How to solve this DE for $v^3$?

I've been looking over this for a while now and can't find where I went wrong in my attempt.

A car's velocity $v$ is related to its displacement $x$ by the differential equation $$v\frac{dv}{dx}=\frac{p}{v}-kv^2$$ where $p$ and $k$ are constants. Show that $$v^3=\frac{1}{k}(p-pe^{-3x})$$

My attempt:

$$\frac{dv}{dx}=\frac{p-kv^3}{v^2}$$ $$\frac{v^2}{p-kv^3}\frac{dv}{dx}=1$$ $$\int\frac{v^2}{p-kv^3}dv=x+c$$ Let $u=p-kv^3$, $du=-3kv^2dv$ $$-\frac{1}{3k}\int\frac{1}{u}du=x+c$$ $$\ln(p-kv^3)=-3k(x+c)$$ $$v^3=\frac{1}{k}(p-e^{-3k(x+c)})$$ After substituting $v=0$, $x=0$ I found that, $$ -3kc=\ln(p)$$ $$ v^3=\frac{1}{k}(p-e^{-3kx}\cdot e^{-3kc})$$ $$v^3=\frac{1}{k}(p-pe^{-3kx})$$ But this does not match up with the required form so where is the mistake I have made? Thanks.

Solution 1:

There is no mistake, the reference solution is wrong in missing the factor $k$ in the exponent.

If you set alternatively $y=v^3$, you get $$ y'=3p-3ky, $$ and in this form as linear DE you can see that the exponential basis solution has a factor $-3k$ in the exponent.