$\frac{dR}{dx}=\frac{2R^3-x^3}{3xR^2} $. Find $R(x)$.

Question

The relationship between Revenue '$R$' and quantity demanded '$x$' is such that

$$\frac{dR}{dx}=\frac{2R^3-x^3}{3xR^2}.$$

Show that the relationship between $R$ and $x$ is given by

$$R^3 = Kx^2 - x^3.$$

Answer 1

$$\frac{dR}{dx}=\frac{2R^3-x^3}{3xR^2}.$$

$$\implies \frac{dR}{dx}=\frac{2R}{x} - \frac{x^2}{3R^2}$$

Let $\frac{R}{x} = v$.

$$\implies R = vx$$

$$\implies \frac{dR}{dx} = x\frac{dv}{dx} + v$$

Substituting it back into our original equation we get,

$$x\frac{dv}{dx} + v = 2v - \frac{1}{3v^2}$$

$$\implies x\frac{dv}{dx} = v - \frac{1}{3v^2}$$

$$\implies \frac{3v^2}{3v^3-1} dv = \frac{dx}{x}$$

Now integrate both sides to obtain $v(x)$ and use the fact that $R = x v(x)$.

Cheers!

Answer 2

Substituting $R=z^{1/3}$ in the differential equation gives $$ x \frac{dz}{dx} -2z = -x^3. $$ This is a first order ODE. Its general solution is $z= K x^2- x^3$, with $K$ constant. This gives the relation between $R$ and $x$.