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

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Solution

Consider the given equation.

$y^{2} = 4 a (x - a)$ …… (1)

Given differential equation is,

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

Differentiate equation (1) with respect to $x$ .

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

$\frac{d y}{d x} = \frac{2 a}{y}$ …… (3)

Now, substitute this value in the $L H S$ of equation (2).

$\Rightarrow y [1 - {(\frac{2 a}{y})}^{2}]$

$\Rightarrow \frac{y^{2} - 4 a^{2}}{y}$

$\Rightarrow \frac{4 a (x + a) - 4 a^{2}}{y}$

$\Rightarrow \frac{4 a x}{y}$

$\Rightarrow 2 \times (\frac{d y}{d x}) \times (x)$

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

$\Rightarrow R H S$

Hence, proved.

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Prove that y2=4a(x+a) is the solution of differential equation y×[1−(dydx)2]=2xdydx

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