Summary and examples of ‘First/Higher Order Differential Equations’

, Equations
2
.
4

-

If f is a 2-variable function , the differential off is :


at
of ax yo
=
+



af = Mdx + NdY
-
A DE (MdX + NdY = 0) is called exact if (Max + Nay) is exact
(Max + Nay is exact if and only
if N
-




-

Note : If (Max + Nay = 0) is exact with /x = M and /Y = N
, then :

df = Max + Nay = 0 => f(x , y) =




Examples
t DE : Cocoax -
sinsinay =

isa constant
(i) (Cocosy) dy + (-sinsing) dy Let M = COSXCOSY (iii) Define f = /Max
-
M N N = -
S1x sing f =
S (cocosy)dX
f = SinX(OSY + 9(Y)
(ii) verify if exact

(iv)
M =2 (COSX(OSY) of
Then
=
-

cosiny
o




ON(sinsing) &) Sinxcosy
= -
Cosysiny
a + g(y)) = -

sinsing o




M EN
exact sing) + g'(y)
sinsing
:
= = -
-




g'(y) = 0

: 9 (y) = < (constant)




Consequentially , if f = sinx cosy then af = Max + Nay = 0 , so that f(x , y) = Sincosy = C

,Examples
-
#2

Solve the DE : (2 x + y) ax + (x + 6y) dy = 0


(iv) Then

(i)((x + y)dX
Of No

+ (x + 6y) = 0 Let M = 2 x + y
-


x
MN N =
X + 6Y YX + (y) = X+


is constant



ciic Verify if exact DE (iii) Define f =
/Max 0 +x+ g'(y) = *+ by

M 2(2x + y) =
= 0+1 = 1 =
((2x + y)dx 9'(y) = Gy
by Gy
=
2x2
-
+ yx + g(y) 9(y) =

(by dy
_
2X
(X + 6) = H =
= x2 + yX + 9(y) =
342 + k
&M EN
=
:: exact De
=




consequentially , if f = x+ xy + 3y2 is such that df = Max + Ndy

Since , af = Max + Ndy = 0 it follows that f(x , y) = * + X Y + 34 = (CER)
,




examples
X 6x/y xy 3/n/y)

As :
-

+ + y + xy -




3/4 + 3
x)ay
(1 -


+ y=



is constant


(i) (1 -
3/y + x) qy + y= 3/x -
1 cis verify if exact (iii) Define f = /m
dX

M =(1-3/
+) =
((1 -
3/x + y)dx

(1 -
3/4 + x) dy = (3/x -
1 -

y)dx =
X -

3/n/x1 + xy + 9(y)

(1-3/ )
=
(1 -
3/y + x) dy -
(3/X -
1 -
y)dX = 0


(iv)
(1 3/x + y)dX + (1 3/y + x) dy
N
-
-
= 0 Then
-- -


M C
M N .: exacte T
=




f(x 9(y)) 1
3
-
3(n(x) + xy + =
-

+ x
-




24
Let M= 1 -
3/x + y

N= 1 -

3/y + X
*+ g'(y) = 1 -

3/y +*

g'(y) = 1 -

3/y

consequentially , if f = x -

3///y + y -

3/nly) is such that df = Max + Nay g(y) =
((1 -

3/y) dy

Since af Max + Ndy = 0 X 3/n/x / + xy + y 31/y) y
= -

3/(y) + k
it follows that f(x , y)
-
-

= =
,
,

= C (CER)

, Examples
A (54-2x)
a
: -
( = 0




(i) (5y (ii) Verify if exact Smax
2x)a (iii) Define f
-

-
24 = 0 =




(( 24)dx
My 2) (y)
-

=
=
-



=
-
2
Y
(5y 2x) dy
- = (2y)dX =
-
2 x y + 9(Y)




-
(5y 2x)dy-
+ ( (y)dX = 0

( 2y)dX
-

+ (54 -
2x)dY = 0 (iv) Then F = N
- -
M N
Y


6) -

2 x y + g(y)) = 5y 2x
-




by
Let M = -

24

N =
5y -
2x -x+ 9/(y) = 54 -X

g'(y) =
54
consequentially , if f =
-2xy + 54c is such that df = Max + Nay g(y) ((5y)ay
=




Since 5)
by
af Max + Ndy = 0 2 xY +
it follows that f(x , y)
-




, = = =
,

= C (CER)




Examples as (exty(ax + (2 + x + yex) dy =

Y(0) = 1




(i) (ex + y)dx yeY) dy
f
+ (2 + x+ = 0 (ii) verify if exact (iv)Then N
=




- -
M N
M G(ex + y)
= = 1

by Gy
o

(e
+ Xy + 9(y)) = 2 + X + Yey

x
xtye
N
Let M =
e +y = L+

N = 2 +x+ yeY *+ g'(y) = 2 + *+ yeY
9(y) Ity or
M EN : exact De =
=




(iii) Define f = Smax g(y) =
((2 + yeY)dy
=
((ex + y)dx =
((dy Sye dy
+

=
ex + xy + 9(y) = (y + (yey eY) + k
-




consequentially , if f = ex + xY + 24 + yeY-eY is such that df = Max + Nay

Since , af = Max + Ndy = 0
, it follows that f(x , y) = ex + xy + 2y + yes ey
-




= C (CER)