In class we had differential equations of the type $$y'=\frac{\left(Ax+By \right)y+ \alpha x + \beta y}{\left(Ax+By \right) x+ax+by},$$ where $A,\alpha,a,B,\beta,b$ are constants. The names of the constants were chosen in a way, so that the 3 different constants for each variable look a bit an a and b.

For example $$y'=\frac{(x-y)y-x-y}{(x-y)x+x+y}$$

I did not fully understand the method of solution. In class the professor called them "Jacobi Differtial calculus" (translated). I did not find anything suitable on the web, except Wolfram Mathworld http://mathworld.wolfram.com/JacobiDifferentialEquation.html There is at least a differential equation named Jacobi. But this does not seem to be the same as in my question. Any help?

My question:

Q1 What are DE's like those called?

Q2 How to solve them?

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    $\begingroup$ By 'solve', do you mean find a closed form solution? e..g, finding an integrating factor? $\endgroup$ – peter a g May 13 '16 at 13:07
  • $\begingroup$ @peterag: yes. How to solve them in closed form. And what are they called? Is there some book about them? PS: Sorry for answering so late. $\endgroup$ – user50224 May 19 '16 at 8:23
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    $\begingroup$ WolframAlpha gives this unimaginable result: wolframalpha.com/input/… $\endgroup$ – doraemonpaul May 22 '16 at 4:04
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    $\begingroup$ In your class, did the method of solution involve Lie theory methods - i.e./e.g., finding a (non-trivial) symmetry of the diff equation? or a vector field that 'normalized' the v.f. arising naturally from the difff eq? When I had asked my first question, I had thought / hoped to do so, but so far, I haven't been imaginative enough. You say that you're having difficulty following the sol of class - can you add a bit about the solution to the question? Thanks - and sorry for not replying more quickly. $\endgroup$ – peter a g May 22 '16 at 14:21

For $y'=\dfrac{(x-y)y-x-y}{(x-y)x+x+y}$ ,

Let $u=\dfrac{y}{x}$ ,

Then $y=xu$


$\therefore x\dfrac{du}{dx}+u=\dfrac{(x-xu)xu-x-xu}{(x-xu)x+x+xu}$








Luckily this becomes a Bernoulli ODE.

Let $v=\dfrac{1}{x}$ ,

Then $x=\dfrac{1}{v}$




I.F. $=e^{-\int\frac{du}{u+1}}=e^{-\ln(u+1)}=\dfrac{1}{u+1}$









Now for $y'=\dfrac{(Ax+By)y+\alpha x+\beta y}{(Ax+By)x+ax+by}$ ,

Let $u=\dfrac{y}{x}$ ,

Then $y=xu$


$\therefore x\dfrac{du}{dx}+u=\dfrac{(Ax+Bxu)xu+\alpha x+\beta xu}{(Ax+Bxu)x+ax+bxu}$

$x\dfrac{du}{dx}=\dfrac{(A+Bu)xu+\alpha+\beta u}{(A+Bu)x+a+bu}-u$





Which also luckily that this becomes a Bernoulli ODE.

  • $\begingroup$ Not at all what I was expecting (see my comment above). I haven't checked - forgive me for asking: does this work for any equation of the form of the OP's question? That (your answer) is quite a bit of algebra (+1). If the method works generally, I assume the algebra won't appear as unmotivated as it is to me, now, "just looking at it, like that." Again, I should like to know what the OP's "method of solution" was in class. $\endgroup$ – peter a g Jun 3 '16 at 2:05
  • $\begingroup$ Thanks for following up with the general case... $\endgroup$ – peter a g Jun 5 '16 at 3:00

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