(1 x^2)dydx+2xy=x√(1 x^2) ⇒dydx+2x(1 x^2)y=x√(1 x^2)(1 x^2) ⇒dydx+2x1 x^2y=x√(1 x^2) ........ (1) Comparing equation (1) by dydx+p ...

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Derivative from First Principle

1-x2dydx+2xy=...

Question

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

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Solution

$(1 - x^{2}) \frac{d y}{d x} + 2 x y = x \sqrt{1 - x^{2}}$
$\Rightarrow \frac{d y}{d x} + \frac{2 x}{(1 - x^{2})} y = \frac{x \sqrt{1 - x^{2}}}{(1 - x^{2})}$
$\Rightarrow \frac{d y}{d x} + \frac{2 x}{1 - x^{2}} y = \frac{x}{\sqrt{1 - x^{2}}}$ ........ $(1)$
Comparing equation $(1)$ by $\frac{d y}{d x} + p y = Q$
$P = \frac{2 x}{1 - x^{2}}, Q = \frac{x}{\sqrt{1 - x^{2}}}$
$∴ I . F = e^{\int \frac{2 x}{1 - x^{2}} d x}$
$\Rightarrow I . F = e^{log \frac{l}{t}}$
$\Rightarrow I . F = \frac{1}{t} = \frac{1}{(1 - x^{2})}$
Multiplying equation $(1)$ by $\frac{1}{(1 - x^{2})}$
$\Rightarrow \frac{1}{(1 - x^{2})} \frac{d y}{d x} + \frac{2 x}{(1 - x^{2})} y$
$= \frac{x}{(1 - x^{2})^{3 / 2}}$
$\Rightarrow \frac{d}{d x} [\frac{1}{(1 - x^{2})} . y] = \frac{x}{(1 - x^{2})^{3 / 2}}$
$\int \frac{d}{d x} [\frac{1}{(1 - x^{2})} y] d x = \int \frac{x}{(1 - x^{2})^{3 / 2}} d x$
$\Rightarrow \frac{1}{(1 - x^{2})} y = \frac{1}{\sqrt{1 - x^{2}}} + C$
$\Rightarrow y = \sqrt{1 - x^{2}} + (1 - x^{2}) C$
This is required solution.

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