       Re: Simple DSolve equation

• To: mathgroup at smc.vnet.net
• Subject: [mg122642] Re: Simple DSolve equation
• From: Rui Rojo <rui.rojo at gmail.com>
• Date: Fri, 4 Nov 2011 06:01:09 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201111030846.DAA15245@smc.vnet.net>

```(not true what I just said about linear combinations being solutions)

On Thu, Nov 3, 2011 at 11:02 AM, Rui Rojo <rui.rojo at gmail.com> wrote:

> What I want to somehow get is the most general result with restrictions
> over k. That is:
>
> {y''[x] == k y[x], y == 0, y == 0} /. {
>    y -> (Sin[2 Pi p/20 #] &),
>    k -> -(1/100) p^2 \[Pi]^2 }
> Out: {True, True, Sin[p \[Pi]] == 0}
> Simplify[%, p \[Element] Integers]
> {True, True, True}
> ...or any linear combination of those solutions. Typical 10m rope tied on
> the extremes
>
>
> 2011/11/3 Bob Hanlon <hanlonr357 at gmail.com>
>
>> soln = DSolve[{y''[x] == k y[x], y == y0, y == y10}, y[x], x][[1,
>> 1]]
>>
>> y[x] -> (E^(20*Sqrt[k])*y0 - E^(2*Sqrt[k]*x)*y0 - E^(10*Sqrt[k])*y10 +
>>        E^(10*Sqrt[k] + 2*Sqrt[k]*x)*
>>     y10)/(E^(Sqrt[k]*x)*(-1 + E^(20*Sqrt[k])))
>>
>> soln /. y0 -> 0 // Simplify
>>
>> y[x] -> ((-1 + E^(2*Sqrt[k]*x))*
>>    y10)/(E^(Sqrt[k]*(-10 + x))*(-1 + E^(20*Sqrt[k])))
>>
>> soln /. y10 -> 0 // Simplify
>>
>> y[x] -> ((E^(20*Sqrt[k]) - E^(2*Sqrt[k]*x))*
>>    y0)/(E^(Sqrt[k]*x)*(-1 + E^(20*Sqrt[k])))
>>
>> soln /. {y0 -> 0, y10 -> 0}
>>
>> y[x] -> 0
>>
>> What solution are you expecting?
>>
>>
>> Bob Hanlon
>>
>>
>> On Thu, Nov 3, 2011 at 4:46 AM, Rui <rui.rojo at gmail.com> wrote:
>> > Why does something like this not give the correct answer with
>> restrictions over k?
>> > How would you go about getting the right general solutions in these
>> kind of basic differential equations?
>> >
>> > Thanks
>> >
>> > DSolve[{y''[x] == k y[x], y == 0, y == 0}, y[x], x]
>> > Out={{y[x] -> 0}}
>> >
>>
>
>

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(not true what I just said about linear combinations being solutions)<br><b=
r><div class="gmail_quote">On Thu, Nov 3, 2011 at 11:02 AM, Rui Rojo <spa=
n dir="ltr">&lt;<a href="mailto:rui.rojo at gmail.com">rui.rojo at gmail.com<=
/a>&gt;</span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1p=
x #ccc solid;padding-left:1ex;">What I want to somehow get is the most gene=
ral result with restrictions over k. That is:<br><br>{y&#39;&#39;[x] ===
k y[x], y == 0, y == 0} /. {<br>
y -&gt; (Sin[2 Pi p/20 #] &amp;), <br>   k -&gt; -(1/100) p^2 \[=
Pi]^2 }<br>
Out: {True, True, Sin[p \[Pi]] == 0}<br>Simplify[%, p \[Element] Intege=
rs]<br>{True, True, True}<br>...or any linear combination of those solution=
s. Typical 10m rope tied on the extremes<div class="HOEnZb"><div class==
"h5">
<br><br><div class="gmail_quote">2011/11/3 Bob Hanlon <span dir="ltr">&=
lt;<a href="mailto:hanlonr357 at gmail.com" target="_blank">hanlonr357@gma=
il.com</a>&gt;</span><br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1p=
x #ccc solid;padding-left:1ex">soln = DSolve[{y&#39;&#39;[x] == k y[x=
], y == y0, y == y10}, y[x], x][[1, 1]]<br>
<br>
y[x] -&gt; (E^(20*Sqrt[k])*y0 - E^(2*Sqrt[k]*x)*y0 - E^(10*Sqrt[k])*y10 +<b=
r>
E^(10*Sqrt[k] + 2*Sqrt[k]*x)*<br>
y10)/(E^(Sqrt[k]*x)*(-1 + E^(20*Sqrt[k])))<br>
<br>
soln /. y0 -&gt; 0 // Simplify<br>
<br>
y[x] -&gt; ((-1 + E^(2*Sqrt[k]*x))*<br>
y10)/(E^(Sqrt[k]*(-10 + x))*(-1 + E^(20*Sqrt[k])))<br>
<br>
soln /. y10 -&gt; 0 // Simplify<br>
<br>
y[x] -&gt; ((E^(20*Sqrt[k]) - E^(2*Sqrt[k]*x))*<br>
y0)/(E^(Sqrt[k]*x)*(-1 + E^(20*Sqrt[k])))<br>
<br>
soln /. {y0 -&gt; 0, y10 -&gt; 0}<br>
<br>
y[x] -&gt; 0<br>
<br>
What solution are you expecting?<br>
<br>
<br>
Bob Hanlon<br>
<br>
<br>
On Thu, Nov 3, 2011 at 4:46 AM, Rui &lt;<a href="mailto:rui.rojo at gmail.co=
m" target="_blank">rui.rojo at gmail.com</a>&gt; wrote:<br>
&gt; Why does something like this not give the correct answer with restrict=
ions over k?<br>
&gt; How would you go about getting the right general solutions in these ki=
nd of basic differential equations?<br>
&gt;<br>
&gt; Thanks<br>
&gt;<br>
&gt; DSolve[{y&#39;&#39;[x] == k y[x], y == 0, y == 0}, =
y[x], x]<br>
&gt; Out={{y[x] -&gt; 0}}<br>
&gt;<br>
</blockquote></div><br>
</div></div></blockquote></div><br>

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