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Derivative from First Principle

Solve 1+x2d...

Question

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

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Solution

We have,

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

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

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

On comparing that,

$\frac{d y}{d x} + P y = Q$

Then,

$P = - \frac{2 x}{1 + x^{2}}, Q = x^{2} + 2$

Now,

$I . F . = e^{\int p d x} = e^{\int \frac{- 2 x}{1 + x^{2}} d x} = e^{- log (1 + x^{2})} = \frac{1}{1 + x^{2}}$

So,

$y . I . F . = \int Q I . F . d x + C$

$\Rightarrow y . \frac{1}{1 + x^{2}} = \int \frac{x^{2} + 2}{1 + x^{2}} d x + C$

$\Rightarrow \frac{y}{1 + x^{2}} = \int \frac{x^{2} + 1}{1 + x^{2}} d x + \int \frac{1}{1 + x^{2}} d x + C$

$\Rightarrow \frac{y}{1 + x^{2}} = x + {tan}^{- 1} x + C$

Hence, this is the answer.

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