
2

















p
a
xro
dx
dp
dx
dp
p
a
x
dx
dp
p
a
xpp
dx
dp
p
a
dx
dp
xp
dx
dy
p
a
p xyEx

23. (parabola)solutionsingle4
22
2
2
2
2
2
222
2
2
222
axy
axa xa, yax
p
a
xpy
p
a
px
p
a
xpy
c
a
x cy
x
a
p&cp






24. 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:-
constantare,.....,,,,
)()........(
3210
2
2
1
10
n
n
nnn
aaaaa
dx
d
D
xfyaDaDaDa

 

25. 1- if roots are real and different each other then the complement
solution is
x
n
xx
c
n
eCeCeCy 
.........21
21 
2- if roots are real and equal each other then complement
solution is
3- if roots are imaginary then complement solution is
).........( 1
21

 r
n
x
c xCxCCey 
)sincos( 21 xCxCey x
c 


26. Examples:-
xx
c eCeCCy
yDD
yy
321
321
23
3
1,1,0
0)1)(1(
0)1(0)(
0)(
0)1










xx
c eCeCy
yDD
yyy
2
21
21
2
2,1
0)2)(1(
0)23(
0)23(
023)2










axCaxCy
ai
a
yaD
yay
c sincos
0)(
0)(
0))(3
21
22
22







xCxCeCy
i
yDD
x
c sincos
,1
0)1)(1(
0)22)(4
321
21
2
2





