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[PDF] Solving Second Order Differential Equations Python

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Given:

Determine:

=f(t,x), with x()=.dx(t) dtt0x0 x(t) for all t≥.t0x(t) t0x(t) x(t)=αeλtα(,)t0x0λ>0tλ<0 tx(t,x) {(,)}tjxkf(,)tjxk x(t)=f(t,x)x()=t0x0(,)t0x0 x(t)=xx(0)=x0x(0)=1/4 x(0)=-1/100 =sin(xt).dx dtsin(xt) x(0)=3x(0)=0x(0)=- f(t,x) x(t)∈ℝn

Given:

Determine:

x(t)=f(t,x), with x()=t0x0 x(t) for all t≥.t0 tf =[]=[]=f(t,x).dx dt x˙1 x˙2 -x1x1x2 -+x2x1x2x1x2 (0)=(0)=1/2x1x2x1x2x1x2t=0

F(t)=ma(t)a(t)

=F(t).xd2 dt2m-1

Given:

Determine:

(t)=f(t,x,), with x()=,()=xd2 dt2x˙t0x0x˙t0y0 x(t) for all t≥.t0x()t0()x˙t0 (t)=x(t)x1(t)=(t)x2x˙ (t)x˙1 (t)x˙2 (t)=(t)x˙x2 (t)=f(t,x,).x¨x˙ x=[,x1x2]T =[]=[]=f(t,x).x˙(t)x˙1 (t)x˙2 (t)x2 f(t,,)x1x2 =[]=[].x0()x1t0 ()x2t0 x()t0 ()x˙t0x(t)x(t) (t)=-x(t),x¨ x(t)=αcos(t)+βsin(t)αβ =[]=[].x˙x˙1 x˙2 x2 -x1 (t)=x¨-x(t) ||x(t)||32x∈ℝ3 ||x(t)|=.|2[(t+(t+(t]x1)2x2)2x3)21/2 x(t)∈ℝ3six f(t,x)(t,x)f′

I∈[-h,+h]x0x0c>h>0

f(t,x) c≥(-1).K

LeL(-)tfinalt0

x∈[,]C1t0tfinal C1

Dt∈[,]t0tfinal

(t)=f(t,x)x˙ x()=t0x0 t>t0 =+khtkt0hh =+htktk-1 xtkxk≈x().xktk =x().x0t0 (,)t0x0t1 (t)=.x˙limh→0 x(t+h)-x(t) ht0h ()≈.x˙t0x(+h)-x()t0t0 (,)t0x0f(,)t0x0h(,)t1x1f(,)t1x1 ≈x()x1t1x()t1x2x1xkx()tk x2x3 .=+hf(,)xk+1xktkxk dtx tx(t)x(t)x(a)=α

Nh[a,b]

h=(b-a)/N x0 si+1 =+hf(,), for i∈[0,N).sitisitf(t,x) ≈x(t)xix(a)=α x(t)=(t+1-0.5)2et N=10 =|x()-|.εabs∑i=1 N tixi N=100 =εrel∑i=1

N|-y()|witi

|y()|ti f(t,y)

2f′titi

+(,x())hn n!f(n-1)titi +(,x())hn+1 (n+1)!fnξiξi x0 xi+1 =+h(,),xiT(n)tixi (,)=f(,)T(n)tixitixi+(,)+⋯+(,).h

2f′tixihn-1

n!f(n-1)tiwi f(t,x(t))=x-+1t2 (t,x(t))f′=(x-+1)d dtt2 =-2t+0x˙ =f(t,x(t))-2t =x-+1-2tt2 (t,x(t))f′′=(x-+1-2t)d dtt2 =-2t+0-2x˙ =f(t,x(t))-2t-2 =x-+1-2t-2t2 =x--2t-1t2 x0 xi+1 =+h[f(t,x(t))+(t,y(t))+(t,x(t))]xih

2f′h2

6f′′

x0 xi+1 =+h[-+1+(-+1-2)+(--2-1)]xixit2ih

2xit2itih2

6xit2iti

=+hf(,),xk+1xktkxkh h h f(,)tkxkf(,)tk+1xk+1 xk =+h[f(,)+f(,)].xk+1xk1

2tkxktk+1xk+1

(t)=f(t,x(t))x′ dt(t)=dtf(t,x(t))∫ tk+1 tk x′ tk+1 tk x()-x()tk+1tk-xk+1xkxk+1 =+h[f(,)+f(+h,+hf(,))].xk+1xk1

2tkxktkxktkxk

f(t,x) x0 xi+1 =+hf(+,+f(,))xitih 2xih 2tixi x0 xi+1 =+[f(,)+f(,+hf(,))]xih

2tixiti+1xitixi

x0 xi+1 =+[f(,)+3f(+h,+hf(,))]xih

4tixiti2

3xi2 3tixi f fh =+hf(+h,+hf(,)),xk+1xktk1 2xk1 2tkxk f =+h(+2+2+),xk+1xk1

6k1k2k3k4

k1 k2 k3 k4 f(,)tkxk f(+h,+h)tk1 2xk1 2k1 f(+h,+h)tk1 2xk1 2k2 f(+h,+h).tkxkk3 x0 k1 k2 k3 k4 xi+1 =hf(,)tixi =hf(+,+)tih 2xi1 2k1 =hf(+,+)tih 2xi1 2k2 =hf(,+)ti+1xik3 =+(+2+2+)xi1

6k1k2k3k4

=-xd2 dt2 x |x|3quotesdbs_dbs19.pdfusesText_25