       Re: Re: help

• To: mathgroup at smc.vnet.net
• Subject: [mg15122] Re: [mg15074] Re: help
• From: Jurgen Tischer <jtischer at col2.telecom.com.co>
• Date: Sat, 12 Dec 1998 03:59:18 -0500
• References: <199812100813.DAA04249@smc.vnet.net.>
• Sender: owner-wri-mathgroup at wolfram.com

```Hi Ron,
here is what I could make out of it. I supposed your problem is of the
form

y'[t] + y[t] == 1 - E^t*Integrate[y[tau]/E^tau, {tau, t0, t}]

where t0 is your initial point (that is, known). Then:

In:= eq1 = y'[t] + y[t] == 1 - E^t*Integrate[y[tau]/E^tau, {tau, t0,
t}];

In:= eq2 = D[eq1, t];

In:= eq3 = Eliminate[{eq1, eq2}, Integrate[y[tau]/E^tau, {tau, t0,
t}]]

Out= y''[t] == -1

In:= {sol} = DSolve[eq3, y, t]

Out= {{y -> (C + C*#1 - #1^2/2 & )}}

In:= Simplify[eq1 /. sol]

Out= -(1/2)*E^(t - t0)*(t0^2 - 2*t0*(-1 + C) - 2*(-1 + C +
C)) == 0

In:= {C2} = Solve[t0^2 - 2*t0*(-1 + C) - 2*(-1 + C + C) ==
0, C]

Out= {{C -> -((-2 - 2*t0 - t0^2 + 2*C)/(2*(1 + t0)))}}

In:= sol = sol /. C2

Out= {y -> (C - ((-2 - 2*t0 - t0^2 + 2*C)*#1)/(2*(1 + t0)) -
#1^2/2 & )}

In:= Simplify[eq1 /. sol]

Out= True

Jurgen

Ron Gooch wrote:
>
> yes mathematica based solution is great.  I'm having trouble relating
> this to mathematica.  and help is appreciated.
>
> Ron
>
> -----Original Message-----
> From: steve at smc.vnet.net <steve at smc.vnet.net> To: mathgroup at smc.vnet.net
To: mathgroup at smc.vnet.net
> Subject: [mg15122] [mg15074] Re: help
>
> >Do you want a Mathematica based solution?  If no, post this to
> >sci.math.
> >
> >Moderator
> >
> >> From gooch at ilinkusa.net Mon Dec  7 11:22:40 1998
> >> To: comp-soft-sys-math-mathematica at mail.uu.net
> >> From: "Ron Gooch" <gooch at ilinkusa.net>
To: mathgroup at smc.vnet.net
> To: mathgroup at smc.vnet.net
> >> Subject: [mg15122] [mg15074] help
> >> Date: Mon, 7 Dec 1998 10:16:28 -0800
> >> Message-ID: <74gva1\$neo at enews4.newsguy.com>
> >> X-MimeOLE: Produced By Microsoft MimeOLE V4.72.3110.3
> >> Content-Length: 118
> >>
> >> can any one solve this eq.
> >>
> >> solve the initial value problem
> >>
> >>  y'(t) + y(t)=1-(integral) e^(t-tau) y(tau)dtau, y(0).
> >>
> >>
> >>

```

• References:
• Re: help