If $A=\left[\begin{array}{rr}1 & 2 \\ -1 & 4\end{array}\right]$ and $A^{-1}=\alpha I+\beta A \alpha, \beta \in R$ where I is the identity matrix of order 2 , then $4(\alpha+\beta)=$

The solution of $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x+y)^2$ is