Question

$\frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1 is \: a \: solution \: of \: the \: d.e$
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Answer 1

Answer:

$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{b^2x}{a^2y}$

Step-by-step explanation:

We have a hyperbolic equation:

$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Since the equation is the solution for a differential equation, hence we can differentiate the equation to get our d.e:

$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \left[\frac{x^2}{a^2}-\frac{y^2}{b^2}\right]= \frac{\mathrm{d}}{\mathrm{d}x} [1]$

$\displaystyle \frac{2x}{a^2}-\frac{\mathrm{d}}{\mathrm{d}x} \left[\frac{y^2}{b^2}\right]=0$

We can use the chain rule to differentiate $y^2$ w.r.t x:

$\displaystyle \frac{2x}{a^2}-\frac{\mathrm{d}y}{\mathrm{d}x}\cdot\frac{\mathrm{d}}{\mathrm{d}y} \left[\frac{y^2}{b^2}\right]=0$

$\displaystyle \frac{2x}{a^2}-\frac{\mathrm{d}y}{\mathrm{d}x}\cdot\frac{2y}{b^2}=0$

$\displaystyle \frac{2x}{a^2}=\frac{\mathrm{d}y}{\mathrm{d}x}\cdot\frac{2y}{b^2}$

$\displaystyle \frac{b^2x}{a^2y}=\frac{\mathrm{d}y}{\mathrm{d}x}$

Thus, we obtain the d.e:

$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{b^2x}{a^2y}$