An equation of the curve satisfying xdy-y dx - -x2 - y2 dx and y1 - 0 is

The equation can be written as x^2 xdy ydx/x^2 = x √(1 (y/x)^2)dx ⇒d (y/x)/√(1 (y/x)^2) = dx/x⇒ Sin^ 1y/x = log|x| + const. Since y(1) = 0 so const = 0. H ...

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Standard XIII

Mathematics

Solving Homogeneous Differential Equations

An equation o...

Question

An equation of the curve satisfying $x d y - y d x = \sqrt{x^{2} - y^{2}}$ dx and $y (1) = 0$ is

y = x^{2} l o g | s i n x |

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y = x s i n (l o g | x |)

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y^{2} = x (x - 1)^{2}

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y = 2 x^{2} (x - 1)

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Solution

The correct option is B $y = x s i n (l o g | x |)$
The equation can be written as $x^{2} \frac{x d y - y d x}{x^{2}} = x \sqrt{1 - (\frac{y}{x})^{2}} d x$
$\Rightarrow \frac{d (\frac{y}{x})}{\sqrt{1 - (\frac{y}{x})^{2}}} = \frac{d x}{x} \Rightarrow S i n^{- 1} \frac{y}{x} = l o g | x | + c o n s t .$
Since y(1) = 0 so const = 0. Hence y = x sin (log |x|)

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