ODR: Does this one have a solution?

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Hi all,

I'm given this one:

<br /> y&#039; = 10^{x+y}<br />

Here's how I went:

<br /> \frac{y&#039;}{10^{y}} = 10^{x}<br />

<br /> \int\left(\frac{1}{10}\right)^{y}\ dy = \int 10^{x}\ dx<br />

<br /> \frac{\left(\frac{1}{10}\right)^{y}}{\log \frac{1}{10}}} = \frac{10^{x}}{\log 10} + C<br />

<br /> \left(\frac{1}{10}\right)^{y} = -10^{x}<br />

From which I conclude that it has no solution (on the right side there is negative number whereas on the left side there always will be a positive one).

Is this right conclusion?

Thank you.
 
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What happened to your constant?
 
StatusX said:
What happened to your constant?

Well, it just...disappeared http://pinkfloyd.o106.com/forum/images/smiles/icon_redface.gif
 
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You're confusing "assumptions" with "conclusions." It is a conclusion to say that there is no solution.

If the constant of integration just disappeared then you cannot just make a general conclusion. Surely, the problem provided an intial condition or you can infer one. :)
 
Tide said:
You're confusing "assumptions" with "conclusions." It is a conclusion to say that there is no solution.
If the constant of integration just disappeared then you cannot just make a general conclusion. Surely, the problem provided an intial condition or you can infer one. :)

Of course, I mistakingly got rid of the constant :)
 
Firstly as mentioned, you cannot just make constants disappear. Secondly, can you elaborate on how you got the negative infront of the 10^x.
 
mezarashi said:
Firstly as mentioned, you cannot just make constants disappear. Secondly, can you elaborate on how you got the negative infront of the 10^x.

Sorry, there was a typo in the original post. Now I corrected it and hope it's clear.
 
Welcoming the constant back, I finished this ODE. Anyway, I got something slightly different than Maples tells me.

<br /> \left(\frac{1}{10}\right)^{y} = -10^{x} + C<br />

<br /> y \log \frac{1}{10} = \log (C - 10^{x})<br />

<br /> y = \frac{\log(C-10^{x})}{\log \frac{1}{10}} = -\frac{\log (C-10^{x})}{\log 10}<br />

Which implies

<br /> C &gt; 0<br />

and for particular C

<br /> 10^{x} &lt; C \leftrightarrow x &lt; \frac{\log C}{\log 10}<br />

so

<br /> y = -\frac{\log (C-10^{x})}{\log 10},\ x \in \left(-\infty, \frac{\log C}{\log 10}\right), C &gt; 0<br />

Proof shows that the equation

<br /> y&#039; = 10^{x+y}<br />

so I think it's correct.

Anyway, Maple tells me that it's

<br /> y(x) = \frac{\log \left(\frac{1}{10^{x} + C \log 10}\right)}{\log 10}<br />

but it seems not to satisfy the equation when I differentiated it.

Where's the truth?

Thank you.
 
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I'm pretty sure the solution is:

y(x)=\text{log}\left[\frac{1}{k-10^x}\right];\quad x&lt;\text{log}(k)

Now . . . does this exists:

\int_0^{\text{log(k)}} y(x)dx

?
 
  • #10
saltydog said:
I'm pretty sure the solution is:

y(x)=\text{log}\left[\frac{1}{k-10^x}\right];\quad x&lt;\text{log}(k)

Now . . . does this exists:

\int_0^{\text{log(k)}} y(x)dx

?

You're right with solution (assuming it is decimal logarithm you use).

Anyway, why is Maple giving wrong answer? I thought it will be good tool to check my own results but it doesn't seem so...
 
  • #11
twoflower said:
You're right with solution (assuming it is decimal logarithm you use).
Anyway, why is Maple giving wrong answer? I thought it will be good tool to check my own results but it doesn't seem so...

Twoflower, I dont' know about Maple. I use Mathematica and I believe software like it are execllent tools to check results. For example, I just finishing working on:

\mathcal{L}^{-1}\left\{\frac{2s-s^2+1}{2(s^3+s-2)}\right\}

Well, that's a mess to do by hand (for me anyway). Have to break it up into several pieces like:

\frac{2s}{s^2+s-2},\quad \frac{1}{s-1},\quad \frac{2-s}{2+s+s^2}

and

\frac{s^2-1}{s^3+s-2}

Tough to keep track of everything but I can use Mathematica to invert each to verfy each step of my work. I've found this approach, using Mathematica to check steps, very useful as a learning tool. Try and learn to do so with Maple or whatever you use.:smile:
 
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