DIFF. EQUATIONSDefinition; A diff equation is an equation
involving differential coefficient and constants
2
3
or examples
cot , sin
y = x ( ) +( )
F
dyx v
dx u vdy dy
dx dx
A diff. equation which involve only one diff.coefficient and one
independent variable is called an ordinary diff. equation
for example
Ordinary Diff. equations
= sinxdy
dx2
2 sindy
x xdx
2 x 2dy
xy xdx
Order of a diff. equation The order of a diff. equation is the highest ordered
derivative present in the diff. equation
for example,
(1) sin order =1 dy
xdx
2
2 (2) = 2xy order = 2
d y
dx2
2 (3) = 2xy - x
d y dy
dx dx
Degree of a diff. equation The degree of a diff. equation is the degree of the highest
ordered derivative present in the diff. equation
for example,
(1) sin order =1 degree = 1dy
xdx
2
2 (2) = 2xy order = 2 degree =1
d y
dx
2
2 (3) = 2xy - x order = 2 degree = 1
d y dy
dx dx
Formation of diff. equations Diff. equation are formed by eleminating arbitrary
constants or functions present in a equation
For example
y = a cos x
a sinx = - sinxcos
dy y
dx x
cos + y sinx = 0dy
xdx
Example
y = e a cos x sinx b x
w.r.t. xDiff
e a cos x sin e - a sin x cosx xdyb x b x
dx
= y + e - a sin x cosx b x
e - a sin x cosxdyy b x
dx
w.r.t. xAgain Diff
2
2+ - y
d y dy dyy
dx dx dx
2
2 +e - a sin x cos - e a cos x sinx xd y dy
b x b xdx dx
2
22 - 2y is the req. sol
d y dy
dx dx
; y = aex xExample be On diff. we get
aex xdybe
dx
2
2
Again diff. we get
aex xd ybe y
dx
2
2 d y
ydx
222
2
2
33 2
3 2
1: Find the order and degree of the following equations.
(a) 0
(b) y = x 1
(c) 4 sin
(d)
Q
d y dyx
dx dx
dy dya
dx dx
d y d y dyxy x
dx dx dx
32 22
2
2 32
2
1 5
(e) 2 4 0
dy d y
dx dx
d y dy
dx dx
2 2 2
2 2
2 2
2 2
2 : Find the diff. equations in the following cases.
(a)
(b) y = e sin 2
(c) 1
(d) 1
(e) y
x
Q
x a y b r
a x
x y
a b
x y
a b
a b x
Equ. of first order and first degree
A diff. equ .of the form ,
or Mdx +Ndy = 0 where M ,N are function
of x,y is called a diff.equ. of first order
and first degree
dyf x y
dx
for ex.
2
or (x+y)dx + dy = 0
dyx y
dx
x
y
A diff. equation can be exprsed as all function of x and dx
on one side and all function of y and dy on another side
then we say the diff.equ.is variable seperable
for example
Variable seperable
1 = dy x
dx y
ydy = xdx
2 2 x 2dx ydy
To solve these diff. equation integrate both sides.
f dy = g dx
f dy g dx
y x
y x c for example,
(1) sin dy = sinx dx dy
xdx
Integrate both side we get
dy = sinx dx
y = - cos x + c
3 2 2 2 (2) x y ydye x e
dx
3 2 2
Sol: The given equ. can be wrriten as
x ydye x e
dx
2 3 2 e dy = y xor e x dx
2 3 2
in tegrate both side we get
e dy = +cy xe x dx 2 3 3e e
= + + c2 3 3
y x x
2 (3) xy 1 1y dy x dx
2
Sol: The given equ. can be wrriten as
1 y 1
xy dy dx
x
2
in tegrate both side we get
1 y 1 dy = +c
xy dx
x
2 1or dy = +cy y x dx
x
3 2 2
= log x + +c3 2 2
y y x
1 cos 2
(4) 01 cos 2
ydy
dx x
Sol: The given equ. can be wrriten as
1 cos 2
1 cos 2
ydy
dx x
2 2
in tegrate both side we get
sec = - cosec xdx + cydy
2 2or = -
cos sin
dy dx
y x
tan y = cot x + c
2 2or sec = - cosec xdxydy
2
2
2cos
2sin
y
x
2
2
cos
sin
y
x
Reduction to variable seperable
Sol: Let = z
then 1+
x y
dy dz
dx dx
or - 1dy dz
dx dx
and the equ. can be wrritten as
- 1 = cos zdz
dx
(1) cosdy
x ydx
2
in tegrate both side we get
1 sec dz = dx + c
2 2
z
21 or sec dz = dx
2 2
z
or 1 cos
dzdx
z
replace the value of z we get
tan = x + c2
x y
tan = x + c2
z
Exercise
2
2
Q1: x
Q2: 0
logQ3:
cos
sin sin 2Q4:
2 log 1
Q5: xy 1
x x
x
dyy y
dx
dy xy y
dx xy x
dy xe x e
dx x y
e x xdy
dx y y
dyx y xy
dx
, A diff. equ .of the form
,
f x ydy
dx g x y
2
for ex. 2
x ydy
dx x y
Homogeneous equations
,where f
,
is called a homogenious diff.equ.
nf x y yx
g x y x
or x ydy
dx x y
Step 1 : Put y = vx so that + xdy dv
vdx dx
Method of solutions
Step 2 : Change all y to vx
Step 3 : Solve the equ.as seperable variable
Step 4 : Replace the value of v as y
x
2 2 2 (1) Solve xdy
x xy ydx
2 2
2
Sol: The given equ. can be wrriten as
x
x xy ydy
dx
Put y = vx so that + xdy dv
vdx dx
2 we get + x 1+ v + vdv
vdx
2 or x 1+ vdv
dx
2
integrate both side we get
= +c1+ v
dv dx
x
1 tan logv x c
2 or
1+ v
dv dx
x
1 tan logy
x cx
y tan logx x c
2 2 (2) xdy - ydx = x y dx
2 2
Sol: The given equ. can be wrriten as
y x ydy
dx x
Put y = vx so that + xdy dv
vdx dx
2 2 2
we get + xdv vx x x v
vdx x
2 x 1dv
vdx
21v v
2 or
1+ v
dv dx
x
2
in tegrate both side we get
= +c1+ v
dv dx
x 2 log 1 log logv v x c 2 log 1 logv v xc
2 1v v xc
2 2 2y x y cx
Reduction to homogenious
where a,b,c,A,B,C are different constants.
If the given equ is in form
then we can reduce it to homogenious equ.
ax by cdy
dx Ax By C
we can reduce it to homogenious equ.
in two diff cases
1Case 2 when = a constant
a bsay
A B m
Case 1 whena b
A B
put x = X - h and y = Y - k
where h, k are arbitrary constant such that ah + bk + c = 0
and Ah + Bk + C = 0
put A = am and B = bm
where m is a arbitrary constant
to solve put ax +by = z
(1) Solve 2 = dy + dx x y dx dy
Sol: The given equ. can be wrriten as
2 1 2 1 0x y dx x y dy
2 1 or
2 1
x ydy
dx x y
here a = 1 , A = 1 and b = -2 , B = -2
so we have a b
A B 1 2
= =1 1 2
put x - 2y = z
1 1 1 and the equ. reduced to - =
2 2 1
dz z
dx z
so that 1- 2 dy dz
dx dx
3 1 or =
1
dz z
dx z
1 or =2 +1
1
dz z
dx z
integrate both side we get
1 dz = dx +c
3 1
z
z
1 4 1
dz = dx +c3 3 3 1z
1 4 z + log z
3 9x c
1dz = dx
3 1
z
z
3x - 3y + c = 2log 3 6 1x y
Exercise
2 2
2 2
2 2 2
Q1: x 2
Q2: x log log 1
Q3: tan
Q4: xdy-ydx
Q5: x +xy x +y
dyxy y
dxdy
y y xdxdy y y
dx x x
x y dx
dy
dx
A diff. equ .of the form dy
Py Qdx
2 for ex. 2 3dy
x x y xdx
Linear diff. equ.
where P,Q are the function of x only or constant
is called a linear diff.equ.
or 5 2 dy
x ydx
Step 1 : write the diff.equ. in standered form.
Method of solutions
Step 2 : Calculate the integrating facter . .
. .Pdx
I F
I F e Step 3 : Multiply both sides by I.F.
Step 4 : Solution is y* I.F.= * I.F.dx + cQ
2 (1) Solve dy y
xdx x
Sol: The given equ. is in the form of
dy
Py Qdx
21 where P = and Q = x
x
I.F.= ePdx
1
edxx log = e x = x
2 (2) 1 2 cosdy
x xy xdx
2 2
Sol: The given equ. can be wrriten as
2 cosx + y=
1 1
dy x
dx x x
2 2
2 cos where P = and Q =
1 1
x x
x x
I.F.= ePdx 2
2
1 exdx
x 2log 1 = e
x
2 = 1 x
2
int egrate both side we get
y 1 = cos +cx xdx
2 1 y = sinxx c
2
2
now multiply both side by 1 we get
1 2 cos
x
dyx xy xdx
22 2
2 1 (3)
1 1
dy xy
dx x x
Sol: The given equ. is in standered form
22 2
2 1 where P = and Q =
1 1
x
x x
I.F.= ePdx 2
2
1 exdx
x 2log 1 = e
x
2 = 1 x
2
2
int egrate both side we get
1 y 1 = +c
1x dx
x
2 1 1 y = tan xx c
2
2
2
now multiply both side by 1 we get
1 1 2
1
x
dyx xydx x
1
we can reduce it to linear equ.as multiply both sides
by y and put y = z .n n
If the given equ is in form
then we can reduce it to linear equ.
ndyPy Qy
dx
Reduction to Linear equations
2 (1) Solve x logdy
y y xdx
2
Sol: The given equ. can be wrriten as
1 1 1 log
dyx
y dx xy x
2
1 1put = z so that
dy dz
y y dx dx
2
1 or we have
dy dz
y dx dx
1the equ.can be log
dz zx
dx x x
1 1loglog 1
the I.F.= e = e = e =dx xx x
x
1log 1
zx c
x x
2
the solution is
1 = log
zxdx c
x x
1 1log 1x c
xy x
Exercise
2 1
2
22
5
2
Q1: x 1 tan
1Q2:
11
Q3: sec tan
Q4:
Q5: x log
dyy x
dx
dy y x x
dx xx
dyy x x
dxdy y
ydx xdy
y y xdx
A diff. equ .is said to be exact if it is formed
by its primitive without any operations.
for ex. x 0
is a exact diff. equ . because it is formed by diff of
xy = c
dyy
dx
Exact diff. equ.
A diff. equ .Mdx +Ndy =0 is said to be exact if it is
satisfy the equ . M N
y x
Step 1 : write the diff.equ. in standered form.
Method of solutions
Step 2 : Campute M,N and satisfy the condition
= M N
y x
Step 3 : integrate M w.r.t.x keeping y as a constant
Step 4 : integrate only those term of N
which is not containing x
Step 5 : Solution is step 3 +step 4 = c
(1) Solve 12 5 9 5 2 4 0x y dx x y dy
Sol: The given equ. is in standerd form
now = 5 and = 5 M N
y x
hence = M N
y x
so M =12x + 5y - 9
and N = 5x + 2y - 4
2 2solution is 6 5 9 y 4x yx x y c
2
not cont x
and 5 2 4 dy = y 4Ndy x y y
2 12 5 9 6 5 9Mdx x y dx x yx x
(2) Solve 1 cos sin 0y ye xdx e xdy
Sol: The given equ. is in standerd form
now = and = y yM Ne cosx e cosx
y x
hence = M N
y x
so M = 1 cosye x
and N = sinye x