Verify that y² = 4a (x + a) is a solution of the differential equations $y \{1 - {(\frac{d y}{d x})}^{2}\} = 2 x \frac{d y}{d x}$ .

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Solution

We have,
$y^{2} = 4 a (x + a)$ ...(1)
Differentiating both sides of (1) with respect to $x$ , we get
$2 y \frac{d y}{d x} = 4 a \Rightarrow y \frac{d y}{d x} = 2 a \Rightarrow \frac{d y}{d x} = \frac{2 a}{y} . . . (2)$
Now,
$y \{1 - {(\frac{d y}{d x})}^{2}\} - 2 x \frac{d y}{d x} = y \{1 - \frac{4 a^{2}}{y^{2}}\} - 2 x (\frac{2 a}{y}) = y \{\frac{y^{2} - 4 a^{2}}{y^{2}}\} - \frac{4 a x}{y} = \frac{y^{2} - 4 a^{2}}{y} - \frac{4 a x}{y} = \frac{(4 a x + 4 a^{2}) - 4 a^{2}}{y} - \frac{4 a x}{y} [Using (1)] = \frac{4 a x}{y} - \frac{4 a x}{y} = 0 \Rightarrow y \{1 - {(\frac{d y}{d x})}^{2}\} = 2 x \frac{d y}{d x}$
Hence, the given function is the solution to the given differential equation.

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