Solutions of the Black-Scholes equation are closed under scaling The following is the Black Scholes equation for a function $V(S,t)$:
$$\frac{∂V}{∂t} +\frac{1}{2}σ^2S^2\frac{∂^2V}{∂S^2}+ (r − y) S\frac{∂V}{∂S} − r V = 0$$
The question asks me to show that if $V(S,t)$ is a solution then so is $V(bS,t)$, where $b>0$. The equation holds $\forall S>0, t<T$, so substituting in $K=bS$ gives still a valid equation from which the result follows.
However, my problem is the following: If we set $bS=K$, then $V(bS,t)=V(K,t)$ and we'll have $\frac{\partial (V(K,t))}{\partial K}=\frac{\partial (V(S,t))}{\partial S}$ (similar results for the second partial derivative) and plugging that in doesn't results in an equality since $K\neq S$. Where is my mistake?
 A: Remember that you are supposed to show that:
$$\frac{\partial}{\partial t}V(bS,t)+\frac{1}{2}\sigma^2S^2\frac{\partial^2 }{\partial S^2}V(bS,t)+(r-y)S\frac{\partial }{\partial S}V(bS,t)-rV(bS,t)=0$$
So you need to figure out how to write $\partial_S V(bS,t)$ in terms of $K$, and not $\partial_S V(S,t)$ as you have done. But you can do this using the chain rule: $\partial_S V(bS,t) = \partial_K V(K,t) \cdot \partial_S K=b\partial_KV(K,t)$. Doing this for the entire equation yields:
$$\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2S^2b^2\frac{\partial^2 V}{\partial K^2}+(r-y)Sb\frac{\partial V}{\partial K} - rV=\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2K^2\frac{\partial^2 V}{\partial K^2}+(r-y)K\frac{\partial V}{\partial K}-rV$$
This is the original Black Scholes equation, for which we know $V(K,t)$ is a solution. So:
\begin{align}
\frac{\partial}{\partial t}V(bS,t)+\frac{1}{2}\sigma^2S^2\frac{\partial^2 }{\partial S^2}V(bS,t)+&(r-y)S\frac{\partial }{\partial S}V(bS,t)-rV(bS,t)\\
&=\frac{\partial V(K,t)}{\partial t}+\frac{1}{2}\sigma^2K^2\frac{\partial^2 V(K,t)}{\partial K^2}+(r-y)K\frac{\partial V(K,t)}{\partial K}-rV(K,t)\\
&= 0
\end{align}
