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As simple as this should be, I can not seem to solve it. I can't classify its type and thus figure out how to solve it. It's the only ODE in my problem sheet that I can't solve (embarrassing).

$$t\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)=x+\sqrt{t^2+x^2}$$

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  • $\begingroup$ What happens if you divide through by $t$? $\endgroup$
    – abiessu
    Commented Aug 19, 2013 at 0:15
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    $\begingroup$ @eXtremiity: Yes, this has been asked before and I answered it. See: math.stackexchange.com/questions/458539/… $\endgroup$
    – Amzoti
    Commented Aug 19, 2013 at 0:27
  • $\begingroup$ @Amzoti: you might want to vote to close as duplicate rather than answering again. $\endgroup$
    – robjohn
    Commented Aug 23, 2013 at 17:30
  • $\begingroup$ @robjohn: I didn't realize it was a dupe after I had answered it, but then found the dup. I then flagged the question as a duplicate. I have a hard time finding anything with the MSE search tool and resort to trying to find it using google with detailed search strings. So, if I would have found it, I would have done that. Regards $\endgroup$
    – Amzoti
    Commented Aug 23, 2013 at 18:02

2 Answers 2

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Hint, let:

$$x = t v(t) \rightarrow x' = v + t v'$$

Substituting into the original equation, doing some algebra and rearranging, eventually leads to the integration:

$$\int \dfrac{dv}{\sqrt{v^2+1}} = \int \dfrac{1}{t} dt$$

Can you take it from there?

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  • $\begingroup$ @Amzoti: Yep, asked and answered by me :-), see: math.stackexchange.com/questions/458539/… $\endgroup$
    – Amzoti
    Commented Aug 19, 2013 at 0:22
  • $\begingroup$ Great hint, Amzoti! +1 $\endgroup$
    – amWhy
    Commented Aug 19, 2013 at 0:23
  • $\begingroup$ @amWhy: thanks, it looked familiar and I had to use google to find it! I was the one that answered, so I flagged my own answer :-). math.stackexchange.com/questions/458539/… $\endgroup$
    – Amzoti
    Commented Aug 19, 2013 at 0:25
  • $\begingroup$ So a change of variable is being applied. Makes perfect sense. I didn't think any of my questions would have required the technique. Bad assumption. Thank you very much for your help @Amzoti. $\endgroup$ Commented Aug 19, 2013 at 0:37
  • $\begingroup$ @eXtremiity: You are welcome and sometimes a problem looks easy, but involves some work and this is an example of that. Glad I could help. Regards $\endgroup$
    – Amzoti
    Commented Aug 19, 2013 at 0:39
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This differential equation is an example of a homogeneous differential equation. The reason that change of variables works is because that is the standard technique for solving Euler-homogeneous differential equations. You should always check if a differential equation is exact (poincare's lemma), if non-exact then find an integrating factor (Frobenius theorem) & theoretically the problem can always be solved. However there are shortcuts or more direct methods in special cases such as separable equations, euler-homogeneous equations, linear equations, linear fractional equations, then a bunch of equations with names attached. After that unless you can find a change of variables (Lie theory) you have to use some form of approximation technique, at least in the case of first degree ode's of first order. You shouldn't really think of it as change of variables technique because it only really applies in the homogeneous case, & homogeneity is used for studying higher degree ode's as well hence it has some importance on it's own. Here's a good book.

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  • $\begingroup$ Thanks for the insight. I am really enjoying this subject ! I hope to learn all the equations you have mentioned in class. $\endgroup$ Commented Aug 19, 2013 at 2:30

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