Can we solve $q(x)p'(x)+2p(x)q'(x)=0$ given constraints?

Question

If we suppose that we want $-p'(x)q(x) = f(x)$ for a given $f(x)$, and

$$q(x)p'(x)+2p(x)q'(x)=0$$

Can we get $p(x)$ and $q(x)$?

Answer 1

Since no answer has been submitted to complete the result. $$ p=Cq^{-2} $$

Then using the expression for q(x) we find $$ p(x) = \frac{C}{\int f(x) dx} $$ We can immediately obtain q(x).

Answer 2

From $qp'+2pq'=0$ you obtain: $$\frac{dp}{dq}=-2\frac{p}{q}$$ that gives $p=Cq^{-2}$, where $C=p(x_0)q^{2}(x_0)$.Substituting in $-p'q=f$ you get $$2Cq^{-2}\frac{dq}{dx}=f$$ $$\frac{dq}{q^2}=\frac{1}{2C}f(x)dx$$ $$-q^{-1}(x)+q^{-1}(x_0)=\frac{1}{2C}\int_{x_0}^{x}f(t)dt\ .$$ $$q(x)=q(x_0)\left(1-\frac{1}{2q(x_0)p(x_0)}\int_{x_0}^{x}f(t)dt\right)^{-1}$$ $$p(x)=p(x_0)\left(1-\frac{1}{2q(x_0)p(x_0)}\int_{x_0}^{x}f(t)dt\right)^{2}$$