How to Reduce the Order of the ODE y'' = c - 2(y')²/y?

[efmsolver]

reduce order of ODE y"=c-2y'2/y

I'm stumped trying to come up with the appropriate substitution for:

y''=c-2y'²/y

c is a constant and the dependent variable does not appear in the equation. According to the paper I am working from the equation can be reduced to a Bernoulli-type problem and solved, but I can't seem to replicate their work. Any hints?

y'=0 at the far boundary and y(0)=y0

 

[Matthew Rodman]

Any second-order equation of the form

y^{\prime \prime} + \alpha(y) y^{\prime 2} + \beta(y) = 0

(where the derivative is with respect to 'x') may be converted into a first order equation of the form

\frac{du}{dy} + 2 \alpha(y) u + 2 \beta(y) = 0

with the simple substitution

u = y^{\prime 2}.

Just plug that into the second equation and you'll see it works, bearing in mind that

\frac{d y^{\prime 2}}{dy} = 2 y^{\prime \prime}

After some tedious calculation, it turns out that your particular equation may be expressed (in first-order terms) as

y^{\prime 2} = \frac{2c}{5} y + \frac{\lambda}{y^4}

(where lambda is some constant). To demonstrate this, differentiate the above with respect to 'y', and you get

2 y^{\prime \prime} = \frac{2c}{5} - \frac{4 \lambda}{y^5}

Now make the substitution

\frac{\lambda}{y^4} = y^{\prime 2} - \frac{2c}{5} y

to get

2 y^{\prime \prime} = \frac{2c}{5} - \frac{4}{y}(y^{\prime 2} - \frac{2c}{5} y)

which expands to

2 y^{\prime \prime} = \frac{2c}{5} - \frac{4}{y} y^{\prime 2} + \frac{8c}{5}

Now add the constant terms and divide by 2 on both sides to get your original equation.