Solution of the differential equation $2\,y\, dx +(3x^2+y^2)\, dy =0$

Question

Please give me some hint to solve first order first degree differential equation:$2\,y\,dx +(3x^2+y^2)\, dy =0$. This cannot be solved by variable separable method. This is not exact or linear equation either. Putting $y=vx $ will not help. So I am clueless now. A small hint will be sufficient.

Answer 1

hint:

A possible approach would be $$ {{dy} \over {dx}} = - {{2y} \over {3x^{\,2} + y^{\,2} }} =- {1 \over {\left( {y - i\sqrt 3 \,x} \right)}} - {1 \over {\left( {y + i\sqrt 3 \,x} \right)}} $$ The component equations $$ \eqalign{ & {{dy} \over {dx}} = - {1 \over {\left( {y + ia\,x} \right)}}\quad \left| {\;a = \pm \sqrt 3 } \right.\quad \Leftrightarrow \cr & \Leftrightarrow \quad {{d\left( {y + ia\,x} \right)} \over {dx}} = - {1 \over {\left( {y + ia\,x} \right)}} - ia\quad \Leftrightarrow \cr & \Leftrightarrow \quad \left( {y + ia\,x} \right)' = f\left( {y + ia\,x} \right) \cr} $$ are "1st order autonomous DE", that falls in the class of the "Chini's equations".

WolframAlpha gives a solution in terms of the Lambert function.

Answer 2

$$2\,y\,dx +(3x^2+y^2)\, dy =0$$ $$\frac{dx}{dy}=-\frac{3}{2y}x^2-\frac12 $$ This is a Riccati ODE.

The method to transform the non-linear first order ODE into a linear second order ODE is shown in : http://mathworld.wolfram.com/RiccatiDifferentialEquation.html , Eqs.$(4-6)$.

Let : $\quad x=\frac{2y}{3} \frac{u'(y)}{u(y)}$

$dx=\frac23 \frac{u'(y)}{u(y)}dy+\frac23 y\frac{u''}{u}dy-\frac23 y\frac{(u')^2}{u^2}dy$ $$2y\left(\frac23 \frac{u'(y)}{u(y)}dy+\frac23 y\frac{u''}{u}dy-\frac23 y\frac{(u')^2}{u^2}dy \right) +\left(3\left(\frac23 y\frac{u'}{u} \right)^2+y^2\right) dy =0$$ After simplification : $$u'' +\frac{1}{y}u'+ \frac34 u =0$$ This is a Bessel ODE. The solutions are : $$u(y)=c_1J_0\left(\frac{\sqrt{3}}{2}y\right)+c_2Y_0\left(\frac{\sqrt{3}}{2}y\right)$$ $J_0$ and $Y_0$ are the Bessel functions of first and second kind respectively. $$u'(y)=c_1\frac{\sqrt{3}}{2}J_1\left(\frac{\sqrt{3}}{2}y\right)+c_2\frac{\sqrt{3}}{2}Y_1\left(\frac{\sqrt{3}}{2}y\right)$$ $$x(y)=\sqrt{\frac23}\:y\:\frac{c_1J_1\left(\frac{\sqrt{3}}{2}y\right)+c_2Y_1\left(\frac{\sqrt{3}}{2}y\right)}{c_1J_0\left(\frac{\sqrt{3}}{2}y\right)+c_2Y_0\left(\frac{\sqrt{3}}{2}y\right)}$$ Note : The function $x(y)$ can be written with only one arbitrary constant $\frac{c_0}{c_1}$ or $\frac{c_1}{c_0}$.

There is probably no closed form for the inverse function $y(x)$.