ODR: Does this one have a solution?

[twoflower]

Hi all,

I'm given this one:

$$y' = 10^{x+y}$$

Here's how I went:

$$\frac{y'}{10^{y}} = 10^{x}$$

$$\int\left(\frac{1}{10}\right)^{y}\ dy = \int 10^{x}\ dx$$

$$\frac{\left(\frac{1}{10}\right)^{y}}{\log \frac{1}{10}}} = \frac{10^{x}}{\log 10} + C$$

$$\left(\frac{1}{10}\right)^{y} = -10^{x}$$

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.

 

[StatusX]

What happened to your constant?

 

[twoflower]

What happened to your constant?

Well, it just...disappeared http://pinkfloyd.o106.com/forum/images/smiles/icon_redface.gif

 

[Tide]

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. :)

 

[twoflower]

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 :)

 

[mezarashi]

Firstly as mentioned, you cannot just make constants disappear. Secondly, can you elaborate on how you got the negative infront of the 10^x.

 

[twoflower]

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.

 

[twoflower]

Welcoming the constant back, I finished this ODE. Anyway, I got something slightly different than Maples tells me.

$$\left(\frac{1}{10}\right)^{y} = -10^{x} + C$$

$$y \log \frac{1}{10} = \log (C - 10^{x})$$

$$y = \frac{\log(C-10^{x})}{\log \frac{1}{10}} = -\frac{\log (C-10^{x})}{\log 10}$$

Which implies

$$C > 0$$

and for particular C

$$10^{x} < C \leftrightarrow x < \frac{\log C}{\log 10}$$

so

$$y = -\frac{\log (C-10^{x})}{\log 10},\ x \in \left(-\infty, \frac{\log C}{\log 10}\right), C > 0$$

Proof shows that the equation

$$y' = 10^{x+y}$$

so I think it's correct.

Anyway, Maple tells me that it's

$$y(x) = \frac{\log \left(\frac{1}{10^{x} + C \log 10}\right)}{\log 10}$$

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

Where's the truth?

Thank you.

 

[saltydog]

I'm pretty sure the solution is:

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

Now . . . does this exists:

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

?

 

[twoflower]

I'm pretty sure the solution is:

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

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...

 

[saltydog]

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.