1. separable equation

This equation is supposedly separable:

$\displaystyle t\frac{dy}{dx} = y + \sqrt{t^2 + y^2}$

but I can't separate it with algebra, and the substitutions I've tried (v=y/t, v=(t^2 + y^2)^1/2) haven't worked. The full problem, if it makes a difference, is to solve for the initial value y(1) = 0.

Could you give me a hint please?

2. Originally Posted by Korgoth28
This equation is supposedly separable:

$\displaystyle t\frac{dy}{dx} = y + \sqrt{t^2 + y^2}$

but I can't separate it with algebra, and the substitutions I've tried (v=y/t, v=(t^2 + y^2)^1/2) haven't worked. The full problem, if it makes a difference, is to solve for the initial value y(1) = 0.

Could you give me a hint please?
I suppose you meant that x to be a t:

$\displaystyle t\frac{dy}{dt} = y + \sqrt{t^2 + y^2}$

$\displaystyle \Rightarrow \frac{dy}{dt} = \frac{y}{t} + \frac{\sqrt{t^2 + y^2}}{t} = \frac{y}{t} + \sqrt{1 + \left( \frac{y}{t}\right)^2}$

and now you make the substitution $\displaystyle y = v t$ etc.

3. That makes sense, thank you. After substituting and rearranging I get this:
$\displaystyle \int \frac{dt}{t} = \int \frac{dv}{\sqrt{1+v^2}}$
And after integrating:
$\displaystyle \ln{t} = arcsinh (\frac{y}{t}) + c$
and plugging in the initial value gives me
$\displaystyle \ln{1} = arcsinh(0) + c$
so c = 0.
However the answer in the back of the book is $\displaystyle \frac{t^2 - 1}{2}$. Could you please tell me where I'm messing up?

Also, out of curiosity, is there any time when it'd be useful to make a substitution other than y/t in a problem like this? For example, t/y?

4. I thought I hit discrete mathematics topic but I guess not. Just happened to look at this and the question that came to mind for me, and forgive my ignorance if not because I have yet to have this subject, would the sq root in the denominator have an impact in following processes?

5. Originally Posted by Korgoth28
That makes sense, thank you. After substituting and rearranging I get this:
$\displaystyle \int \frac{dt}{t} = \int \frac{dv}{\sqrt{1+v^2}}$
And after integrating:
$\displaystyle \ln{t} = arcsinh (\frac{y}{t}) + c$
and plugging in the initial value gives me
$\displaystyle \ln{1} = arcsinh(0) + c$
so c = 0.
However the answer in the back of the book is $\displaystyle \frac{t^2 - 1}{2}$. Could you please tell me where I'm messing up?

[snip]
You have not messed up. You're expected to know that $\displaystyle \text{arcsinh} \, (z) = \ln (z + \sqrt{1 + z^2})$ and hence simplify your answer accordingly.