ODEs and classification 1 1
.
-contains equal signg
D
Differential Equations - > contains derivatives F=
Independent vs .
dependent
ma
↳
general solution ra
ODE - ordinary >
-
PDE- partial differentials 2xy" 3xy' + 12y
-
= 12
GOAL:
example Ex)
Purpose : shows a RELATIONSHIP
- # Y
T2
=
*
not a solution ! IVP = initial
value
Edifa is a solution
without an
initial condition
differential
I
LASSIFICATIONS
yl" = third
problem
order
·
y= ex is a solution equation has l order derivative present
infinite solutions !
>
-
largest
2 .
Type >
-
ODE Or PDE
A function is a solution . Linear
3 order, type, linearing
!
vs . Nonlinear DE ex .
to a differential equation, definition ↳ all L ODE 1st
"
powers , ,
of one
and when of a solution
plugging in Land NONLINEAR
NL, ODE , 2MD
derivatives) into the equation 3) NL ODE 2nd
⑭ , ,
it's satisfied 4) ,
L ODE ,
4th
X= any -: any 5) NL PDE, und
,
dy 6) NL , PDE 2nd
2e2)
= 2y ,
9t with
↳ ex
the + c
works with
y
=
e
is this
·
-
Gy' 2e2 ↓
A solution
Ce
31
because general
Six-
=
31 to the solution
x12
cauchy-ewler is linear!
C ,x
-
32
377x - y(t) = +
zx
y
-
=
312y 9x
y
= -
(2x"2
4x2y" 12xy' + 3y
-
= -
y =
-
zy + = 0
5/2y) 3m -
-
2x
=
Ey
12(x)) 2x51
y"
-
Y= Ex =
4x2(5x /)
*
+
+ 3(x1) = 0
y"
12
= =
*
generalnation 15 x
- 32 -
18x 43x *O
3
y(H) = y'(M)= - j the function is
Is this a solution ?
-
y(t) = ( ,x +
solution
(2x"2
- at that
4xy" + 12xy' 3y +
= 0
Y = 3 2
+
It T
point
32
-
y(x)
(x2y'(x) ·
-3
-
y +
=
y = -
2ct
general solutions
=
can represent very
ty'
I
· ()
4y = 3
<(2) 2
G
+
+ different functions
= C (1/32) 2(( %) -
,
-
=
E The # of initial conditions
needed depends on the order.
E
& y(t)
=
Ez =Sty +
-
+ Yy = 3y(1)
=
4
C = 1 1 . We need to find something to plug into
Y 2ct
PREF C= 0 -
y) = -
X-3 2t( 2ct 3) + y(t +
3) 3
general solution
=
so
-
=
y =
-
nc + 4) + 3 = 3r
,PRACTICE - just do the derivatives and plug it in. "
y COSX + x -y" + x+ y2x5 let t be
y =
y
=
↓
64e5t
y sinx +7xb 5e4t 55ty" (y y) zy
-
=
+
- -
yox+ 42x5 + cox+x /+42x5
= -
-
= y= -
5
y"= -
cosx + 42x ~ y 20e4t 2595t
= -
y" 80e4t 12595t
= -
y Sinx
=
+ Xyl +
y = x9 + 72x)
· ve"t-125est (ge4t 5e5t)(20e" = 25e5t) 2(54E5e5t) 6yest
ax
- -
+ = -
y cosX +
x9+ 7(xv
Finx + / +x + xi
=
-
=
y" 1258/10e
**
Sinx + 72x
**
-64e5t
=
-
80e4t 125est
-
-
(100e -
125e * 100e9t +
- 10e5t =
-
100eot + 225e9t 125elot
-
let o be
y
27ebt
2e4t 5esty"
-
(yy' +
5y =
-
y =
y =
094t 15237 -
y" = 32e4t 4593t -
32e _
45e[et get) (gent est)1 _ .
_ + 5(2e4t je3t)
-
=
-
27e3t
3 zelt -
usetyyeot-goet -40e <sett)
*
+ 10e4t 2523t -
27eX
3t
=
-
ebxY zy2 3x2+ 1x
yx(6y 6x6xy Gy
+
y
=
+
+ +
= ↓
(
* its
for implicit , do basically here
everything inx to differentiate
X asking you
x6(y + and when
you come
sides of the
DE
+ by = 6X across a
y, multiply
both
then re-arrange
and
by dy/dx
see if it matches
g6x46(y xx) by = 6x
-
+ +
, separable ode
·
solving 1st order ODEs du = f(x , y)
dX
if a function is separable
then it can be written like if the equation is nonlinear,
acceptable
=(49(x) +
goal is factors there is no
method
universally
d
.
y Fy
* to
get
dx. 2- 2 .2
Sydy SxdX
-
=
an
explicit
solution S2y -
2 dy =
f3x + 4x + 2dx
multiply yucky
·
way
Y y + 2x2 2x + C
-
-
zy = +
complete
IMPLICIT
the
! OR JUST LEAVE IT
square C iS THE
SOLVE FOR
dy
2 . ety) =e 1 +
BEST SOMETIMES !
te
-
isolate
-
factor
ety() e (e 2t)(e-3)=
+
Sedy Set de =
u=
= 3 de
3t -
-
separate
use negative
y(i)( e2t)
exponents
-e
et .
+
=
=
22
n
& du =
-2
Se y dy Set (l e2)
+ de
-
I
gebe
*
dx .
14434
= x -
y
y2 + 4 ↓
= edx &(x + 9) dy -
5xdx =
0
niy
+ Stan (x-u)=
Y
= 8y -
My-9, =
d3YyT (32 4)dy
+ =
y(e3x)(e39)dx
, =xtd
&d
↓ wiin t +
=
x Ed =
(x + 5xy"(dx + ex2y3dy = 0
*
x(1 + 5y)dx + e
ydy = 0 t+
xdx
Xe
=
x(1 +5y)dx= -
(eX(()d Xe
Y
)
I exu = 7X
- Sxe* dx =
Sedt
ex
i
Se-t
& +
Se +
xe* AfeE
-
-
-
Xe -
Ye
↓ ye t
-
tex + 5 = C
.
-contains equal signg
D
Differential Equations - > contains derivatives F=
Independent vs .
dependent
ma
↳
general solution ra
ODE - ordinary >
-
PDE- partial differentials 2xy" 3xy' + 12y
-
= 12
GOAL:
example Ex)
Purpose : shows a RELATIONSHIP
- # Y
T2
=
*
not a solution ! IVP = initial
value
Edifa is a solution
without an
initial condition
differential
I
LASSIFICATIONS
yl" = third
problem
order
·
y= ex is a solution equation has l order derivative present
infinite solutions !
>
-
largest
2 .
Type >
-
ODE Or PDE
A function is a solution . Linear
3 order, type, linearing
!
vs . Nonlinear DE ex .
to a differential equation, definition ↳ all L ODE 1st
"
powers , ,
of one
and when of a solution
plugging in Land NONLINEAR
NL, ODE , 2MD
derivatives) into the equation 3) NL ODE 2nd
⑭ , ,
it's satisfied 4) ,
L ODE ,
4th
X= any -: any 5) NL PDE, und
,
dy 6) NL , PDE 2nd
2e2)
= 2y ,
9t with
↳ ex
the + c
works with
y
=
e
is this
·
-
Gy' 2e2 ↓
A solution
Ce
31
because general
Six-
=
31 to the solution
x12
cauchy-ewler is linear!
C ,x
-
32
377x - y(t) = +
zx
y
-
=
312y 9x
y
= -
(2x"2
4x2y" 12xy' + 3y
-
= -
y =
-
zy + = 0
5/2y) 3m -
-
2x
=
Ey
12(x)) 2x51
y"
-
Y= Ex =
4x2(5x /)
*
+
+ 3(x1) = 0
y"
12
= =
*
generalnation 15 x
- 32 -
18x 43x *O
3
y(H) = y'(M)= - j the function is
Is this a solution ?
-
y(t) = ( ,x +
solution
(2x"2
- at that
4xy" + 12xy' 3y +
= 0
Y = 3 2
+
It T
point
32
-
y(x)
(x2y'(x) ·
-3
-
y +
=
y = -
2ct
general solutions
=
can represent very
ty'
I
· ()
4y = 3
<(2) 2
G
+
+ different functions
= C (1/32) 2(( %) -
,
-
=
E The # of initial conditions
needed depends on the order.
E
& y(t)
=
Ez =Sty +
-
+ Yy = 3y(1)
=
4
C = 1 1 . We need to find something to plug into
Y 2ct
PREF C= 0 -
y) = -
X-3 2t( 2ct 3) + y(t +
3) 3
general solution
=
so
-
=
y =
-
nc + 4) + 3 = 3r
,PRACTICE - just do the derivatives and plug it in. "
y COSX + x -y" + x+ y2x5 let t be
y =
y
=
↓
64e5t
y sinx +7xb 5e4t 55ty" (y y) zy
-
=
+
- -
yox+ 42x5 + cox+x /+42x5
= -
-
= y= -
5
y"= -
cosx + 42x ~ y 20e4t 2595t
= -
y" 80e4t 12595t
= -
y Sinx
=
+ Xyl +
y = x9 + 72x)
· ve"t-125est (ge4t 5e5t)(20e" = 25e5t) 2(54E5e5t) 6yest
ax
- -
+ = -
y cosX +
x9+ 7(xv
Finx + / +x + xi
=
-
=
y" 1258/10e
**
Sinx + 72x
**
-64e5t
=
-
80e4t 125est
-
-
(100e -
125e * 100e9t +
- 10e5t =
-
100eot + 225e9t 125elot
-
let o be
y
27ebt
2e4t 5esty"
-
(yy' +
5y =
-
y =
y =
094t 15237 -
y" = 32e4t 4593t -
32e _
45e[et get) (gent est)1 _ .
_ + 5(2e4t je3t)
-
=
-
27e3t
3 zelt -
usetyyeot-goet -40e <sett)
*
+ 10e4t 2523t -
27eX
3t
=
-
ebxY zy2 3x2+ 1x
yx(6y 6x6xy Gy
+
y
=
+
+ +
= ↓
(
* its
for implicit , do basically here
everything inx to differentiate
X asking you
x6(y + and when
you come
sides of the
DE
+ by = 6X across a
y, multiply
both
then re-arrange
and
by dy/dx
see if it matches
g6x46(y xx) by = 6x
-
+ +
, separable ode
·
solving 1st order ODEs du = f(x , y)
dX
if a function is separable
then it can be written like if the equation is nonlinear,
acceptable
=(49(x) +
goal is factors there is no
method
universally
d
.
y Fy
* to
get
dx. 2- 2 .2
Sydy SxdX
-
=
an
explicit
solution S2y -
2 dy =
f3x + 4x + 2dx
multiply yucky
·
way
Y y + 2x2 2x + C
-
-
zy = +
complete
IMPLICIT
the
! OR JUST LEAVE IT
square C iS THE
SOLVE FOR
dy
2 . ety) =e 1 +
BEST SOMETIMES !
te
-
isolate
-
factor
ety() e (e 2t)(e-3)=
+
Sedy Set de =
u=
= 3 de
3t -
-
separate
use negative
y(i)( e2t)
exponents
-e
et .
+
=
=
22
n
& du =
-2
Se y dy Set (l e2)
+ de
-
I
gebe
*
dx .
14434
= x -
y
y2 + 4 ↓
= edx &(x + 9) dy -
5xdx =
0
niy
+ Stan (x-u)=
Y
= 8y -
My-9, =
d3YyT (32 4)dy
+ =
y(e3x)(e39)dx
, =xtd
&d
↓ wiin t +
=
x Ed =
(x + 5xy"(dx + ex2y3dy = 0
*
x(1 + 5y)dx + e
ydy = 0 t+
xdx
Xe
=
x(1 +5y)dx= -
(eX(()d Xe
Y
)
I exu = 7X
- Sxe* dx =
Sedt
ex
i
Se-t
& +
Se +
xe* AfeE
-
-
-
Xe -
Ye
↓ ye t
-
tex + 5 = C
