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We are given the differential equation: \[ (1 + x^2)^2 y'' + 2x(1 + x^2) y' + 4y = 0. \] To solve this numerically, we will follow these steps: ### Step 1: Change of Independent Variable We introduce a new variable: \[ t = \tan^{-1}(x) \Rightarrow x = \tan t, \quad \frac{dx}{dt} = \sec^2 t. \] Using the chain rule, \[ \frac{dy}{dx} = \frac{dy}{dt} _____ ______ ______ __________ __________ _______.
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