Solve the differential equation x d^2 y/dx^2 + 2y = 0

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We want to find the general solution of the differential equation:

x y'' + 2 y = 0

We can try to bring this differential equation to a more standard form by the change of variables x^p = t. We then have:

d/dx = dt/dx d/dt = px^(p-1) d/dt = p t^[(p-1)/p] d/dt

d^2/dx^2 = p t^[(p-1)/p] d/dt {p t^[(p-1)/p] d/dt} = p(p-1) t^[(p-2)/p] d/dt + p^2 t^[2(p-1)/p] d^2/dt^2

The differential equation x y'' + 2 y = 0 then becomes:

p(p-1) t^[(p-1)/p] dy/dt + p^2 t^[(2p-1)/p] d^2y/dt^2 +2 y = 0

We can expect things to simplify if we choose p to ...