Non-Linear Second Order Differential Equation Regarding Elasticity

Question

What methods can I use to solve the following equation?

$$q''(p)=\frac{q'(p)^2}{q(p)}+\frac{q'(p)}{p}$$

I know from wolfram alpha that the solution is $q(p)=c_1p^{c_2}$.

Answer 1

The solution given by Wolfram Alpha is for the problem $$ q''(p)=\frac{q'(p)^2}{q(p)} - \frac{q'(p)}{p} \, . $$

Here is the argument for the case $p > 0$.

Let $w(p) = \log q(p)$. Then $w'= \frac{q'}{q}$ and $w'' = \frac{q''}{q} - \frac{(q')^2}{q^2}$.

Divide the differential equation by $q$. It becomes $$ \frac{q''(p)}{q(p)} = \frac{(q'(p))^2}{q^2(p)} + \frac{q'(p)}{q(p) p} $$ or $$ w''(p) = \frac{w'(p)}{p} \, . $$ Thus $\frac{w''(p)}{w'(p)} = \frac{1}{p}$, implying $\log |w'(p)| = c_0 + \log p$. Then $w'(p) = c_1 p$, thus $w(p) = c_2 + c_3p^2$ and consequently $$ \boxed{q(p) = e^{c_2 + c_3p^2} = c_4e^{c_3 p^2}} $$

Answer 2

You can also divide through by $q'(p)$ to get $$ \frac{q''}{q'} = \frac{q'}{q} + \frac{1}{p}$$ $$ \ln{q'} = \ln{q} + \ln{p} + c_1 $$ $$ q' = c_2qp $$

This gives $$ q(p) = ae^{bp^2}$$