Ode
6
Example: Initial conditions 0 0 F x x t t 0 t from = + = → = t cos B t sin A x H + = 0 p 0 p F x F C 0 C x = ∴ = + ∴ = 0 F t cos B t sin A ) t ( x + + = 0 0 F B 0 F B ) 0 ( x − = = + = ) t cos 1 ( F t sin A ) t ( x 0 − + = A 0 ) 0 ( x ) t (sin F t cos A ) t ( x 0 = = + = t sin F ) t ( x 0 = 0 F 2 x = π 0 ) 2 ( F x = π 2 t π ≥ t cos C t sin C ) t ( x 2 1 + = 1 0 C F ) 2 ( x = = π ) t cos t (sin F ) t ( x 0 − = ) t cos 1 ( F ) t ( x 0 − = 0 ) 0 ( x ) 0 ( x = = ) cos 1 ( ) ( 0 t F t x − = 0 2 ) 2 ( F C x = = π 2 to 0 t from π = ∴ 2 t π ≥ ) t ( F x x = +
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ode
Transcript of Ode
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Example:
Initial conditions
00 Fxxtt0tfrom =+==
tcosBtsinAxH +=
0p0p FxFC0Cx ==+=
0FtcosBtsinA)t(x ++=
00 FB0FB)0(x ==+=
)tcos1(FtsinA)t(x 0 +=
A0)0(x)t(sinFtcosA)t(x 0 ==+=
tsinF)t(x 0=
0F2x = 0)2
( Fx =
2t
tcosCtsinC)t(x 21 +=
10 CF)2(x ==
)tcost(sinF)t(x 0 =
)tcos1(F)t(x 0 =
t
0)0(x)0(x ==
)cos1()( 0 tFtx =
02)2( FCx ==
2to0tfrom =
2t
)t(Fxx =+
-
)tcost(sinF)t(x 0 =
0201 F)2(xF)
2(x ===
00 F)10(F)(x +=+=
}tcostsin)12/{(f)t(x 0 =
2t =
00 f}1{f)(x ==
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Superposition:
)tsint(f)t(x 0 =
= )
2tsin()
2t(f)t(U)tsint(f)t(x 00
0000 fff2f
2)(x =+=
Example:
0)0(x)0(x)t(fxx==
=+
Solution:
0tFxx =+
)12
(F)2
(x 0 =
0F)2(x =
0F2xx =+
0F2tcosBtsinA)t(x ++=
tsinBtcosA)t(x =
00 FBBf)2(x ===
+= 1
2F)(x 0
t
oF2
2
t
F(t)
2t
)tsint(F)t(x 0 =
2t
000
0
FA)12
(F)2
(FA)2
(x
)tcos2
(FtsinA)t(x
==+=
+=
2t
2t
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Superposition:
2t
= )
2tsin()
2t(F)tsint(F)t(x 00
)12
(F)12
(F)(F)(x 000 +==
Example: )t(fxx =+
Solution:
)tsint(F)t(x o =
00 F)2(x),1
2(F)
2(x ==
2t
tFF)t(Fxx 000 ==+ tFFxtCCx 00p21p =+=
000
0
F2AffA)2
(x
)t(FtsinBtcosA)t(x
===
++=
0000
0
FBF2
F)2
(fB)2
(x
)tcos2t(FtsinB)t(x
==+=
+=
0)(x)tcostsin21(F)t(x)2(F)(x
))tsintcos2(t(F)t(x
0
0
0
=+=
=
+=
0F2)(x = 0)('x =
>t tcosCt sinC)t(x 43 +=
oF2
2
t
F(t)
t
F(t)
= +
t 2
F(t)
t
-
04 F2C)(x == 04 F2C =
tsinCt cos C)t('x 43 =
>t )t(cosF2)t(x 0=
0F2)2(x =
Superposition:
OK 0)2
(F2FF(o) )( at check O ==
0F )2
2(2F 2)F(2 )(2at 0O =+= F
0F 2)2
3(F2F 3)F(3 )(3 at 000 =+=
OK 0) - 4(F)2
4(2F 4)F(4 )(4at 0O =+= F
))tsin()t((F))2
-sin(t-)2
-((tF2t) sint(F)t(x 000
+=
)(tF
t
tFtF 0)( =
2
)(tF
t
tFtF 02)( =
)(tF
t
)(0 = tF
)(tF
t + +
oF2
2
2
oF2
