Implicit Solution to a differential equation

Question

I'm looking at the ODE:

$\frac{dY}{dX} - \frac{ X^2 + 2 Y^2 - 1 }{ ( Y - 2 X )X } = 0$

I'm looking for an $implicit$ solution to the above. Meaning, I want to find a relation $F(X,Y)=0$, where $\frac{\partial}{\partial x} (F(x,y)) = ( Y - 2 X )X \frac{dY}{dX} - ( X^2 + 2 Y^2 - 1) $.

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For example, if we would consider the simpler DE $dY/dX = -X/Y$, the implicit solution is $X^2 + Y^2 = \mathrm{constant}$.

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I think that the implicit solution will be a cubic polynomial of $X$ and $Y$, but I am troubling identifying the function.

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I thought about setting $F(x,y)= a Y^3 + b Y^2 X + c Y X^2 + d X^3 + e Y^2 + f X Y + g X^2 + h Y + j X + k$, and differentiating, but this gets complicated too fast.

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How can I solve this problem?

Answer 1

Reaarange your equation $$(yx-2x^2)\ dy-(x^2+2y^2-1)\ dx=0$$ And we know that $$F(x,y)=0 \Rightarrow dF(x,y)=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy$$ It follows that $$\frac{\partial F}{\partial x}=-(x^2+2y^2-1)$$ $$\frac{\partial F}{\partial y}=yx-2x^2$$

Answer 2

$\dfrac{dY}{dX}-\dfrac{X^2+2Y^2-1}{(Y-2X)X}=0$

$\dfrac{dY}{dX}=\dfrac{X^2+2Y^2-1}{(Y-2X)X}$

$(Y-2X)\dfrac{dY}{dX}=\dfrac{X^2+2Y^2-1}{X}$

Let $U=Y-2X$ ,

Then $Y=U+2X$

$\dfrac{dY}{dX}=\dfrac{dU}{dX}+2$

$\therefore U\left(\dfrac{dU}{dX}+2\right)=\dfrac{X^2+2(U+2X)^2-1}{X}$

$U\dfrac{dU}{dX}+2U=\dfrac{2U^2+8XU+9X^2-1}{X}$

$U\dfrac{dU}{dX}=\dfrac{2U^2}{X}+6U+9X-\dfrac{1}{X}$

This belongs to an Abel equation of the second kind.

Follow the method in http://eqworld.ipmnet.ru/en/solutions/ode/ode0126.pdf:

Let $U=X^2W$ ,

Then $\dfrac{dU}{dX}=X^2\dfrac{dW}{dX}+2XW$

$\therefore X^2W\left(X^2\dfrac{dW}{dX}+2XW\right)=2X^3W^2+6X^2W+9X-\dfrac{1}{X}$

$X^4W\dfrac{dW}{dX}+2X^3W^2=2X^3W^2+6X^2W+9X-\dfrac{1}{X}$

$X^4W\dfrac{dW}{dX}=6X^2W+9X-\dfrac{1}{X}$

$W\dfrac{dW}{dX}=\dfrac{6W}{X^2}+\dfrac{9}{X^3}-\dfrac{1}{X^5}$

Let $T=-\dfrac{6}{X}$ ,

Then $X=-\dfrac{6}{T}$

$\dfrac{dW}{dX}=\dfrac{dW}{dT}\dfrac{dT}{dX}=\dfrac{6}{X^2}\dfrac{dW}{dT}$

$\therefore\dfrac{6W}{X^2}\dfrac{dW}{dT}=\dfrac{6W}{X^2}+\dfrac{9}{X^3}-\dfrac{1}{X^5}$

$W\dfrac{dW}{dT}=W+\dfrac{3}{2X}-\dfrac{1}{6X^3}$

$W\dfrac{dW}{dT}-W=\dfrac{T^3}{1296}-\dfrac{T}{4}$

This belongs to an Abel equation of the second kind in the canonical form.

Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf