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Differential Equations
Unit-7
Exact Differential Equations: 0M d x N d y+ =
Verify the condition M N
y x
∂ ∂=
∂ ∂
Then integrate M d x with respect to x as if y were constants,
then integrate the terms in N d y which do not contain terms in x and equate sum of
these two integrals to a constant.
Example: Solve
2 2 2
2 2 2 2 2
( 2 ) ( ) 0
2 ; ( ) 2
2 2 ; 2 2
a xy y dx x y dy
M a xy y N x y x y xy
M N M Nx y x y
y x y x
− − − + =
= − − = − + = − − −
∂ ∂ ∂ ∂∴ = − − = − − ∴ =
∂ ∂ ∂ ∂
∴ The given equation is an exact differential equation.∴ Solution is given by
2 2 2( 2 ) ( ) 0a x y y d x y d y− − + − =∫ ∫
2 2 2 33 3 3 3a x x y y x y c⇒ − − − =
Example:
Solve ( 1 ) c o s s i n 0y y
e d x e x d y+ + =
cos ; cosy yM N
e x e xy x
∂ ∂= =
∂ ∂
Solution is : ( )1 cosy
e x dx c+ =∫
( ) cxey =+ sin1
Integrating Factors:
Sometimes a differential equations which is not exact may become exact on multiplication by a
suitable function known as an Integrating Factor.
For example the differential equation 0ydx xdy− =
is not exact, but when multiplied by 2 2
1 1 1o r o r b y
y x x y
it becomes on exact equation. The number of integrating factor is infinite for a given equation.
Rules For Finding The Integrating Factors::
Rule I: when Mx + N y ≠ 0 and the equation is homogeneous then NyMx +
1 is an Integrating
factor of Mdx + Ndy = 0
Example: Solve: ( ) 0332 =+− dyyxydxx
Solution: Mx+Ny=-y4 I.F.=-1/y4
Multiply Given equation By I.F. 0
14
3
3
2
=
++− dy
yy
xdx
y
x
On integration we have cyy
x=+− log
3 3
3
Rule II: When Mx - Ny ≠ 0 and the equation has the form f1(xy)ydx+f2(xy)xdy=0
then Integrating factor is NyMx −
1
Example: Solve ( ) ( ) 011 =−++ xdyxyydxxy
Mx - Ny = 2x2y2 I.F.=1/2x2y2
Multiply given equation by I.F. we get 22yx
xdyydx
y
dy
x
dx ++− on intergration
cxyy
xlog
1log =−
Rule III: WhenN
x
N
y
M
∂
∂−
∂
∂
is a function x alone say f(x) then ∫ dxxf
e)(
is an Integrating factor
Problem:
Solve (x2+y2+2x)dx + 2ydy = 0
x
y
ye
N
x
N
y
M
=∫===∂
∂−
∂
∂
1dx
2
2eI.F. 1
Multiplying given equation by I.F. and it becomes exact
Solution of given equation is given by ( ) cdxxyxe x =++∫ 222
ceyexxx =+ 22
Rule IV: WhenM
y
M
x
N
∂
∂−
∂
∂
is a function y alone say F(y) then ∫ dyyF
e)(
is an Integrating
factor
Example: solve ( ) ( ) 0422 434 =−+++ dyxyxydxyy
( )( ) yyy
y
M
y
M
x
N
3
2
232
2 −=
+
+−=
∂
∂−
∂
∂
3
)( 1..
−=∫=
yeFI
dyyF
Multiply given equation by I.F. and it becomes
04
22
32=
−++
+ dy
y
xyxdx
yy
Solution is given by cyxy
y =+
+ 2
2
2
Linear Differential Equation:
A differential equation is said to be linear if the dependent variable and its derivatives appear
only in the first degree.
The first order linear differential equation is of the form QPydx
dy=+ where P and Q are
functions of x only or constants.
Solution of Linear differential equation: ( ) ( )
∫=
+= ∫Pdx
eI.F.
....
where
cdxFIQFIy
Example: Solve 241
=+ ydx
dy
xgiven the boundary conditions x =0 when y = 4.
Solution : Rearranging gives xxydx
dy24 =+
Integrating factor
22 xpdx
ee =∫.
Solution gives: ( ) .2
12
222 222cedxxeye
xxx +== ∫
Using B.C. Obtain C=7/2.Hence particular solution gives ( ).712
1 22 xey
−+=
Example: Solve the differential equation θθθ
tansec yd
dy+=
given the boundary conditions y = 1 when θ = 0.
Solution: Rearranging gives ( ) θθθ
sectan =− yd
dy
Integrating factor ( ) θθ coscosln ==∫ ee
pdx
.
Solution gives: ( ) .seccoscos cdy +== ∫ θθθθθ
When θ = 0, y = 1, thus c=1. Hence particular solution is ( )
.cos
1
θ
θ +=y
Example: Solve the differential equation 2xx
y
dx
dy=+
Solution:
Integrating factor xee xpdx
==∫ ln.
Solution gives: c
xy
dxxxy
+=
= ∫
4
x
3
2
Equations Reducible to linear form Or (Bernoulli’s equation):
nQyPy
dx
dy=+ Dividing by yn
( )11 −−−−−=+ +−− QPydx
dyy nn
put v=y-1+n we get ( ) ( )
dx
dv
dx
dyyn
n =+− −+− 111
i.e ( )
( ) dx
dv
ndx
dyy n
1
1
+−=−
( )QPv
dx
dv
n=+
+− 1
1
( ) ( )QnPvndx
dv11 +−=+−+
Which is linear in v
Example: Solve ( )2y
x
y
dx
dy=+
Solution: ( ) ( ) ( )11
112 −−=+ −−
xy
dx
dyy
Put y-1=v , ( )
dx
dv
dx
dyy =− 2
1
From (1)
11
.
11
−=−
=+−
xv
dx
dvei
xv
dx
dv
which is linear in v
Integrating factor x
eedx
xpdx 1
1
=∫
=∫ −
Solution is cdxxx
v +−= ∫1
11
cxxy
+−= log1
Example: solve xyxydx
dy 22 tantan2 =−
Solution: ( )12
tantan1
22
−−=−
−−
xydx
dyy
Put y-1=v dx
dv
dx
dyy =− − 2
Fron (1) xxvdx
dv 2tantan2 −=+ which is linear in v
Integrating factor xx
exdx
epdx
e2
secseclog2tan2
==∫
=∫
∴ Solution is cxdxxxv +∫ −=2
sec2
tan2
sec
cx
xy +−=−
3
tansec
321
Orthogonal trajectory
Introduction:
The word “Orthogonal” comes from the Greek “right angle” and the word “trajectory”
comes from Latin “cut-across”. Hence the curve that curve that cut across the other at
right angle is called orthogonal trajectory.
Rules to find the equation of orthogonal trajectories of a family of cartesian
curves:
• Diff f(x ,y , c)=0 and eliminate c.
• Resulting equation gives the DE of the family f(x ,y , c)=0
• i.e.
• Replace
• The differential equation of orthogonal trajectory is
• Integrate to get the equation of the required orthogonal
trajectory
0dxdy
y.x,F =
0y,x,Fin dvdx-by =
dx
dy
dxdy
0dydxy,-x,F =
0,,F =−
dydxyx
Rules to find the equation of orthogonal trajectories of a family of polar curves:
Let the equation of given family of curves be f(r , θθθθ , c)= 0
• Diff f(r , θθθθ , c)= 0 and eliminate c.
• Resulting equation gives the DE of the family f(r , θθθθ , c)= 0
• i.e.
• Replace
• The DE of OT is
• Integrate to get the equation of the required
orthogonal trajectory
Example :Find the orthogonal trajectories of parabolas y=c x2
Solution: Differentiate given equation we get
Replace we get is the DE of OT.
On integration we get
Example: Find the orthogonal trajectories of the series of hyperbolas xy= k2
Solution: Differention given equation w.r.t x we get y + x y1=0 Replace y1 by
The differntial equation of the OT is
0ddr.r,F =
θθ
0,r,Fin drd2r-by =
θθθ
θ d
dr
ddr
0dr
d2,-rr,F =
θθ
02,,F =−
drdrr θθ
x
yx
x
yCx
dx
dy 2
22 2 ===
dydx
dxdy −= y
xdxdy
2−=
−= V.S 2 dxxdyy
2
22 cxy =+
1
1y
−
0x-or 0 =−⇒−=−⇒−==− ydyxdxydyxdxydy
dx
dy
dxxy
Which the required OT
Example: Find the orthogonal trajectories of circles
x2+(y-c)2=c2 (1)
solution: Differentiation w.r.t x , 2x + 2(y-c)y1=0
or x + (y-c)y1 = 0 (2)
From (1) x2+y2 – 2cy = 0 or c = (x2+y2) / 2y Substitute this in (2) we get
(3)
Replace y1 by The differential equation of the OT is
(4)
(5) Put y2/2 = z Diff w. r . t. x
Equation (5) becomes
linear in z I.F.=1/x solution is
1222 x
2
2
2
2ccyc
yx ==−⇒=−
012
22or 0
12
22=−+=+−
+ yyxy
xyyyx
yx
222xy
1y x
12
22 x -
12
22
yxy
yyx
yy
xy
−=⇒=
−⇒=−
1
1y
−
0221
2 12
22
22
2
1
1 =+−=−
⇒−
=− xyxyyoryxy
xy
yx
xyy
y
x
x
yxy
dx
dyxy
22dx
dy2x or 2xy by dividen 0222
−=−⇒=+−
22
2
dx
dy x
x
yy
−=−
dx
dzy =
dx
dy
2dx
dz x
x
z −=−
0222yor 22
2y
or 22
11=−++−=+−=∫ +−= cxxc
x
xc
xcdx
xz
• Or be the equation of OT
Example: The orthogonal trajectories of cardiod r=a(1+cos(θθθθ)) where a is a parameter.
Solution: Given equation r = a(1+cos(θθθθ)) Diff. w.r.t. θθθθ
Which is the DE of family r=a(1+cos(θθθθ))
Replace Integrate this we get
on simplification r = c(1- cos(θθθθ))
( ) 222 cycx =−−
θθθ
θ ddra
ddr
sin1 a sin −=⇒−=
( )
−=−
=+
−=⇒+−=2
tan
22cos2
2cos
2sin2
cos1sincos1
sin1 θ
θ
θθ
θθ
θθ
θθr
rr
ddr
ddrr
θθθθ
dr
drddr
=⇒=2
cot2
cot
cr +=
21
2sin
loglogθ