Can this differential equation be solved manually?

Question

$ x\;\ddot{y} - 2\dot{x} + 9y + \frac{2}{3} = 0$

I tried making use of wolfram alpha but it was unable to provide me with a solution. I also wanted to apply the Laplace transform method, however I am not sure how to apply it given the first term of the equation.

Answer 1

$$xy''-2x'+9y+\frac23=0 \tag 1$$ $y''=\frac{d^2y}{dt^2}$ and $x'=\frac{dx}{dt}$ .

Let $x(t)=u(t)e^{y'/2}\quad\implies\quad x'=u'e^{y'/2}+\frac12 ue^{y'/2}y'' $

$$(ue^{y'/2})y''-2(u'e^{y'/2}+\frac12 ue^{y'/2}y'')+9y+\frac23=0$$ $$-2u'e^{y'/2}+9y+\frac23=0$$ $$u'=\left(\frac92 y+\frac13\right)e^{-y'/2}$$ $$u(t)= \int \left(\frac92 y+\frac13\right)e^{-y'/2}dt+c$$ $c=$ arbitrary constant. $$\boxed{x(t)=c\:e^{y'/2}+e^{y'/2}\int \left(\frac92 y+\frac13\right)e^{-y'/2}dt}$$ This is the general solution of equation $(1)$ for any arbitrary function $y(t)$ , insofar supposed integrable.

Answer 2

Note that if $y_0(t)$ is fixed, the equation is an ODE of 1° order without initial condition, and if $x_0(t)$ is fixed, the equation is an ODE of 2° order without initial condition.

Answer 3

The equation is linear, first order, in $x$.

The solution of the homogeneous equation is easily found to be $$x=ce^{\dot y/2}.$$

Then by variation of the constant, we find $$2c'=\left(9y+\frac23\right)e^{-\dot y/2}$$

and

$$x=e^{\dot y/2}\left(\frac12\int\left(9y+\frac23\right)e^{-\dot y/2}dt+c\right).$$

where $y$ is an arbitrary differentiable function of $t$.

Answer 4

There are two unknowns with only one equation, it is undetermined and wont take off and cannot be solved.

But as an aid to insight, a second arbitrary supporting ode function .. here an accompanying Sine function is assumed with the following BC:

$ (x',x)=(-0.25,-2), (y,y') = (0,0.25) \;;$

$$ x''+ \dfrac{2 \pi x }{\lambda} =0 ;\quad \lambda =0.2 ;\text{ Numerical solution looks like: } $$

EDIT1:

With coupled DE s the wavelength is seen $ \lambda\approx 1$ whereas given uncoupled wave length is only $0.2 ;$ it makes the arbitrary function choice important too.