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**Ackbeet** Because your second-order ODE operator is still linear, though your quadratic equation is not linear. If Ly is the LHS of your ODE, where L represents the differential operator

$\displaystyle L=D^{2} + 3D-4,$

then you can convince yourself that if $\displaystyle y_{1}$ and $\displaystyle y_{2}$ are both solutions to the homogeneous ODE $\displaystyle Ly=0,$ then any linear combination $\displaystyle ay_{1}+by_{2}$ is also a solution to $\displaystyle Ly=0.$

This fact is not true of your quadratic.