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8/12/2019 First Order Eq Ns
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Examples for obtaining a set of first-order ODE from a higher order ODE
1) )2cos(3)1()1( 2 x x y x y x ⋅⋅⋅⋅=⋅++′′′⋅− π
First rewrite the equation and re-label the derivatives as follows:
)2cos(3)1(00)1( 2 x x y x y y y x ⋅⋅⋅⋅=⋅++′⋅+′′⋅+′′′⋅− π
Note: the higher order derivative is labeled differently.
The set of first-order differential equations is:
20
2
21
10
)1(
)1()2cos(3
−
⋅+−⋅⋅⋅⋅=
′
=′
=′
x
z x x x z
z z
z z
π
with initial conditions
)0()0(
)0()0(
)0()0(
2
1
0
y z
y z
y z
′′=
′=
=
Note: the number of first-order differential equations is three because this is a third orderdifferential equation.
Note: there will be three initial conditions because this is a third order differentialequation.
In Mathcad notation we can write:
−
⋅+−⋅⋅⋅⋅
=
20
2
1
)1(
)1()2cos(3
:),(
x
z x x x
z
z
z x D
π
and
′′
′=
)0(
)0(
)0(
:
y
y
y
zinitial
Note: actual numbers should be substituted in the expression for the initial conditions.
z2’ z2 z1 z0
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2) )cos(2120)sin(5
t e
t x xt
t ⋅⋅++=⋅+⋅
⋅ ω
This equation is already a first order differential equation ! Solving for the first order
derivative, xD :
)sin(
20)cos(215
t
xt e
t
xt
⋅−⋅⋅++
=⋅
ω
D with initial condition )0( x
Note: the number of first-order differential equations is one because this is a first order
differential equation.
Note: there is one initial condition because this is a first order differential equation.
In Mathcad notation we can write:
)sin(
20)cos(21
:),(5
t
xt e
t
xt Dt
⋅−⋅⋅++
=⋅
ω
and )0(: x xinitial =
Note: actual numbers must be substituted in the expression for the initial condition.
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3) t t y y yt yt ⋅+=+⋅+⋅⋅+⋅+ 9)cos()sin(5)3sin()2( 2
First rewrite the equation and re-label the derivatives as follows:
t t y y yt y yt ⋅+=+⋅+⋅⋅+⋅+⋅+
9)cos()sin(5)3sin(0)2(
2
Note: )sin( y becomes )sin( 0w .
Note: the higher order derivative is labeled differently.
The set of first-order differential equations is:
)2(
9)cos()sin(5)3sin(
2
0012
21
10
+
⋅++−⋅−⋅⋅−=
=
=
t
t t wwwt w
ww
ww
D
D
D
with initial conds.
)0()0(
)0()0(
)0()0(
2
1
0
yw
yw
yw
DD
D
=
=
=
Note: the number of first-order differential equations is three because this is a third order
differential equation.
Note: there will be three initial conditions because this is a third order differential
equation.
In Mathcad notation we can write:
+
⋅++−⋅−⋅⋅−
=
)2(
9)cos()sin(5)3sin(
:),(
2
001
2
1
t
t t wwwt
w
w
wt D and
=
)0(
)0(
)0(
:
y
y
y
winitial
Note: actual numbers should be substituted in the expression for the initial conditions.
2w w2 w1 w0 w0
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4) Here we have two differential equations that are “coupled” (connected):
0)(
)sin()(
2
211
=−−⋅
⋅⋅=⋅+−+⋅+⋅
y xk ym
t F xk y xk xb xm
ω
First rewrite the first equation and re-label the derivatives as follows:
)sin()( 211 t F yk xk k xb xm ⋅⋅=⋅−⋅++⋅+⋅ ω
Now rewrite the second equation and re-label the derivatives.
002 =⋅−⋅+⋅+⋅ xk yk y ym
Note: notice how the labels for the second equation are a continuation of the labels for the
first equation.
The set of first-order differential equations is:
[ ]
[ ]022
3
32
2021111
10
1
)sin()(1
zk zk m
z
z z
t F zk zk k zbm
z
z z
⋅+⋅−⋅=
=
⋅⋅+⋅+⋅+−⋅−⋅=
=
ω
and initial conds
)0()0(
)0()0(
)0()0(
)0()0(
3
2
1
0
y z
y z
x z
x z
=
=
=
=
Note: the number of first-order differential equations is four because two second order
differential equations are used.
Note: there will be four initial conditions. Two for each second order differential
equation.
1 z z1 z0
z3 3 z z2
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In Mathcad notation we can write:
[ ]
[ ]
⋅+⋅−⋅
⋅⋅+⋅+⋅+−⋅−⋅
=
022
3
202111
1
1
)sin()(1
:),(
zk zk m
z
t F zk zk k zbm
z
zt D
ω
and
=
)0(
)0()0(
)0(
:
y
y x
x
zinitial
Note: actual numbers should be substituted in the expression for the initial conditions.
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