DIFFERENTIAL
EQUATIONS
,OLD SEPARABLE DIFFERENTIAL EQUATIONS
sold separable differential equations
separable separated be to find
equations can be
by their variables and
integrated
their solutions .
Example :
① dd- =
22
1-
y2
i. ( 1-
yz) ¥ = x2
i.
11-921 dy = x2 dx
S S
dy
1- x2 dx
: y2 =
g- ¥ =
¥ to
x3
3g y3 t e
-
=
23
3g -
y3
- = C
Solution ! !
② dy Initial conditions :
910) =
-1
DI
=
3221-4×1-2
)
2cg -
l
(
2g 2)
d×
-
= 3×2+4×+2
Zy dx
-2
dy = 3×2+4×+2
S2y -2
dy = 53×2 1-4×1-2 dx
23
y2 2y= t 2×2 1- 2x 1- C
-
x3 1-2×2 1- 2X 1- c
y
=
-2
y
%
-
I =
-
C =3
TZX
i.
YZ -2cg = 23 1-2×2 1- 3
↳ solution ! !
,③
digg Initial condition
ylo)
'
yc
: =
=
11-292
-1292 dy dx
I
ycosx
=
tgtzy dy = Cosx dx
Sty +2g dy
=
Scosxdx
lnyt y2 sinx +
=
c
lnli ) t 12 =
sinloltc
0 1- I = 0 1- C
÷ c. = I
i.
lny tyz
=
sinx 1- 1
CHECKING SOLUTIONS IN DIFF EQNS
① y=3é2✗
'
y t
2g 0 ;
=
2x
i.
be be -250=0
-
+
-
'
. .
0=0 it
②
'
"
Y' =
9g ; y ,
=
e ; yz=é3✗
' '
Yi
=
9e3✗
"= 9 e-
3✗
Yz
i. for qe3✗=9e3✗
y , :
i. True ✓
: for 9é3✗ =
qe
-
3✗
yz
:
-
i. True ✓
yz=xe-2✗
" ' "
③ y t 4y t
4g =o
; y , = e- ,
2e-2✗
'
i.
Yi = -
" -
2x
4e
y , =
for 4e
2✗
-14C Ze
Zx
) 4e
2x
-
i. +
-
y
-
: -
,
=
qe
-
2x -
8e→× + 4E
-
H
=
0
i. True ✓
, yz
'
=
e-
2x -
2xe-2✗
" -4 "
Ze 2C t 4xé2✗
-
=
yz
- -
= -
4e-2✗ + 4xe-2✗
( e- 2x Zxe -2×7
2x
4e -2×1-4 >ce
2x
for yz 1- 4 axe
-
-
i. +
-
-
:
4e_2✗ 4xe-2✗ 1- 4e
2x 2x
4xe
2X
- -
8xe t
-
=
+
-
-
= 0
i. True .
✓
④ xzy
"
t
xy
'
-
y
= dna
① Yi
= x -
lnx
i.
y ,
'
= I -
¥
y
" =
¥2
i. for y, : xz( ¥2) t sell -
¥) -
octlnx
=
I t X -
I -
x + lnx
= lnx
i. True ✓
② Yz = Éc -
lnx
i.
Yz
'
=
-
¥2 -
¥
¥3 ¥2
"
lfz t
=
i. x2( ¥3 + ¥2 ) txt -
¥2 -
¥) -
¥ thnx
2
= I + I -
¥ -
I -
÷ tens
= lnx
i. True ✓
⑤ x2y
" '
✗
y 1-29=0
-
① Yi = rxcosllnx )
y ,
'
=
cosclnxltxc-sinllnxl.sc )
= cos Clnx) -
sinllnx )
EQUATIONS
,OLD SEPARABLE DIFFERENTIAL EQUATIONS
sold separable differential equations
separable separated be to find
equations can be
by their variables and
integrated
their solutions .
Example :
① dd- =
22
1-
y2
i. ( 1-
yz) ¥ = x2
i.
11-921 dy = x2 dx
S S
dy
1- x2 dx
: y2 =
g- ¥ =
¥ to
x3
3g y3 t e
-
=
23
3g -
y3
- = C
Solution ! !
② dy Initial conditions :
910) =
-1
DI
=
3221-4×1-2
)
2cg -
l
(
2g 2)
d×
-
= 3×2+4×+2
Zy dx
-2
dy = 3×2+4×+2
S2y -2
dy = 53×2 1-4×1-2 dx
23
y2 2y= t 2×2 1- 2x 1- C
-
x3 1-2×2 1- 2X 1- c
y
=
-2
y
%
-
I =
-
C =3
TZX
i.
YZ -2cg = 23 1-2×2 1- 3
↳ solution ! !
,③
digg Initial condition
ylo)
'
yc
: =
=
11-292
-1292 dy dx
I
ycosx
=
tgtzy dy = Cosx dx
Sty +2g dy
=
Scosxdx
lnyt y2 sinx +
=
c
lnli ) t 12 =
sinloltc
0 1- I = 0 1- C
÷ c. = I
i.
lny tyz
=
sinx 1- 1
CHECKING SOLUTIONS IN DIFF EQNS
① y=3é2✗
'
y t
2g 0 ;
=
2x
i.
be be -250=0
-
+
-
'
. .
0=0 it
②
'
"
Y' =
9g ; y ,
=
e ; yz=é3✗
' '
Yi
=
9e3✗
"= 9 e-
3✗
Yz
i. for qe3✗=9e3✗
y , :
i. True ✓
: for 9é3✗ =
qe
-
3✗
yz
:
-
i. True ✓
yz=xe-2✗
" ' "
③ y t 4y t
4g =o
; y , = e- ,
2e-2✗
'
i.
Yi = -
" -
2x
4e
y , =
for 4e
2✗
-14C Ze
Zx
) 4e
2x
-
i. +
-
y
-
: -
,
=
qe
-
2x -
8e→× + 4E
-
H
=
0
i. True ✓
, yz
'
=
e-
2x -
2xe-2✗
" -4 "
Ze 2C t 4xé2✗
-
=
yz
- -
= -
4e-2✗ + 4xe-2✗
( e- 2x Zxe -2×7
2x
4e -2×1-4 >ce
2x
for yz 1- 4 axe
-
-
i. +
-
-
:
4e_2✗ 4xe-2✗ 1- 4e
2x 2x
4xe
2X
- -
8xe t
-
=
+
-
-
= 0
i. True .
✓
④ xzy
"
t
xy
'
-
y
= dna
① Yi
= x -
lnx
i.
y ,
'
= I -
¥
y
" =
¥2
i. for y, : xz( ¥2) t sell -
¥) -
octlnx
=
I t X -
I -
x + lnx
= lnx
i. True ✓
② Yz = Éc -
lnx
i.
Yz
'
=
-
¥2 -
¥
¥3 ¥2
"
lfz t
=
i. x2( ¥3 + ¥2 ) txt -
¥2 -
¥) -
¥ thnx
2
= I + I -
¥ -
I -
÷ tens
= lnx
i. True ✓
⑤ x2y
" '
✗
y 1-29=0
-
① Yi = rxcosllnx )
y ,
'
=
cosclnxltxc-sinllnxl.sc )
= cos Clnx) -
sinllnx )
