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

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Solution

Given, $y^{2} = 4 a (x + a)$

Differentiating both side w.r.t. x, we get,

$\frac{d}{d x} y^{2} = \frac{d}{d x} 4 a (x + a)$

$2 y \frac{d y}{d x} = \frac{d}{d x} 4 a x + \frac{d}{d x} 4 a^{2}$

$2 y \frac{d y}{d x} = 4 a + 0$

$\frac{d y}{d x} = \frac{2 a}{y}$

Now,

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

$L . H . S = y {1 - {(\frac{d y}{d x})}^{2}}$

Put the value of $\frac{d y}{d x} = \frac{2 a}{y}$

$= y {1 - {(\frac{2 a}{y})}^{2}}$

$= y {1 - (\frac{4 a^{2}}{y^{2}})}$

$= y (\frac{y^{2} - 4 a^{2}}{y^{2}})$

Simplifying

$= (\frac{y^{2} - 4 a^{2}}{y})$

Given, $y^{2} = 4 a (x + a)$

$= (\frac{4 a (x + a) - 4 a^{2}}{\sqrt{4 a (x + a)}})$

$= (\frac{4 a x + 4 a^{2} - 4 a^{2}}{\sqrt{4 a (x + a)}})$

$= (\frac{4 a x}{\sqrt{4 a (x + a)}})$

$= 2 x (\frac{2 a}{\sqrt{4 a (x + a)}})$

$= 2 x (\frac{2 a}{y})$

$= 2 x \frac{d y}{d x}$

$= R . H . S$

$∴ y^{2} = 4 a (x + a)$ is a solution for the differential equation $y {1 - {(\frac{d y}{d x})}^{2}} = 2 x \frac{d y}{d x}$

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