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

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

Then $y=xu$

$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$

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

$x\dfrac{du}{dx}=\dfrac{(1-u)xu-1-u}{(1-u)x+1+u}-u$

$x\dfrac{du}{dx}=\dfrac{(x-1)u-xu^2-1}{(1-x)u+x+1}-u$

$x\dfrac{du}{dx}=\dfrac{xu^2-(x-1)u+1}{(x-1)u-x-1}-u$

$x\dfrac{du}{dx}=\dfrac{xu^2-(x-1)u+1-(x-1)u^2+(x+1)u}{(x-1)u-x-1}$

$x\dfrac{du}{dx}=\dfrac{u^2+2u+1}{(u-1)x-u-1}$

$\dfrac{dx}{du}=\dfrac{(u-1)x^2-(u+1)x}{(u+1)^2}$

$\dfrac{dx}{du}+\dfrac{x}{u+1}=\dfrac{(u-1)x^2}{(u+1)^2}$

Luckily this becomes a Bernoulli ODE.

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

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

$\dfrac{dx}{du}=-\dfrac{1}{v^2}\dfrac{dv}{du}$

$\therefore-\dfrac{1}{v^2}\dfrac{dv}{du}+\dfrac{1}{(u+1)v}=\dfrac{(u-1)}{(u+1)^2v^2}$

$\dfrac{dv}{du}-\dfrac{v}{u+1}=-\dfrac{(u-1)}{(u+1)^2}$

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

$\therefore\dfrac{d}{du}\left(\dfrac{v}{u+1}\right)=-\dfrac{(u-1)}{(u+1)^3}$

$\dfrac{v}{u+1}=-\int\dfrac{(u-1)}{(u+1)^3}du$

$\dfrac{1}{(u+1)x}=\int\left(-\dfrac{1}{(u+1)^2}+\dfrac{2}{(u+1)^3}\right)du$

$\dfrac{1}{(u+1)x}=\dfrac{1}{u+1}-\dfrac{1}{(u+1)^2}+c$

$\dfrac{1}{x}=\dfrac{u}{u+1}+c(u+1)$

$\dfrac{1}{x}=\dfrac{\dfrac{y}{x}}{\dfrac{y}{x}+1}+c\left(\dfrac{y}{x}+1\right)$

$\dfrac{xy}{x+y}+c(x+y)=1$

$\dfrac{xy}{x+y}-1=C(x+y)$

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

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

Then $y=xu$

$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$

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

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

$x\dfrac{du}{dx}=\dfrac{Bxu^2+(Ax+\beta)u+\alpha}{(Bx+b)u+Ax+a}-u$

$x\dfrac{du}{dx}=\dfrac{Bxu^2+(Ax+\beta)u+\alpha-(Bx+b)u^2-(Ax+a)u}{(Bx+b)u+Ax+a}$

$x\dfrac{du}{dx}=\dfrac{-bu^2+(\beta-a)u+\alpha}{(A+Bu)x+a+bu}$

$\dfrac{dx}{du}=-\dfrac{(A+Bu)x^2+(a+bu)x}{bu^2+(a-\beta)u-\alpha}$

Which also luckily that this becomes a Bernoulli ODE.

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  • $\begingroup$ 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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