DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

The equation of the family of curves is

y2 - 2ay + 2+ = a² ...(1)

 where a is a parameter.

This equation contains only one arbitrary constant, so we shall get a differential equation of first order.

Differentiating equation (1) with respect to x, we get

$2\text{y}\frac{\text{dy}}{\text{dx}}-2\text{a}\frac{\text{dy}}{\text{dx}}+2\text{x}=0$

$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}+2\text{x}=2\text{a}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\text{y}+\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}}=\text{a}$

Substituting the value of a in equation (2), we get

$\text{y}^2-2\Bigg(\text{y}+\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}}\Bigg)\text{y}+\text{x}^2=\Bigg(\text{y}+\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}}\Bigg)^2$

$\Rightarrow\frac{\text{y}^2\frac{\text{dy}}{\text{dx}}-2\Big(\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}\Big)\text{y}+\text{x}^2\frac{\text{dy}}{\text{dx}}}{\frac{\text{dy}}{\text{dx}}}=\frac{\Big(\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}\Big)}{\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}$

$\Rightarrow\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-2\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-2\text{xy}\Big(\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\\=\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+2\text{xy}\Big(\frac{\text{dy}}{\text{dx}}\Big)+\text{x}^2$

$\Rightarrow(\text{x}^2-2\text{y}^2)\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-4\text{xy}\Big(\frac{\text{dy}}{\text{dx}}\Big)-\text{x}^2=0$

It is the required differential equation.