I have found a solution to the differential equation:
$$\frac{dx}{dt} = \frac{x^2 +t\sqrt{t^2 + x^2}}{tx}$$ where $t, x > 0$
to be:
$$x = t\ln{t} + ct$$
I calculated this by using the fact that the given differential equation can be written in the form: $$M(x, y)dx + N(x, y)dy = 0$$
and the fact that both co-efficient functions M(x, y) and N(x, y) are homogeneous functions of the same degree, that is:
$$M(tx, ty) = t^\alpha M(x, y)$$
and
$$N(tx, ty) = t^\alpha N(x, y)$$
To check whether my solution is correct, I substituted my solution back into the differential equation to yield 0 = 1, that is, a contradiction. The only two possible reasons I have to this is either:
- My solution is indeed incorrect; or
- Re-substituting my solution back into the differential equation and simplifying encountered an error in the process
I have also tried a few Computer Algebra Systems (CAS), including MuPad, to check this solution is correct by either:
Symbolically finding the solution to the differential equation; or
Substituting my solution to the differential equation
Interestingly, the former case couldn't find a solution and the latter case couldn't simplify the LHS and RHS.
Naturally, the fact that the checking process found a contradiction means the solution must be wrong, but I've sense checked it so many times I cant find anything erroneous.
My suspicion is that the condition t, x > 0 must have caused me to make a wrong step somewhere.
My question is, is my solution incorrect or is my checking process incorrect?
The equation being
$$x' = \frac{x^2 +t\sqrt{t^2 + x^2}}{tx}$$ first let $x=t y$ to make $$t y'=\frac{\sqrt{y^2+1}}{y}$$ which is already separable if you write it as $$\frac t {t'}=\frac{\sqrt{y^2+1}}{y}\implies \frac {t'}t =\frac y{\sqrt{y^2+1}}$$ Integrate both sides $$\log(t)+c=\sqrt{y^2+1}\implies y=\sqrt{(\log (t)+c-1) (\log (t)+c+1)}$$ and then $$x=t\sqrt{(\log (t)+c-1) (\log (t)+c+1)}$$ which is a bit more complex than $$x = t\log{t} + ct$$