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Higher Order Derivatives

1+x 2 dydx-2x...

Question

$(1 + x^{2}) \frac{d y}{d x} - 2 x y = (x^{2} + 2) (x^{2} + 1)$

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Solution

$We have, (1 + x^{2}) \frac{d y}{d x} - 2 x y = (x^{2} + 2) (x^{2} + 1) \Rightarrow \frac{d y}{d x} - \frac{2 x y}{x^{2} + 1} = x^{2} + 2 . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - \frac{2 x}{x^{2} + 1} Q = x^{2} + 2 ∴ I . F . = e^{\int P d x} = e^{- \int \frac{2 x}{x^{2} + 1} d x} = e^{- \log |x^{2} + 1|} = \frac{1}{x^{2} + 1} Multiplying both sides of (1) by \frac{1}{x^{2} + 1}, we get \frac{1}{x^{2} + 1} (\frac{d y}{d x} - \frac{2 x y}{x^{2} + 1}) = \frac{1}{x^{2} + 1} (x^{2} + 2) \Rightarrow \frac{1}{x^{2} + 1} \frac{d y}{d x} - \frac{2 x y}{{(x^{2} + 1)}^{2}} = \frac{x^{2} + 2}{x^{2} + 1} Integrating both sides with respect to x, we get \frac{1}{x^{2} + 1} y = \int \frac{x^{2} + 2}{x^{2} + 1} d x + C \Rightarrow \frac{y}{x^{2} + 1} = \int \frac{x^{2} + 1 + 1}{x^{2} + 1} d x + C \Rightarrow \frac{y}{x^{2} + 1} = \int d x + \int \frac{1}{x^{2} + 1} d x + C \Rightarrow \frac{1}{x^{2} + 1} y = x + \tan^{- 1} x + C \Rightarrow y = (x + \tan^{- 1} x + C) (x^{2} + 1) Hence, y = (x + \tan^{- 1} x + C) (x^{2} + 1) is the required sol$

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