Question

Verify that y² = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]

Answer 1

We have,

\[y^2 = 4a\left( x + a \right)...........(1)\]

Differentiating both sides of (1) with respect to x, we get

\[2y\frac{dy}{dx} = 4a\]

\[ \Rightarrow y\frac{dy}{dx} = 2a\]

\[ \Rightarrow \frac{dy}{dx} = \frac{2a}{y} ..........(2)\]

Now,

\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} - 2x\frac{dy}{dx}\]

\[ = y\left\{ 1 - \frac{4 a^2}{y^2} \right\} - 2x\left( \frac{2a}{y} \right)\]

\[ = y\left\{ \frac{y^2 - 4 a^2}{y^2} \right\} - \frac{4ax}{y}\]

\[ = \frac{y^2 - 4 a^2}{y} - \frac{4ax}{y}\]

\[ = \frac{\left( 4ax + 4 a^2 \right) - 4 a^2}{y} - \frac{4ax}{y} ...........\left[\text{Using }\left( 1 \right) \right]\]

\[ = \frac{4ax}{y} - \frac{4ax}{y} = 0\]

\[ \Rightarrow y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]

Hence, the given function is the solution to the given differential equation.